RD SHARMA SOLUTION CHAPTER- 31 Probability I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 31 Probability Exercise Ex. 31.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Chapter 31 Probability Exercise Ex. 31.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is probability that both drawn balls are black?Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Chapter 31 Probability Exercise Ex. 31.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 5(iv)

If A and B are two events such that

Solution 5(iv)

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls? Given that

(i) the youngest is a girl

(ii) at least one is girl.Solution 27

(i) Let ‘A’ be the event that both the children born are girls.

Let ‘B’ be the event that the youngest is a girl.

We have to find conditional probability P(A/B).

begin mathsize 12px style table attributes columnalign left end attributes row cell straight P left parenthesis straight A divided by straight B right parenthesis equals fraction numerator straight P left parenthesis straight A intersection straight B right parenthesis over denominator straight P left parenthesis straight B right parenthesis end fraction end cell row cell straight A subset of straight B rightwards double arrow straight A intersection straight B equals straight A end cell row cell rightwards double arrow straight P left parenthesis straight A intersection straight B right parenthesis equals straight P left parenthesis straight A right parenthesis equals straight P left parenthesis GG right parenthesis equals 1 half cross times 1 half equals 1 fourth end cell row cell straight P left parenthesis straight B right parenthesis equals straight P left parenthesis BG right parenthesis plus straight P left parenthesis GG right parenthesis equals 1 half cross times 1 half plus 1 half cross times 1 half equals 1 fourth plus 1 fourth equals 1 half end cell row cell text Hence ,  end text straight P left parenthesis straight A divided by straight B right parenthesis equals fraction numerator 1 divided by 4 over denominator 1 divided by 2 end fraction equals 1 half end cell end table end style

(ii) Let ‘A’ be the event that both the children born are girls.

Let ‘B’ be the event that at least one is a girl.

We have to find the conditional probability P(A/B).

begin mathsize 12px style table attributes columnalign left end attributes row cell straight P left parenthesis straight A divided by straight B right parenthesis equals fraction numerator straight P left parenthesis straight A intersection straight B right parenthesis over denominator straight P left parenthesis straight B right parenthesis end fraction end cell row cell straight A subset of straight B rightwards double arrow straight A intersection straight B equals straight A end cell row cell rightwards double arrow straight P left parenthesis straight A intersection straight B right parenthesis equals straight P left parenthesis straight A right parenthesis equals straight P left parenthesis GG right parenthesis equals 1 half cross times 1 half equals 1 fourth end cell row cell straight P left parenthesis straight B right parenthesis equals 1 minus straight P left parenthesis BB right parenthesis equals 1 minus 1 half cross times 1 half equals 1 minus 1 fourth equals 3 over 4 end cell row cell text Hence ,  end text straight P left parenthesis straight A divided by straight B right parenthesis equals fraction numerator 1 divided by 4 over denominator 3 divided by 4 end fraction equals 1 third end cell end table end style

Chapter 31 Probability Exercise Ex. 31.4

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   probabilities   of   two   students   end text straight A text   and   end text straight B text   coming   end text end cell row cell text to   the   school   in   time   are end text fraction numerator text 3 end text over denominator text 7 end text end fraction text   and   end text fraction numerator text 5 end text over denominator text 7 end text end fraction text   respectively .  end text end cell row cell text Assuming   that   the   events , ' end text straight A text   coming   in   time '  and  ' end text straight B end cell row cell text coming   in   time '  are   independent ,  find   the   probability   of   end text end cell row cell text only   one   of   them   coming   to   the   school   in   time .  end text end cell row cell text Write   atleast   one   advantage   of   coming   to   school   in   time. end text end cell end table end style

Solution 23

Given that the events ‘A coming in time’ and ‘B coming in time’ are independent.

begin mathsize 12px style table attributes columnalign left end attributes row cell text Let  ' A '  denote   the   event   of  ' A   coming   in   time '.  end text end cell row cell text Then , ' end text stack text A end text with bar on top apostrophe text   denotes   the   complementary   event   of   A. end text end cell row cell text Similarly   we   define   B   and   end text stack text B end text with bar on top. end cell end table end style
begin mathsize 12px style table attributes columnalign left end attributes row cell straight P left parenthesis only text   one   coming   in   time end text right parenthesis equals straight P left parenthesis straight A intersection straight B with bar on top right parenthesis plus straight P left parenthesis straight A with bar on top intersection straight B right parenthesis end cell row cell equals straight P left parenthesis straight A right parenthesis cross times straight P left parenthesis straight B with bar on top right parenthesis plus straight P left parenthesis straight A with bar on top right parenthesis cross times straight P left parenthesis straight B right parenthesis... left parenthesis text since   A   and   B   are   independent   events end text right parenthesis end cell row cell equals 3 over 7 cross times 2 over 7 plus 4 over 7 cross times 5 over 7 equals 6 over 49 plus 20 over 49 equals 26 over 49 end cell end table end style

The advantage of coming to school in time is that you will not miss any part of the lecture and will be able to learn more.Question 24

Two dice are thrown together and the total score is noted. The event E, F and G are “a total 4”, “a total of 9 or more”, and “a total divisible by 5”, respectively. Calculate P (E), P(F) and P(G) and decide which pairs of events, if any, are independent.Solution 24

Question 25

Let A and B be two independent events such that P (A) = p1 and P (B) = p2. Describe in words the events whose probabilities are:

(i) p1p2 (ii) (1 – p1)p2 (iii) 1-(1- p1) (1 – p2) (iv) p1 + p2 = 2p1p2Solution 25

Chapter 31 Probability Exercise Ex. 31.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

In a hockey match, both teams A and B scored same number of goals upto the end of the game, so to decide the winner, the refree asked both the captains to throw a die alternately and decide that the team, whose captain gets a first six, will be declared the winner. If the captain of team A was asked to start, find their respective probabilities of winning the match and state whether the decision of the refree was fair or not.Solution 35

begin mathsize 12px style table attributes columnalign left end attributes row cell text Probability   of   getting   six   in   any   toss   of   a   dice end text equals 1 over 6 end cell row cell text Probability   of   not   getting   six   in   any   toss   of   a   dice end text equals 5 over 6 end cell row cell text A   and   B   toss   the   die   alternatively .  end text end cell row cell text Hence   probability   of   A ' s   win end text end cell row cell equals straight P left parenthesis straight A right parenthesis plus straight P left parenthesis straight A with bar on top straight B with bar on top straight A right parenthesis plus straight P left parenthesis straight A with bar on top straight B with bar on top straight A with bar on top straight B with bar on top straight A right parenthesis plus straight P left parenthesis straight A with bar on top straight B with bar on top straight A with bar on top straight B with bar on top straight A with bar on top straight B with bar on top straight A right parenthesis plus....... end cell row cell equals 1 over 6 plus 5 over 6 cross times 5 over 6 cross times 1 over 6 plus 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 1 over 6 plus 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 1 over 6 plus..... end cell row cell equals 1 over 6 plus open parentheses 5 over 6 close parentheses squared 1 over 6 plus open parentheses 5 over 6 close parentheses to the power of 4 1 over 6 plus open parentheses 5 over 6 close parentheses to the power of 6 1 over 6 plus..... end cell row cell equals fraction numerator 1 divided by 6 over denominator 1 minus open parentheses 5 divided by 6 close parentheses squared end fraction equals 1 over 6 cross times 36 over 11 equals 6 over 11 end cell row cell Similarly comma text   probability   of   B ' s   win end text end cell row cell equals straight P left parenthesis straight A with bar on top straight B right parenthesis plus straight P left parenthesis straight A with bar on top straight B with bar on top straight A with bar on top straight B right parenthesis plus straight P left parenthesis straight A with bar on top straight B with bar on top straight A with bar on top straight B with bar on top straight A with bar on top straight B right parenthesis plus...... end cell row cell equals 5 over 6 cross times 1 over 6 plus 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 1 over 6 plus 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 1 over 6 plus..... end cell row cell equals 5 over 6 cross times 1 over 6 plus open parentheses 5 over 6 close parentheses squared 5 over 6 cross times 1 over 6 plus open parentheses 5 over 6 close parentheses to the power of 4 5 over 6 cross times 1 over 6 plus open parentheses 5 over 6 close parentheses to the power of 6 5 over 6 cross times 1 over 6 plus..... end cell row cell equals fraction numerator 5 over 6 cross times 1 over 6 over denominator 1 minus open parentheses 5 over 6 close parentheses squared end fraction equals 5 over 36 cross times 36 over 11 equals 5 over 11 end cell row cell text Since   the   probabilities   are   not   equal , end text end cell row cell text the   decision   of   the   refree   was   not   a   fair   one. end text end cell end table end style

Chapter 31 Probability Exercise Ex. 31.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

There machines E1, E2, E3 in a certain factory produce 50%, 25% and 25% respectively, of the total daily output of electric bulbs. It is known that 4% of the tubes produced one each of machines E1 and E2 are defective, and that 5% of those produced on E3 are defective. If one tube is picked up at random from a day’s production, calculate the probability that it is defective.Solution 13

Chapter 31 Probability Exercise Ex. 31.7

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3,4, 5 or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, what is the probability that she threw 3, 4, 5 or 6 with the die?Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

An item is manufactured by three machine A, B and C. out of the total number of items manufactured during a specified period, 50% are manufacture on machine A 30% on B and 20% on C. 2% of the items produced on A and 2% of items produced on B are defective and 3% of these produced on C are defective. All the items stored at one godown. One items is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?Solution 14

Question 15

There are three coins. One is two-headed coin (having head on both faces), another is biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tail 40% of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?Solution 15

begin mathsize 12px style table attributes columnalign left end attributes row cell text Let   end text straight E subscript 1 comma straight E subscript 2 comma straight E subscript 3 text   be   the   events   that   we   choose   the   first   coin , end text end cell row cell text second   coin ,  and   third   coin   respectively   in   a   random   toss. end text end cell row cell straight P left parenthesis straight E subscript 1 right parenthesis equals 1 third comma straight P left parenthesis straight E subscript 2 right parenthesis equals 1 third comma straight P left parenthesis straight E subscript 3 right parenthesis equals 1 third end cell row cell text Let   A   denote   the   event   when   the   toss   shows   heads. end text end cell row cell text It   is   given   that end text end cell row cell straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis equals 1 comma straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis equals 0.75 comma straight P left parenthesis straight A divided by straight E subscript 3 right parenthesis equals.60 end cell row cell text We   have   to   find   end text straight P left parenthesis straight E subscript 1 divided by straight A right parenthesis. end cell row cell By text   Baye ' s   theorem end text end cell row cell straight P left parenthesis straight E subscript 1 divided by straight A right parenthesis equals fraction numerator straight P left parenthesis straight E subscript 1 right parenthesis straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis plus straight P left parenthesis straight E subscript 2 right parenthesis straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis plus straight P left parenthesis straight E subscript 3 right parenthesis straight P left parenthesis straight A divided by straight E subscript 3 right parenthesis end fraction equals end cell row cell equals fraction numerator 1 third left parenthesis 1 right parenthesis over denominator 1 third left parenthesis 1 right parenthesis plus 1 third left parenthesis 0.75 right parenthesis plus 1 third left parenthesis 0.60 right parenthesis end fraction equals fraction numerator 1 divided by 3 over denominator left parenthesis 1 divided by 3 right parenthesis plus left parenthesis 1 divided by 4 right parenthesis plus left parenthesis 1 divided by 5 right parenthesis end fraction end cell row cell equals fraction numerator 1 divided by 3 over denominator 47 divided by 60 end fraction equals 20 over 47 end cell end table end style

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

In a group of 400 people, 160 are smokers and non-vegetarian, 100 are smokers and vegetarian and the remaining are non-smokers and vegetarian. The probabilities of getting a special chest disease are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the disease. What is the probability that the selected person is a smoker and non-vegetarian?Solution 19

begin mathsize 12px style table attributes columnalign left end attributes row cell text Let   end text straight E subscript 1 comma straight E subscript 2 comma straight E subscript 3 text   be   the   events   that   the   people   are   end text end cell row cell text smokers   and   non-vegetarian ,  smokers   and   vegetarian ,  end text end cell row cell text and   non-smokers   and   vegetarian   respectively. end text end cell row cell straight P left parenthesis straight E subscript 1 right parenthesis equals 2 over 5 comma straight P left parenthesis straight E subscript 2 right parenthesis equals 1 fourth comma straight P left parenthesis straight E subscript 3 right parenthesis equals 7 over 20 end cell row cell text Let   A   denote   the   event   that   the   person   has   the   special   chest   disease. end text end cell row cell text It   is   given   that end text end cell row cell straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis equals 0.35 comma straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis equals 0.20 comma straight P left parenthesis straight A divided by straight E subscript 3 right parenthesis equals 0.10 end cell row cell text We   have   to   find   end text straight P left parenthesis straight E subscript 1 divided by straight A right parenthesis. end cell row cell By text   Baye ' s   theorem end text end cell row cell straight P left parenthesis straight E subscript 1 divided by straight A right parenthesis equals fraction numerator straight P left parenthesis straight E subscript 1 right parenthesis straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis plus straight P left parenthesis straight E subscript 2 right parenthesis straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis plus straight P left parenthesis straight E subscript 3 right parenthesis straight P left parenthesis straight A divided by straight E subscript 3 right parenthesis end fraction equals end cell row cell equals fraction numerator 2 over 5 left parenthesis 0.35 right parenthesis over denominator 2 over 5 left parenthesis 0.35 right parenthesis plus 1 fourth left parenthesis 0.20 right parenthesis plus 7 over 20 left parenthesis 0.10 right parenthesis end fraction equals fraction numerator 7 divided by 50 over denominator left parenthesis 7 divided by 50 right parenthesis plus left parenthesis 1 divided by 20 right parenthesis plus left parenthesis 7 divided by 200 right parenthesis end fraction end cell row cell equals fraction numerator 7 divided by 50 over denominator 9 divided by 40 end fraction equals 28 over 45 end cell end table end style

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

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RD SHARMA SOLUTION CHAPTER- 30 Linear Programming I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 30 Linear programming Exercise Ex. 30.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Chapter 30 Linear programming Exercise Ex. 30.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13 Old

Solution 13 Old

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Find graphically, the maximum value of z = 2x + 5y, subject to constraints given below:

2x + 4y £ 8

3x + y £ 6

X + y £ 4

X ³ 0, y ³ 0Solution 26

Converting the inequations into equations, we obtain the lines

2x + 4y = 8, 3x + y = 6, x + y = 4, x = 0, y = 0.

These lines are drawn on a suitable scale and the feasible region of the LPP is shaded in the graph.

From the graph we can see the corner points as (0, 2) and (2, 0).

Chapter 30 Linear programming Exercise Ex. 30.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. food p costs Rs. 60 kg and Food Q costs Rs. 80 kg. Food P contains 3 units / kg of Vitamin A and 5 units/ kg of Vitamin B while food Q contains 4 unit / kg of Vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.Solution 10

Question 11

One kind of cake requires 200 g of flour and 25 g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes. Solution 11

Question 12

A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires at least 240units of calcium, atleasy 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to minimize the amount of vitamin A in the diet? What is the minimum amount of vitamin A?Solution 12

Question 13

A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs. 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs. 200 per bag contains 1.5 unit of nutritional element A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture pre bag?Solution 13

Note: Answer given in the book is incorrect.Question 14

A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:

FoodVitamin AVitamin BVitamin C
X123
Y221

One kg of food X costs Rs. 16 and one kg of food costs Rs. 20. Find the least cost of the mixture which will produce the required diet?Solution 14

Question 15

A fruit grower can use two types of fertilizer in his garden, brand P and Q. The amounts (in kg) of nitrogen, acid potash and chlorine in a bag of each brand are given in the Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.

Kg per bag
 Brand PBrand Q
Nitrogen33.5
Phosphoric acid12
Potash31.5
Chlorine1.52

If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?Solution 15

Chapter 30 Linear programming Exercise Ex. 30.4

Question 1

If a young man drives his vehicle at 25km/hr, he has to spend Rs 2per km on petrol. if he drives it as a fast of 40 km/hr, the petrol cost increase to Rs 5 per km. He has Rs 100 to speed on petrol and travel a maximum distance in one hour time with less polution . Express this problem as an LPP and solve it graphically. What value do you find hear?Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

A firm makes items A and B and the total number of items it can make in a day is 24. It takes one hour to make an item of A and half an hour to make an item B. The maximum time available per day is 16 hours. The profit on an item of A is Rs 300 and on one item of B is Rs 160. How many items of each type should be produced to maximize the profit? Solve the problem graphically.Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

A manufacturer makes two products, A and B. Product A sells at Rs. 200 each and takes ½ hour to make. Product B sells at Rs.300 each and takes 1 hour to make. There is a permanent order for 14 units of product A and 16 units of product B. A working week consist of 40 hours of production and the weekly turn over must not be less than Rs. 1000. If the profit on each of product A is Rs. 20 and an product B is Rs. 30, then how many of each should be produced so that the profit is maximum? Also find the maximum profit.Solution 35

Question 36

If a young man drives his vehicle at 25 km/hr, he has to spend Rs. 2 per km on petrol. If he drives is at a faster speed of 40 km/hr, the petrol cost increases to Rs. 5/ per km. He has Rs. 100 to spend on petrol and travel within one hour. Express this as an LPP solve the same.Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

A factory makes tennis rackets and cricket bats A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hour of machine time and 24 hours of craftman’s time. If the profit on racket and on a bat is Rs 20 and Rs 10 respectively, find the number of tennis rackets and cricket bats that the factory must manufacture to earn the maximum profit. Make it as an LPP and solve it graphically.    Solution 40

Question 41

A merchant plans to sell two types of personal computers-a desktop model and a portable model that will cost Rs 25,000 and Rs 40,000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and his profit on the desktop model is Rs 4500 and on the portable model is RS 5000. Make an LPP and solve it graphically.    Solution 41

Question 42

A cooperative society of formers has 50 hectare of land to grow two crops X and Y. The profit from crops X and Y per hectare are estimated as Rs. 10,500 and Rs. 9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at rates of 20 litres and 10 litres per hectare. Further, no more than 800 litres of herbicide should be used in order to protect fish and wild life using a pond which collects drainage from this land. How much land should be allocated to each crop so at to maximise the total profit of the society?Solution 42

Question 43

A manufacturing company makes two models A and B of a product. Each piece of Model A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each piece of Model B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of Rs. 8000 on each piece of model A and Rs. 1200 on each piece of Model B. How many pieces of Model A and Model B should be manufactured per week to realise a maximum profit? What is the maximum profit per week?Solution 43

Question 44

A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftrnan’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.

  1. What number of rackets and bats must be made if the factory is to Work at full capacity?
  2. If the profit on a racket and on a bat is Rs. 20 and Rs. 10 respectively, find the maximum profit of the factory when it works at full capacity.

Solution 44

Question 45

A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost Rs. 25000 and Rs. 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs. 70 lakhs and if his profit on the desktop model is Rs. 4500 and on portable model is Rs. 5000.Solution 45

Question 46

A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs. 12 and Rs. 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?Solution 46

Question 47

There are two types of fertilisers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs Rs. 6/kg and F2 costs Rs. 5 /kg, determine how much of each type of fertiliser should be used so that nutrient requirement are met at a minimum cost. What is the minimum cost?Solution 47

Question 48

A manufacturer has three machines I, II and III installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas machine III must be operated for atleast 5 hours a day. She produces only two items M and N each requiring the use of all the three machines.

The number of hours required for producing 1 unit of each of M and N on the three machines are given in the following table:

ItemsNumber of hours required on machines
 IIIIII
M121
N211.25

She makes a profit of Rs. 600 and Rs. 400 on items M and N respectively. How many of each item should she produce so as to maximize her profit assuming that she can sell all the items that she produced? What will be the maximum profit?Solution 48

Question 49

There are two factories located one at place P and the other at place Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are respectively 8 and 6 units. The cost of transportation per unit is given below:

To/fromCost (in Rs.)
 ABC
P160100150
Q100120100

How many units should be transported from each factory to each depot in order that the transportation cost is minimum. What will be the minimum transportation cost?Solution 49

Let x and y units of commodity be transported from factory P to the depots at A and B respectively.

Then (8 – x – y) units will be transported to depot at C.

The flow is shown below.

Question 50

A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:

Types of ToysMachines
 IIIIII
A12186
B609

Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs. 7.50 and that on each toy of type B is Rs.5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.Solution 50

Question 51

An aeroplane can carry a maximum of 200 passengers. A profit of Rs. 1000 is made on each executive class ticket and a profit of Rs. 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. What is the maximum profit?Solution 51

Question 52

A manufacturer considers that men and women workers are equally efficient and so he pays them at the same rate. He has 30 and 17 units of workers (male and female) and capital respectively, which he uses to produce two types of goods A and B. To produce one unit of A, 2 workers and 3 units of capital are required while 3 workers and 1 unit of capital is required to produce one unit of B. If A and B are priced at Rs. 100 and Rs. 120 per unit respectively, how should he use his resources to maximize the total revenue? Form the above as an LPP and solve graphically. Do you agree with this view of the manufacturer that men and women workers are equally efficient and so should be paid at the same rate?Solution 52

Chapter 30 – Linear programming Exercise Ex. 30.5

Question 1

Solution 1

Question 2

Solution 2

Read More

RD SHARMA SOLUTION CHAPTER- 29 The Plane I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 29 The plane Exercise Ex. 29.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Chapter 29 The plane Exercise Ex. 29.2

Question 1

Solution 1

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 29 The plane Exercise Ex. 29.3

Question 1

Solution 1

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 3

Solution 3

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13(i)

Solution 13(i)

Question 13(ii)

Solution 13(ii)

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Find the vector and Cartesian equations of the plane which passes through the point (5, 2, -4) and perpendicular to the line with direction ratios 2, 3, -1.Solution 18

begin mathsize 12px style table attributes columnalign left end attributes row cell text Vector   equation   of   the   plane : end text end cell row cell text Given   that   the   required   plane   passes   through   the end text end cell row cell text point   end text left parenthesis text 5 , 2 , end text minus 4 right parenthesis text   having   the   position   vector end text end cell row cell stack text a end text with rightwards arrow on top equals 5 straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top end cell row cell Also text   given   that   the   required   plane   is   perpendicular end text end cell row cell text to   the   line   with   direction   ratios   2 ,  3   and   end text minus 1. end cell row cell Thus text   the   vector   equation   of   the   normal   vector   to   the end text end cell row cell text plane   is   end text stack text n end text with rightwards arrow on top equals 2 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top. end cell row cell text We   know   that   the   vector   equation   of   the   plane   passing end text end cell row cell text through   a   point   having   position   vector   end text stack text a end text with rightwards arrow on top text   and   normal   to end text end cell row cell text vector   end text stack text n end text with rightwards arrow on top text   is   given   by   end text left parenthesis stack text r end text with rightwards arrow on top minus stack text a end text with rightwards arrow on top right parenthesis times stack text n end text with rightwards arrow on top equals 0 text   or ,  end text stack text r end text with rightwards arrow on top times stack text n end text with rightwards arrow on top equals stack text a end text with rightwards arrow on top times stack text n end text with rightwards arrow on top. end cell row cell Thus text   the   required   equation   of   the   required   plane   is end text end cell row cell stack text r end text with rightwards arrow on top times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top right parenthesis equals left parenthesis 5 straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top right parenthesis times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top right parenthesis end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top right parenthesis equals 10 plus 6 plus 4 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top right parenthesis equals 20 end cell end table end style
begin mathsize 12px style table attributes columnalign left end attributes row cell text The   Cartesian   equation   of   the   plane   is   end text end cell row cell stack text r end text with rightwards arrow on top times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top right parenthesis equals 20 end cell row cell rightwards double arrow left parenthesis straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top right parenthesis times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top right parenthesis equals 2 end cell row cell rightwards double arrow 2 straight x plus 3 straight y minus straight z equals 20 end cell end table end style

Question 19

If O be the origin and the coordinates of P be (1, 2, -3), then find the equation of the plane passing through P and perpendicular to OP.Solution 19

begin mathsize 12px style table attributes columnalign left end attributes row cell text Consider   the   point    P end text left parenthesis text 1 , 2 , end text minus 3 right parenthesis. end cell row cell text Thus   the   position   vector   of   the   point   P   is end text end cell row cell stack text a end text with rightwards arrow on top equals straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top end cell row cell Direction text   ratios   of   the   line   OP ,  where   O   is   the   end text end cell row cell text origin ,  are   1 , 2   and   end text minus 3 end cell row cell Thus text   the   vector   equation   of   the   normal   vector ,  OP ,  to   the end text end cell row cell text plane   is   end text stack text n end text with rightwards arrow on top equals straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top. end cell row cell text We   know   that   the   vector   equation   of   the   plane   passing end text end cell row cell text through   a   point   having   position   vector   end text stack text a end text with rightwards arrow on top text   and   normal   to end text end cell row cell text vector   end text stack text n end text with rightwards arrow on top text   is   given   by   end text left parenthesis stack text r end text with rightwards arrow on top minus stack text a end text with rightwards arrow on top right parenthesis times stack text n end text with rightwards arrow on top equals 0 text   or ,  end text stack text r end text with rightwards arrow on top times stack text n end text with rightwards arrow on top equals stack text a end text with rightwards arrow on top times stack text n end text with rightwards arrow on top. end cell row cell Thus text   the   required   equation   of   the   required   plane   is end text end cell row cell stack text r end text with rightwards arrow on top times left parenthesis straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals left parenthesis straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis times left parenthesis straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals 1 plus 4 plus 9 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals 14 end cell row cell rightwards double arrow left parenthesis straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top right parenthesis times left parenthesis straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals 14 end cell row cell rightwards double arrow straight x plus 2 straight y minus 3 straight z equals 14 end cell end table end style

Question 20

If O is the origin and the coordinates of A are (a, b, c). Find the direction cosines of OA and the equation of the plane through A at right angles to OA.Solution 20

Chapter 29 The plane Exercise Ex. 29.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   vector   equation   of   the   plane   which   end text end cell row cell text is   at   a   distance   of   end text fraction numerator text 6 end text over denominator square root of text 29 end text end root end fraction text   from   the   origin   and   its   end text end cell row cell text normal   vector   from   the   origin   is   end text 2 straight i with hat on top minus 3 straight j with hat on top plus 4 straight k with hat on top. text   end text end cell row cell text Also ,  find   its   cartesian   form. end text end cell end table end style

Solution 10

begin mathsize 12px style table attributes columnalign left end attributes row cell We text   know   that   the   vector   equation   of   a   plane   at   a   distance end text end cell row cell text ' p '  from   the   origin   and   normal   to   the   unit   vector   end text stack text n end text with hat on top text   is   end text stack text r end text with rightwards arrow on top times stack text n end text with hat on top equals straight p end cell row cell Vector text   normal   to   the   plane   is   end text stack text n end text with rightwards arrow on top equals 2 straight i with hat on top minus 3 straight j with hat on top plus 4 straight k with hat on top end cell row cell The text    end text unit text   vector   normal   to   the   plane   is end text end cell row cell stack text n end text with hat on top equals fraction numerator 2 over denominator square root of 2 squared plus left parenthesis negative 3 right parenthesis squared plus 4 squared end root end fraction straight i with hat on top minus fraction numerator 3 over denominator square root of 2 squared plus left parenthesis negative 3 right parenthesis squared plus 4 squared end root end fraction straight j with hat on top plus fraction numerator 4 over denominator square root of 2 squared plus left parenthesis negative 3 right parenthesis squared plus 4 squared end root end fraction straight k with hat on top end cell row cell rightwards double arrow stack text n end text with hat on top equals fraction numerator 2 over denominator square root of 4 plus 9 plus 16 end root end fraction straight i with hat on top minus fraction numerator 3 over denominator square root of 4 plus 9 plus 16 end root end fraction straight j with hat on top plus fraction numerator 4 over denominator square root of 4 plus 9 plus 16 end root end fraction straight k with hat on top end cell row cell rightwards double arrow stack text n end text with hat on top equals fraction numerator 2 over denominator square root of 29 end fraction straight i with hat on top minus fraction numerator 3 over denominator square root of 29 end fraction straight j with hat on top plus fraction numerator 4 over denominator square root of 29 end fraction straight k with hat on top end cell row cell Here comma text   given   that   p = end text fraction numerator text 6 end text over denominator square root of text 29 end text end root end fraction end cell row cell Thus comma text   the   vector   equation   of   the   plane   is end text end cell row cell stack text r end text with rightwards arrow on top times open parentheses fraction numerator 2 over denominator square root of 29 end fraction straight i with hat on top minus fraction numerator 3 over denominator square root of 29 end fraction straight j with hat on top plus fraction numerator 4 over denominator square root of 29 end fraction straight k with hat on top close parentheses equals fraction numerator text 6 end text over denominator square root of text 29 end text end root end fraction end cell row cell The text   Cartesian   equation   of   the   plane   is end text end cell row cell open parentheses straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top close parentheses times open parentheses fraction numerator 2 over denominator square root of 29 end fraction straight i with hat on top minus fraction numerator 3 over denominator square root of 29 end fraction straight j with hat on top plus fraction numerator 4 over denominator square root of 29 end fraction straight k with hat on top close parentheses equals fraction numerator text 6 end text over denominator square root of text 29 end text end root end fraction end cell row cell rightwards double arrow open parentheses fraction numerator 2 straight x over denominator square root of 29 end fraction minus fraction numerator 3 straight y over denominator square root of 29 end fraction plus fraction numerator 4 straight z over denominator square root of 29 end fraction close parentheses equals fraction numerator text 6 end text over denominator square root of text 29 end text end root end fraction end cell row cell rightwards double arrow open parentheses fraction numerator 2 straight x minus 3 straight y plus 4 straight z over denominator square root of 29 end fraction close parentheses equals fraction numerator text 6 end text over denominator square root of text 29 end text end root end fraction end cell row cell rightwards double arrow 2 straight x minus 3 straight y plus 4 straight z equals text 6 end text end cell end table end style

Question 11

Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin.Solution 11

begin mathsize 12px style table attributes columnalign left end attributes row cell The text   Cartesian   equation   of   the   given   plane   is end text end cell row cell text 2 x end text minus 3 straight y plus 4 straight z minus 6 equals 0. end cell row cell The text   above   equation   can   be   rewritten   as end text end cell row cell text 2 x end text minus 3 straight y plus 4 straight z equals 6 end cell row cell Therefore comma text   end text the text   vector   equation   of   the   plane   is end text end cell row cell open parentheses straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top close parentheses times open parentheses 2 straight i with hat on top minus 3 straight j with hat on top plus 4 straight k with hat on top close parentheses equals 6 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times open parentheses 2 straight i with hat on top minus 3 straight j with hat on top plus 4 straight k with hat on top close parentheses equals 6.... left parenthesis 1 right parenthesis end cell row cell We text   know   that   the   vector   equation   of   a   plane   at   a   distance end text end cell row cell text ' p '  from   the   origin   and   normal   to   unit   vector   end text stack text n end text with hat on top text   is   end text stack text r end text with rightwards arrow on top times stack text n end text with hat on top equals straight p end cell row blank row cell We text   have ,  end text stack text n end text with rightwards arrow on top equals 2 straight i with hat on top minus 3 straight j with hat on top plus 4 straight k with hat on top. end cell row cell Thus text   end text vertical line stack text n end text with rightwards arrow on top vertical line equals square root of 2 squared plus left parenthesis negative 3 right parenthesis squared plus 4 squared end root equals square root of 29 end cell row blank row cell Dividing text   the   equation  ( 1 )  by   end text vertical line stack text n end text with rightwards arrow on top vertical line equals square root of text 29 end text end root comma text   we   have , end text end cell row cell stack text r end text with rightwards arrow on top times open parentheses fraction numerator 2 straight i with hat on top minus 3 straight j with hat on top plus 4 straight k with hat on top over denominator square root of 29 end fraction close parentheses equals fraction numerator 6 over denominator square root of 29 end fraction end cell row cell Hence text   the   normal   form   of   the   equation   of   the   plane   is end text end cell row cell stack text r end text with rightwards arrow on top times open parentheses fraction numerator 2 over denominator square root of 29 end fraction straight i with hat on top minus fraction numerator 3 over denominator square root of 29 end fraction straight j with hat on top plus fraction numerator 4 over denominator square root of 29 end fraction straight k with hat on top close parentheses equals fraction numerator 6 over denominator square root of 29 end fraction end cell row cell text Hence   the   perpendicular   distance   of   the   end text end cell row cell text origin   from   the   plane   is   p = end text fraction numerator text 6 end text over denominator square root of text 29 end text end root end fraction. end cell end table end style

Chapter 29 The plane Exercise Ex. 29.5

Question 1

Solution 1

Question 2

Find the vector equation of the plane passing through the points P(2, 5, -3), Q(-2, -3, 5) and R(5, 3, -3).Solution 2

begin mathsize 12px style table attributes columnalign left end attributes row cell Let text   P end text left parenthesis text 2 , 5 , end text minus text 3 end text right parenthesis comma text   Q end text left parenthesis negative 2 comma negative 3 comma 5 right parenthesis text   and   R end text left parenthesis text 5 , 3 , end text minus 3 right parenthesis text   be   the   three end text end cell row cell text points   on   a   plane   having   position   vectors   end text stack text p end text with rightwards arrow on top text ,  end text stack text q end text with rightwards arrow on top text   and   end text stack text s end text with rightwards arrow on top end cell row cell text respectively .  Then   the   vectors   end text stack text PQ end text with rightwards arrow on top text   and   end text stack text PR end text with rightwards arrow on top text   are   in   the   same   plane. end text end cell row cell text Therefore ,  end text stack text PQ end text with rightwards arrow on top cross times stack text PR end text with rightwards arrow on top text   is   a   vector   perpendicular   to   the   plane. end text end cell row cell text Let   end text stack text n end text with rightwards arrow on top text  =  end text stack text PQ end text with rightwards arrow on top cross times stack text PR end text with rightwards arrow on top end cell row blank row cell PQ with rightwards arrow on top equals left parenthesis negative 2 minus 2 right parenthesis straight i with hat on top plus left parenthesis negative 3 minus 5 right parenthesis straight j with hat on top plus left parenthesis 5 minus left parenthesis negative 3 right parenthesis right parenthesis straight k with hat on top end cell row cell rightwards double arrow PQ with rightwards arrow on top equals negative 4 straight i with hat on top minus 8 straight j with hat on top plus 8 straight k with hat on top end cell row cell Similarly comma end cell row cell PR with rightwards arrow on top equals left parenthesis 5 minus 2 right parenthesis straight i with hat on top plus left parenthesis 3 minus 5 right parenthesis straight j with hat on top plus left parenthesis negative 3 minus left parenthesis negative 3 right parenthesis right parenthesis straight k with hat on top end cell row cell rightwards double arrow PR with rightwards arrow on top equals 3 straight i with hat on top minus 2 straight j with hat on top plus 0 straight k with hat on top end cell row cell Thus text   end text end cell row cell stack text n end text with rightwards arrow on top equals stack text PQ end text with rightwards arrow on top cross times stack text PR end text with rightwards arrow on top end cell row cell text     end text equals open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row cell negative 4 end cell cell negative 8 end cell 8 row 3 cell negative 2 end cell 0 end table close vertical bar end cell row cell text     end text equals 16 straight i with hat on top plus 24 straight j with hat on top plus 32 straight k with hat on top end cell row cell The text   plane   passes   through   the   point   P   with   end text end cell row cell text position   vector   end text stack text p end text with rightwards arrow on top equals 2 straight i with hat on top plus 5 straight j with hat on top minus 3 straight k with hat on top end cell row cell text Thus ,  its   vector   equation   is end text end cell row cell left curly bracket stack text r end text with rightwards arrow on top minus left parenthesis 2 straight i with hat on top plus 5 straight j with hat on top minus 3 straight k with hat on top right parenthesis right curly bracket times left parenthesis 16 straight i with hat on top plus 24 straight j with hat on top plus 32 straight k with hat on top right parenthesis equals 0 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis 16 straight i with hat on top plus 24 straight j with hat on top plus 32 straight k with hat on top right parenthesis minus left parenthesis 32 plus 120 minus 96 right parenthesis equals 0 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis 16 straight i with hat on top plus 24 straight j with hat on top plus 32 straight k with hat on top right parenthesis minus 56 equals 0 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis 16 straight i with hat on top plus 24 straight j with hat on top plus 32 straight k with hat on top right parenthesis equals 56 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus 4 straight k with hat on top right parenthesis equals 7 end cell end table end style

Question 3

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   vector   equation   of   the   plane   passing   end text end cell row cell text through   the   points   end text straight A left parenthesis straight a text , 0 , 0 end text right parenthesis comma straight B left parenthesis 0 comma straight b comma 0 right parenthesis text   and   end text straight C left parenthesis 0 comma 0 comma straight c right parenthesis. text   end text end cell row cell text Reduce   it   to   normal   form .  end text end cell row cell text If   plane   end text ABC text   is   at   a   distance   end text straight p text   from   the   origin ,  end text end cell row cell text prove   that   end text fraction numerator text 1 end text over denominator straight p to the power of text 2 end text end exponent end fraction equals fraction numerator text 1 end text over denominator straight a to the power of text 2 end text end exponent end fraction plus fraction numerator text 1 end text over denominator straight b to the power of text 2 end text end exponent end fraction plus fraction numerator text 1 end text over denominator straight c to the power of text 2 end text end exponent end fraction. end cell end table end style

Solution 3

begin mathsize 12px style table attributes columnalign left end attributes row cell Let text   A end text left parenthesis text a , 0 , end text 0 right parenthesis comma text   B end text left parenthesis 0 comma straight b comma 0 right parenthesis text   and   C end text left parenthesis 0 comma 0 comma straight c right parenthesis text   be   three end text end cell row cell text points   on   a   plane   having   their   position   vectors   end text stack text a end text with rightwards arrow on top text ,  end text stack text b end text with rightwards arrow on top text   and   end text stack text c end text with rightwards arrow on top end cell row cell text respectively .  Then   vectors   end text stack text AB end text with rightwards arrow on top text   and   end text stack text AC end text with rightwards arrow on top text   are   in   the   same   plane. end text end cell row cell text Therefore ,  end text stack text AB end text with rightwards arrow on top cross times stack text AC end text with rightwards arrow on top text   is   a   vector   perpendicular   to   the   plane. end text end cell row cell text Let   end text stack text n end text with rightwards arrow on top text = end text stack text AB end text with rightwards arrow on top cross times stack text AC end text with rightwards arrow on top end cell row blank row cell stack text AB end text with rightwards arrow on top equals left parenthesis 0 minus straight a right parenthesis straight i with hat on top plus left parenthesis straight b minus 0 right parenthesis straight j with hat on top plus left parenthesis 0 minus 0 right parenthesis straight k with hat on top end cell row cell rightwards double arrow stack text AB end text with rightwards arrow on top equals negative straight a straight i with hat on top plus straight b straight j with hat on top plus 0 straight k with hat on top end cell row cell Similarly comma end cell row cell stack text AC end text with rightwards arrow on top equals left parenthesis 0 minus straight a right parenthesis straight i with hat on top plus left parenthesis 0 minus 0 right parenthesis straight j with hat on top plus left parenthesis straight c minus 0 right parenthesis straight k with hat on top end cell row cell rightwards double arrow stack text AC end text with rightwards arrow on top equals negative straight a straight i with hat on top plus 0 straight j with hat on top plus straight c straight k with hat on top end cell row cell Thus text   end text end cell row cell stack text n end text with rightwards arrow on top equals stack text AB end text with rightwards arrow on top cross times stack text AC end text with rightwards arrow on top end cell row cell text     end text equals vertical line table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row cell negative straight a end cell straight b 0 row cell negative straight a end cell 0 straight c end table vertical line end cell row cell stack text n end text with rightwards arrow on top equals bc straight i with hat on top plus ac straight j with hat on top plus ab straight k with hat on top end cell row cell rightwards double arrow straight n with hat on top equals fraction numerator bc straight i with hat on top plus ac straight j with hat on top plus ab straight k with hat on top over denominator square root of straight b squared straight c squared plus straight a squared straight c squared plus straight a squared straight b squared end root end fraction end cell row cell The text   plane   passes   through   the   point   P   with   end text end cell row cell text position   vector   end text stack text a end text with rightwards arrow on top equals straight a straight i with hat on top plus 0 straight j with hat on top plus 0 straight k with hat on top end cell end table end style
begin mathsize 12px style table attributes columnalign left end attributes row cell text Thus ,  the   vector   equation   in   the   normal   form   is end text end cell row cell open curly brackets stack text r end text with rightwards arrow on top minus open parentheses left parenthesis straight a straight i with hat on top plus 0 straight j with hat on top plus 0 straight k with hat on top close parentheses close curly brackets times open parentheses fraction numerator bc straight i with hat on top plus ac straight j with hat on top plus ab straight k with hat on top over denominator square root of straight b squared straight c squared plus straight a squared straight c squared plus straight a squared straight b squared end root end fraction close parentheses equals 0 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times fraction numerator left parenthesis bc straight i with hat on top plus ac straight j with hat on top plus ab straight k with hat on top right parenthesis over denominator square root of straight b squared straight c squared plus straight a squared straight c squared plus straight a squared straight b squared end root end fraction equals fraction numerator abc over denominator square root of straight b squared straight c squared plus straight a squared straight c squared plus straight a squared straight b squared end root end fraction end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times fraction numerator open parentheses bc straight i with hat on top plus ac straight j with hat on top plus ab straight k with hat on top close parentheses over denominator square root of straight b squared straight c squared plus straight a squared straight c squared plus straight a squared straight b squared end root end fraction equals fraction numerator 1 over denominator square root of fraction numerator straight b squared straight c squared plus straight a squared straight c squared plus straight a squared straight b squared over denominator straight a squared straight b squared straight c squared end fraction end root end fraction end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times fraction numerator left parenthesis bc straight i with hat on top plus ac straight j with hat on top plus ab straight k with hat on top right parenthesis over denominator square root of straight b squared straight c squared plus straight a squared straight c squared plus straight a squared straight b squared end root end fraction equals fraction numerator 1 over denominator square root of 1 over straight a squared plus 1 over straight b squared plus 1 over straight c squared end root end fraction... left parenthesis 1 right parenthesis end cell row cell The text   vector   equation   of   a   plane   normal   to   the   unit   vector end text end cell row cell stack text n end text with hat on top text   and   at   a   distance  ' d '  from   the   origin   is   end text stack text r end text with rightwards arrow on top times stack text n   end text with hat on top text =  d end text... text ( 2 ) end text end cell row cell Given text   that   the   plane   is   at   a   distance  ' p '  from   the end text end cell row cell text origin. end text end cell row cell text Comparing   equations  ( 1 )  and  ( 2 ),  we   have , end text end cell row cell text d  =  p  =  end text fraction numerator 1 over denominator square root of 1 over straight a squared plus 1 over straight b squared plus 1 over straight c squared end root end fraction end cell row cell rightwards double arrow 1 over straight p squared equals 1 over straight a squared plus 1 over straight b squared plus 1 over straight c squared end cell end table end style

Question 4

Find the vector equation of the plane passing through the points (1, 1, -1), (6, 4, -5) and (-4, -2, 3).Solution 4

begin mathsize 12px style table attributes columnalign left end attributes row cell Let text   P end text left parenthesis text 1 , 1 , end text minus 1 right parenthesis comma text   Q end text left parenthesis 6 comma 4 comma negative 5 right parenthesis text   and   R end text left parenthesis negative text 4 , end text minus text 2 , end text 3 right parenthesis text   be   three end text end cell row cell text points   on   a   plane   having   position   vectors   end text stack text p end text with rightwards arrow on top text ,  end text stack text q end text with rightwards arrow on top text   and   end text stack text s end text with rightwards arrow on top end cell row cell text respectively .  Then   the   vectors   end text stack text PQ end text with rightwards arrow on top text   and   end text stack text PR end text with rightwards arrow on top text   are   in   the   same   plane. end text end cell row cell text Therefore ,  end text stack text PQ end text with rightwards arrow on top cross times stack text PR end text with rightwards arrow on top text   is   a   vector   perpendicular   to   the   plane. end text end cell row cell text Let   end text stack text n end text with rightwards arrow on top text  =  end text stack text PQ end text with rightwards arrow on top cross times stack text PR end text with rightwards arrow on top end cell row blank row cell PQ with rightwards arrow on top equals left parenthesis 6 minus 1 right parenthesis straight i with hat on top plus left parenthesis 4 minus 1 right parenthesis straight j with hat on top plus left parenthesis negative 5 minus left parenthesis negative 1 right parenthesis right parenthesis straight k with hat on top end cell row cell rightwards double arrow PQ with rightwards arrow on top equals 5 straight i with hat on top plus 3 straight j with hat on top minus 4 straight k with hat on top end cell row cell Similarly comma end cell row cell PR with rightwards arrow on top equals left parenthesis negative 4 minus 1 right parenthesis straight i with hat on top plus left parenthesis negative 2 minus 1 right parenthesis straight j with hat on top plus left parenthesis 3 minus left parenthesis negative 1 right parenthesis right parenthesis straight k with hat on top end cell row cell rightwards double arrow PR with rightwards arrow on top equals negative 5 straight i with hat on top minus 3 straight j with hat on top plus 4 straight k with hat on top end cell row cell Thus text   end text end cell row cell Here comma text   end text stack text PQ end text with rightwards arrow on top equals negative stack text PR end text with rightwards arrow on top text   end text end cell row cell Therefore comma text   the   given   points   are   collinear .  end text end cell row cell text Thus ,  end text stack text n end text with rightwards arrow on top equals straight a straight i with hat on top plus straight b straight j with hat on top plus straight c straight k with hat on top text   where ,  5 a + 3 b end text minus 4 straight c equals 0 end cell row cell The text   plane   passes   through   the   point   P   with   end text end cell row cell text position   vector   end text stack text p end text with rightwards arrow on top equals straight i with hat on top plus straight j with hat on top minus straight k with hat on top end cell row cell text Thus ,  its   vector   equation   is end text end cell row cell left curly bracket stack text r end text with rightwards arrow on top minus left parenthesis straight i with hat on top plus straight j with hat on top minus straight k with hat on top right parenthesis right curly bracket times left parenthesis straight a straight i with hat on top plus straight b straight j with hat on top plus straight c straight k with hat on top right parenthesis equals 0 comma text   where ,  5 a + 3 b end text minus 4 straight c equals 0 end cell end table end style

Question 5

Solution 5

Chapter 29 The plane Exercise Ex. 29.6

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Find the angle between the planes:

2x + y – 2z = 5 and 3x – 6y – 2z = 7Solution 2(v)

begin mathsize 12px style table attributes columnalign left end attributes row cell We text   know   that   the   angle   between   the   planes end text end cell row cell text a end text subscript text 1 end text end subscript text x  +  b end text subscript text 1 end text end subscript text y  +  c end text subscript text 1 end text end subscript text z  +  d end text subscript text 1 end text end subscript text  =  0   and   a end text subscript text 2 end text end subscript text x  +  b end text subscript text 2 end text end subscript text y  +  c end text subscript text 2 end text end subscript text z  +  d end text subscript text 2 end text end subscript text  =  0   is   given   by end text end cell row cell text cos end text straight theta equals fraction numerator straight a subscript 1 straight a subscript 2 plus straight b subscript 1 straight b subscript 2 plus straight c subscript 1 straight c subscript 2 over denominator square root of straight a subscript 1 squared plus straight b subscript 1 squared plus straight c subscript 1 squared end root times square root of straight a subscript 2 squared plus straight b subscript 2 squared plus straight c subscript 2 squared end root end fraction end cell row cell Therefore comma text   the   angle   between   2 x + y end text minus 2 straight z equals 5 text   and   3 x end text minus text 6 y end text minus text 2 z = 7 end text end cell row cell text cos end text straight theta equals fraction numerator 2 cross times 3 plus 1 cross times left parenthesis negative 6 right parenthesis plus left parenthesis negative 2 right parenthesis cross times left parenthesis negative 2 right parenthesis over denominator square root of 2 squared plus 1 squared plus left parenthesis negative 2 right parenthesis squared end root times square root of 3 squared plus left parenthesis negative 6 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared end root end fraction end cell row cell rightwards double arrow text cos end text straight theta equals fraction numerator 6 minus 6 plus 4 over denominator square root of 9 times square root of 9 plus 36 plus 4 end root end fraction end cell row cell rightwards double arrow text cos end text straight theta equals fraction numerator 4 over denominator 3 cross times 7 end fraction end cell row cell rightwards double arrow straight theta equals cos to the power of negative 1 end exponent open parentheses 4 over 21 close parentheses end cell end table end style

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.Solution 11

begin mathsize 12px style table attributes columnalign left end attributes row cell The text   equation   of   the   plane   parallel   to   ZOX   is   y  =  constant. end text end cell row cell text Given   that   the   y-intercept   is   3. end text end cell row cell text Thus   the   equation   of   the   plane   is   y  =  3. end text end cell end table end style

Question 12

Find the equation of the plane that contains the point (1, -1, 2) and is perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8.Solution 12

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   equation   of   any   plane   passing   through   end text left parenthesis text 1 , end text minus 1 comma 2 right parenthesis end cell row cell is text   a end text left parenthesis text x end text minus 1 right parenthesis plus straight b left parenthesis straight y plus 1 right parenthesis plus straight c left parenthesis straight z minus 2 right parenthesis equals 0.... left parenthesis 1 right parenthesis end cell row cell Given text   that ,  p end text lane text  ( 1 )  is   perpendicular   to   the   planes end text end cell row cell text 2 x + 3 y end text minus text 2 z  =  5 end text end cell row cell text and end text end cell row cell text x + 2 y end text minus text 3 z  =  8   end text end cell row cell Therefore comma text   we   have , end text end cell row cell text 2 a + 3 b end text minus text 2 c  =  0 end text... text ( 2 ) end text end cell row cell text and   end text end cell row cell text a + 2 b end text minus text 3 c  =  0 end text... text ( 3 ) end text end cell row cell text Solving   equations  ( 2 )  and  ( 3 )  by   cross   multiplication ,  we   have , end text end cell row cell fraction numerator text a end text over denominator text 3 end text cross times left parenthesis negative text 3 end text right parenthesis minus 2 cross times left parenthesis negative 2 right parenthesis end fraction equals fraction numerator text b end text over denominator 1 cross times left parenthesis negative 2 right parenthesis minus 2 cross times left parenthesis negative 3 right parenthesis end fraction equals fraction numerator text c end text over denominator 2 cross times 2 minus 1 cross times 3 end fraction equals straight lambda left parenthesis say right parenthesis end cell row cell rightwards double arrow fraction numerator text a end text over denominator negative 9 plus 4 end fraction equals fraction numerator text b end text over denominator negative 2 plus 6 end fraction equals fraction numerator text c end text over denominator 4 minus 3 end fraction equals straight lambda end cell row cell rightwards double arrow fraction numerator text a end text over denominator negative 5 end fraction equals fraction numerator text b end text over denominator 4 end fraction equals fraction numerator text c end text over denominator 1 end fraction equals straight lambda end cell row cell Thus comma text   we   have , end text end cell row cell text a  = end text minus 5 straight lambda comma straight b equals 4 straight lambda text   and   c  =  end text straight lambda end cell row cell Substituting text   the   above   values   in   equation  ( 1 ),  we   have , end text end cell row cell negative 5 straight lambda left parenthesis text x end text minus 1 right parenthesis plus 4 straight lambda left parenthesis straight y plus 1 right parenthesis plus straight lambda left parenthesis straight z minus 2 right parenthesis equals 0 end cell row cell Since text   end text straight lambda not equal to text 0 ,  we   have , end text end cell row cell negative 5 left parenthesis text x end text minus 1 right parenthesis plus 4 left parenthesis straight y plus 1 right parenthesis plus left parenthesis straight z minus 2 right parenthesis equals 0 end cell row cell rightwards double arrow negative 5 straight x plus 5 plus 4 straight y plus 4 plus straight z minus 2 equals 0 end cell row cell rightwards double arrow negative 5 straight x plus 4 straight y plus straight z plus 7 equals 0 end cell row cell rightwards double arrow 5 straight x minus 4 straight y minus straight z minus 7 equals 0 end cell row cell rightwards double arrow 5 straight x minus 4 straight y minus straight z equals 7 end cell row cell text Thus   the   required   equation   of   the   plane   is   end text 5 straight x minus 4 straight y minus straight z equals 7 end cell end table end style

Question 13

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   equation   of   the   plane   passing   through   end text end cell row cell left parenthesis straight a comma straight b comma straight c right parenthesis text   and   parallel   to   the   plane   end text stack text r end text with rightwards arrow on top. left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis equals 2. end cell end table end style

Solution 13

begin mathsize 12px style table attributes columnalign left end attributes row cell Given text   that   the   equation   of   the   required   end text end cell row cell text p end text lane text   is   parallel   to   the   plane end text end cell row cell straight r with rightwards arrow on top times left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis equals 2... left parenthesis 1 right parenthesis end cell row cell therefore Vector text   equation   of   any   plane   parallel   to  ( 1 )  is end text end cell row cell stack text r end text with rightwards arrow on top times left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis text = end text straight k... left parenthesis 2 right parenthesis end cell row cell Since text   the   given   plane   passes   through   end text left parenthesis text a ,  b ,  c end text right parenthesis comma then end cell row cell left parenthesis straight a straight i with hat on top plus straight b straight j with hat on top plus straight c straight k with hat on top right parenthesis times left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis text = end text straight k end cell row cell rightwards double arrow straight a plus straight b plus straight c equals straight k... left parenthesis 3 right parenthesis end cell row blank row cell Substituting text   the   above   value   of   k   in   equation  ( 2 ),  we   have , end text end cell row cell stack text r end text with rightwards arrow on top times left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis text = end text straight a plus straight b plus straight c end cell row cell text Thus   the   required   equation   of   the   plane   is   end text straight x plus straight y plus straight z text   =  end text straight a plus straight b plus straight c end cell end table end style

Question 14

Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.Solution 14

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   equation   of   any   plane   passing   through   end text left parenthesis negative 1 comma 3 comma 2 right parenthesis end cell row cell is left parenthesis text x  +  end text 1 right parenthesis plus straight b left parenthesis straight y minus 3 right parenthesis plus straight c left parenthesis straight z minus 2 right parenthesis equals 0.... left parenthesis 1 right parenthesis end cell row cell Given text   that ,  end text Plane text  ( 1 )  is   perpendicular   to   the   planes end text end cell row cell text x  +  2 y  +  3 z  =  5 end text end cell row cell text and end text end cell row cell text 3 x  +  3 y end text plus text z  =  0 end text end cell row cell Therefore comma text   we   have , end text end cell row cell text a  +  2 b end text plus 3 text c  =  0 end text... text ( 2 ) end text end cell row cell text and   end text end cell row cell text 3 a  +  3 b end text plus text c  =  0 end text... text ( 3 ) end text end cell row cell text Solving   equations  ( 2 )  and  ( 3 )  by   cross   multiplication ,  we   have , end text end cell row cell fraction numerator text a end text over denominator text 2 end text cross times 1 minus 3 cross times 3 end fraction equals fraction numerator text b end text over denominator 3 cross times 3 minus 1 cross times 1 end fraction equals fraction numerator text c end text over denominator 1 cross times 3 minus 3 cross times 2 end fraction equals straight lambda left parenthesis say right parenthesis end cell row cell rightwards double arrow fraction numerator text a end text over denominator 2 minus 9 end fraction equals fraction numerator text b end text over denominator 9 minus 1 end fraction equals fraction numerator text c end text over denominator 3 minus 6 end fraction equals straight lambda end cell row cell rightwards double arrow fraction numerator text a end text over denominator negative 7 end fraction equals fraction numerator text b end text over denominator 8 end fraction equals fraction numerator text c end text over denominator negative 3 end fraction equals straight lambda end cell row cell Thus comma text   we   have , end text end cell row cell text a  = end text minus 7 straight lambda comma straight b equals 8 straight lambda text   and   c  = end text minus text 3 end text straight lambda end cell row cell Substituting text   the   above   values   in   equation  ( 1 ),  we   have , end text end cell row cell negative 7 straight lambda left parenthesis text x  +  end text 1 right parenthesis plus 8 straight lambda left parenthesis straight y minus 3 right parenthesis minus text 3 end text straight lambda left parenthesis straight z minus 2 right parenthesis equals 0 end cell row cell Since text   end text straight lambda not equal to text 0 ,  we   have , end text end cell row cell negative 7 left parenthesis text x + end text 1 right parenthesis plus 8 left parenthesis straight y minus 3 right parenthesis minus text 3 end text left parenthesis straight z minus 2 right parenthesis equals 0 end cell row cell rightwards double arrow negative 7 straight x minus 7 plus 8 straight y minus 24 minus 3 straight z plus 6 equals 0 end cell row cell rightwards double arrow negative 7 straight x plus 8 straight y minus 3 straight z minus 25 equals 0 end cell row cell rightwards double arrow 7 straight x minus 8 straight y plus 3 straight z plus 25 equals 0 end cell row blank row cell text Thus   the   required   equation   of   the   plane   is   end text 7 straight x minus 8 straight y plus 3 straight z plus 25 equals 0 end cell end table end style

Question 15

Find the vector equation of the plane through the points (2, 1, -1) and (-1, 3, 4) and perpendicular to the plane x – 2y + 4z = 10.Solution 15

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   equation   of   any   plane   passing   through   end text left parenthesis 2 comma 1 comma negative 1 right parenthesis end cell row cell is text   a end text left parenthesis text x end text minus 2 right parenthesis plus straight b left parenthesis straight y minus 1 right parenthesis plus straight c left parenthesis straight z plus 1 right parenthesis equals 0.... left parenthesis 1 right parenthesis end cell row cell Also comma text   the   above   plane   passes   through   the   point   end text left parenthesis negative 1 comma 3 comma 4 right parenthesis. end cell row cell Thus comma text   equation  ( 1 ),  becomes , end text end cell row cell text a end text left parenthesis negative text 1 end text minus 2 right parenthesis plus straight b left parenthesis 3 minus 1 right parenthesis plus straight c left parenthesis 4 plus 1 right parenthesis equals 0 end cell row cell rightwards double arrow negative 3 text a end text plus 2 straight b plus 5 straight c equals 0... left parenthesis 2 right parenthesis end cell row cell Given text   that ,  end text Plane text  ( 1 )  is   perpendicular   to   the   plane end text end cell row cell text x end text minus text 2 y + 4 z = 10 end text end cell row cell Therefore comma text   we   have , end text end cell row cell text a end text minus text 2 b end text plus 4 text c = 0 end text... text ( 3 ) end text end cell row blank row cell text Solving   equations  ( 2 )  and  ( 3 )  by   cross   multiplication ,  we   have , end text end cell row cell fraction numerator text a end text over denominator text 2 end text cross times 4 minus 5 cross times left parenthesis negative 2 right parenthesis end fraction equals fraction numerator text b end text over denominator 1 cross times 5 minus left parenthesis negative 3 right parenthesis cross times 4 end fraction equals fraction numerator text c end text over denominator left parenthesis negative 3 right parenthesis cross times left parenthesis negative 2 right parenthesis minus 1 cross times 2 end fraction equals straight lambda left parenthesis say right parenthesis end cell row cell rightwards double arrow fraction numerator text a end text over denominator 8 plus 10 end fraction equals fraction numerator text b end text over denominator 5 plus 12 end fraction equals fraction numerator text c end text over denominator 6 minus 2 end fraction equals straight lambda end cell row cell rightwards double arrow fraction numerator text a end text over denominator 18 end fraction equals fraction numerator text b end text over denominator 17 end fraction equals fraction numerator text c end text over denominator 4 end fraction equals straight lambda end cell row cell Thus comma text   we   have , end text end cell row cell text a = end text 18 straight lambda comma straight b equals 17 straight lambda text   and   c = end text 4 straight lambda end cell row cell Substituting text   the   above   values   in   equation  ( 1 ),  we   have , end text end cell row cell 18 straight lambda left parenthesis text x end text minus 2 right parenthesis plus 17 straight lambda left parenthesis straight y minus 1 right parenthesis plus 4 straight lambda left parenthesis straight z plus 1 right parenthesis equals 0 end cell row cell Since text   end text straight lambda not equal to text 0 ,  we   have , end text end cell row cell 18 left parenthesis text x end text minus 2 right parenthesis plus 17 left parenthesis straight y minus 1 right parenthesis plus 4 left parenthesis straight z plus 1 right parenthesis equals 0 end cell row cell rightwards double arrow 18 straight x minus 36 plus 17 straight y minus 17 plus 4 straight z plus 4 equals 0 end cell row cell rightwards double arrow 18 straight x plus 17 straight y plus 4 straight z minus 49 equals 0 end cell row blank row cell text Thus   the   required   equation   of   the   plane   is   end text 18 straight x plus 17 straight y plus 4 straight z minus 49 equals 0 end cell end table end style

Chapter 29 The plane Exercise Ex. 29.7

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Chapter 29 The plane Exercise Ex. 29.8

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   vector   equation   of   the   plane   passing   end text end cell row cell text through   the   intersection   of   the   planes   end text end cell row cell stack text r end text with rightwards arrow on top. left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis equals 6 text   and   end text stack text r end text with rightwards arrow on top. left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus 4 straight k with hat on top right parenthesis equals negative 5 text   and   the   point   end text left parenthesis text 1 ,  1 ,  1 end text right parenthesis. end cell end table end style

Solution 13

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   equation   of   a   plane   passing   through   the   intersection   of end text end cell row cell stack text r end text with rightwards arrow on top times left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis equals 6 text   and   end text stack text r end text with rightwards arrow on top times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus 4 straight k with hat on top right parenthesis equals negative 5 text   is end text end cell row cell left square bracket stack text r end text with rightwards arrow on top times left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis minus 6 right square bracket plus straight lambda left square bracket stack text r end text with rightwards arrow on top times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus 4 straight k with hat on top right parenthesis plus 5 right square bracket equals 0 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times left square bracket left parenthesis 1 plus 2 straight lambda right parenthesis straight i with hat on top plus left parenthesis 1 plus 3 straight lambda right parenthesis straight j with hat on top plus left parenthesis 1 plus 4 straight lambda right parenthesis straight k with hat on top right square bracket equals left parenthesis 6 minus 5 straight lambda right parenthesis... left parenthesis 1 right parenthesis end cell row cell rightwards double arrow left square bracket straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top right square bracket times left square bracket left parenthesis 1 plus 2 straight lambda right parenthesis straight i with hat on top plus left parenthesis 1 plus 3 straight lambda right parenthesis straight j with hat on top plus left parenthesis 1 plus 4 straight lambda right parenthesis straight k with hat on top right square bracket equals left parenthesis 6 minus 5 straight lambda right parenthesis end cell row cell rightwards double arrow left square bracket straight x left parenthesis 1 plus 2 straight lambda right parenthesis plus straight y left parenthesis 1 plus 3 straight lambda right parenthesis plus straight z left parenthesis 1 plus 4 straight lambda right parenthesis right square bracket equals left parenthesis 6 minus 5 straight lambda right parenthesis... left parenthesis 2 right parenthesis end cell row cell text The   requried   plane   also   passes   through   the   point end text left parenthesis text 1 ,  1 ,  1 end text right parenthesis. end cell row cell text Substiuting   x  =  1 , y  =  1 , z  =  1   in   equation  ( 2 ),  we   have , end text end cell row cell 1 cross times left parenthesis 1 plus 2 straight lambda right parenthesis plus 1 cross times left parenthesis 1 plus 3 straight lambda right parenthesis plus 1 cross times left parenthesis 1 plus 4 straight lambda right parenthesis equals left parenthesis 6 minus 5 straight lambda right parenthesis end cell row cell rightwards double arrow 1 plus 2 straight lambda plus 1 plus 3 straight lambda plus 1 plus 4 straight lambda equals 6 minus 5 straight lambda end cell row cell rightwards double arrow 3 plus 9 straight lambda equals 6 minus 5 straight lambda end cell row cell rightwards double arrow 14 straight lambda equals 6 minus 3 end cell row cell rightwards double arrow 14 straight lambda equals 3 end cell row cell rightwards double arrow straight lambda equals 3 over 14 end cell end table end style
begin mathsize 12px style table attributes columnalign left end attributes row cell text Substituting   the   value   end text straight lambda equals 3 over 14 text   in   equation  ( 1 ),  we   have , end text end cell row cell stack text r end text with rightwards arrow on top times open square brackets open parentheses 1 plus 2 open parentheses 3 over 14 close parentheses close parentheses space straight i with hat on top plus open parentheses 1 plus 3 open parentheses 3 over 14 close parentheses close parentheses space straight j with hat on top plus open parentheses 1 plus 4 open parentheses 3 over 14 close parentheses close parentheses space straight k with hat on top close square brackets equals open parentheses 6 minus 5 open parentheses 3 over 14 close parentheses close parentheses end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times open square brackets 20 over 14 straight i with hat on top plus 23 over 14 straight j with hat on top plus 26 over 14 straight k with hat on top close square brackets equals 69 over 14 end cell row cell rightwards double arrow stack text r end text with rightwards arrow on top times open square brackets 20 straight i with hat on top plus 23 straight j with hat on top plus 26 straight k with hat on top close square brackets equals 69 end cell end table end style

Question 14

Solution 14

Question 15

Find the equation of the plane through the intersection of the planes 3x – y 2z = 4 and x + y + z = 2 and the point (2, 2, 1).Solution 15

Question 16

Find the vector equation of the plane through the line of intersection of the plane x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.Solution 16

Question 17

Find the equation of the plane passing through (a, b, c) and parallel to the plane

Solution 17

Chapter 29 The plane Exercise Ex. 29.9

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Find the distance between the point (7, 2, 4) and the plane determined by the points A (2, 5, -3), B (-2, -3, 5) and C (5, 3, -3)Solution 11

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   equation   of   any   plane   passing   through   A end text left parenthesis 2 comma 5 comma negative 3 right parenthesis end cell row cell is text   a end text left parenthesis text x end text minus 2 right parenthesis plus straight b left parenthesis straight y minus 5 right parenthesis plus straight c left parenthesis straight z plus 3 right parenthesis equals 0.... left parenthesis 1 right parenthesis end cell row cell The text   above   plane   passes   through   the   point   B end text left parenthesis negative text 2 , end text minus text 3 , 5 end text right parenthesis end cell row cell and text   hence ,  we   have , end text end cell row cell straight a left parenthesis negative text 2 end text minus 2 right parenthesis plus straight b left parenthesis negative 3 minus 5 right parenthesis plus straight c left parenthesis 5 plus 3 right parenthesis equals 0 end cell row cell rightwards double arrow negative 4 straight a minus 8 straight b plus 8 straight c equals 0... left parenthesis 2 right parenthesis end cell row cell Again text   the   required   plane   passes   through   the   point   C end text left parenthesis text 5 , 3 , end text minus 3 right parenthesis end cell row cell and text   hence ,  we   have , end text end cell row cell straight a left parenthesis 5 minus 2 right parenthesis plus straight b left parenthesis 3 minus 5 right parenthesis plus straight c left parenthesis negative 3 plus 3 right parenthesis equals 0 end cell row cell rightwards double arrow 3 straight a minus 2 straight b plus 0 straight c equals 0... left parenthesis 3 right parenthesis end cell row cell Solving text   equations  ( 2 )  and  ( 3 )  by   cross   multiplication ,  we   have , end text end cell row cell fraction numerator text a end text over denominator left parenthesis negative text 8 end text right parenthesis cross times 0 minus left parenthesis negative 2 right parenthesis cross times 8 end fraction equals fraction numerator straight b over denominator 3 cross times 8 minus left parenthesis negative 4 right parenthesis cross times 0 end fraction equals fraction numerator straight c over denominator left parenthesis negative 4 right parenthesis cross times left parenthesis negative 2 right parenthesis minus 3 cross times left parenthesis negative 8 right parenthesis end fraction equals straight lambda left parenthesis say right parenthesis end cell row cell rightwards double arrow fraction numerator text a end text over denominator 0 plus 16 end fraction equals fraction numerator straight b over denominator 24 plus 0 end fraction equals fraction numerator straight c over denominator 8 plus 24 end fraction equals straight lambda end cell row cell rightwards double arrow fraction numerator text a end text over denominator 16 end fraction equals straight b over 24 equals straight c over 32 equals straight lambda end cell row cell rightwards double arrow fraction numerator text a end text over denominator 2 end fraction equals straight b over 3 equals straight c over 4 equals straight lambda end cell row cell rightwards double arrow straight a equals 2 straight lambda comma straight b equals 3 straight lambda text   and   c = 4 end text straight lambda end cell row cell Substituting text   the   above   values   in   equation  ( 1 ),  we   have , end text end cell row cell 2 straight lambda left parenthesis text x end text minus 2 right parenthesis plus 3 straight lambda left parenthesis straight y minus 5 right parenthesis plus text 4 end text straight lambda left parenthesis straight z plus 3 right parenthesis equals 0 end cell row cell Since text   end text straight lambda not equal to text 0 ,  we   have , end text end cell row cell 2 left parenthesis text x end text minus 2 right parenthesis plus 3 left parenthesis straight y minus 5 right parenthesis plus text 4 end text left parenthesis straight z plus 3 right parenthesis equals 0 end cell row cell rightwards double arrow 2 straight x minus 4 plus 3 straight y minus 15 plus text 4 z + 12 end text equals 0 end cell row cell rightwards double arrow 2 straight x plus 3 straight y plus text 4 z end text minus text 7 end text equals 0 end cell end table end style
begin mathsize 12px style table attributes columnalign left end attributes row cell Thus text   the   equation   of   the   plane   is   end text end cell row cell 2 straight x plus 3 straight y plus text 4 z end text minus text 7 end text equals 0 end cell row cell The text   distance   from   the   point   P end text left parenthesis text 7 , 2 , 4 end text right parenthesis text   to   the   plane   is end text end cell row cell text d = end text open vertical bar fraction numerator text ax end text subscript text 1 end text end subscript plus b y subscript 1 plus c z subscript 1 plus straight d over denominator square root of straight a squared plus straight b squared plus straight c squared end root end fraction close vertical bar end cell row cell therefore text Distance ,  d = end text open vertical bar fraction numerator 2 straight x plus 3 straight y plus text 4 z end text minus text 7 end text over denominator square root of 2 squared plus 3 squared plus 4 squared end root end fraction close vertical bar end cell row cell rightwards double arrow straight d subscript left parenthesis 7 comma 2 comma 4 right parenthesis end subscript equals open vertical bar fraction numerator 2 cross times 7 plus 3 cross times 2 plus text 4 end text cross times text 4 end text minus text 7 end text over denominator square root of 2 squared plus 3 squared plus 4 squared end root end fraction close vertical bar end cell row cell rightwards double arrow straight d subscript left parenthesis 7 comma 2 comma 4 right parenthesis end subscript equals open vertical bar fraction numerator 29 over denominator square root of 29 end fraction close vertical bar end cell row cell rightwards double arrow straight d subscript left parenthesis 7 comma 2 comma 4 right parenthesis end subscript equals square root of 29 text   units end text end cell end table end style

Question 12

A plane makes intercepts -6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it.Solution 12

begin mathsize 12px style table attributes columnalign left end attributes row cell Given text   that   a   plane   is   making   intercepts   end text minus 6 comma 3 text   and   4 end text end cell row cell text respectively   on   the   coordinate   axes. end text end cell row cell text Thus   the   equation   of   the   plane   is end text end cell row cell fraction numerator text x end text over denominator negative text 6 end text end fraction plus straight y over 3 plus straight z over 4 equals 1... left parenthesis 1 right parenthesis end cell row cell We text   need   to   find   the   length   of   the   perpendicular end text end cell row cell text from   the   origin   on   the   plane. end text end cell row cell text If   the   plane   end text fraction numerator text x end text over denominator straight a end fraction plus straight y over straight b plus straight z over straight c equals 1 text   is   at   a   distance  ' p ',  then end text end cell row cell fraction numerator text 1 end text over denominator text p end text to the power of text 2 end text end exponent end fraction equals 1 over straight a squared plus 1 over straight b squared plus 1 over straight c squared... left parenthesis 2 right parenthesis end cell row cell Comparing text   equation  ( 1 )  with   the   end text end cell row cell text general   equation ,  we   get , end text end cell row cell text a = end text minus 6 comma straight b equals 3 text   and   c = 4 end text end cell row cell text Thus ,  equation  ( 2 )  becomes , end text end cell row cell fraction numerator text 1 end text over denominator text p end text to the power of text 2 end text end exponent end fraction equals 1 over left parenthesis negative 6 right parenthesis squared plus 1 over 3 squared plus 1 over 4 squared end cell row cell rightwards double arrow fraction numerator text 1 end text over denominator text p end text to the power of text 2 end text end exponent end fraction equals fraction numerator 4 plus 16 plus 9 over denominator 144 end fraction end cell row cell rightwards double arrow fraction numerator text 1 end text over denominator text p end text to the power of text 2 end text end exponent end fraction equals 29 over 144 end cell row cell rightwards double arrow text p end text to the power of text 2 end text end exponent equals 144 over 29 end cell row cell rightwards double arrow text p end text equals fraction numerator 12 over denominator square root of 29 end fraction text   units end text end cell end table end style

Chapter 29 The plane Exercise Ex. 29.10

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Chapter 29 The plane Exercise Ex. 29.11

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

begin mathsize 12px style table attributes columnalign left end attributes row cell Find space the space vector space and space cartesian space forms space of space the space equation space end cell row cell of space the space plane space passing space through space the space point space left parenthesis 1 comma space 2 comma space minus 4 right parenthesis space and space end cell row cell parallel space to space the space lines space straight r with rightwards arrow on top equals left parenthesis straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top right parenthesis plus straight lambda left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus 6 straight k with hat on top right parenthesis space and end cell row cell straight r with rightwards arrow on top equals left parenthesis straight i with hat on top minus 3 straight j with hat on top plus 5 straight k with hat on top right parenthesis plus straight mu left parenthesis straight i with hat on top plus straight j with hat on top minus straight k with hat on top right parenthesis .  Also comma space find space the space distance space of space end cell row cell the space point space left parenthesis 9 ,- 8 ,- 10 right parenthesis space from space the space plane space thus space obtained. end cell end table end style

Solution 18

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   plane   passes   through   the   point   end text stack text a end text with rightwards arrow on top open parentheses 1 comma space 2 comma space minus 4 close parentheses end cell row cell text A   vector   in   a   direction   perpendicular   to   end text end cell row cell stack text r end text with rightwards arrow on top equals open parentheses straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top close parentheses plus straight lambda open parentheses 2 straight i with hat on top plus 3 straight j with hat on top plus 6 straight k with hat on top close parentheses text   and   end text stack text r end text with rightwards arrow on top equals open parentheses straight i with hat on top minus 3 straight j with hat on top plus 5 straight k with hat on top close parentheses plus straight mu open parentheses straight i with hat on top plus straight j with hat on top minus straight k with hat on top close parentheses end cell row cell text is   end text stack text n end text with rightwards arrow on top equals left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus 6 straight k with hat on top right parenthesis cross times left parenthesis straight i with hat on top plus straight j with hat on top minus straight k with hat on top right parenthesis end cell row cell rightwards double arrow stack text n end text with rightwards arrow on top equals open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 2 3 6 row 1 1 cell negative 1 end cell end table close vertical bar equals negative 9 straight i with hat on top plus 8 straight j with hat on top minus straight k with hat on top end cell row cell text Equation   of   the   plane   is   end text open parentheses straight r with rightwards arrow on top minus straight a with rightwards arrow on top close parentheses. straight n with rightwards arrow on top equals 0 end cell row cell open parentheses straight r with rightwards arrow on top minus open parentheses straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top close parentheses close parentheses. open parentheses negative 9 straight i with hat on top plus 8 straight j with hat on top minus straight k with hat on top close parentheses equals 0 end cell row cell rightwards double arrow straight r with rightwards arrow on top. open parentheses negative 9 straight i with hat on top plus 8 straight j with hat on top minus straight k with hat on top close parentheses equals 11 end cell row cell text Substituting   end text straight r with rightwards arrow on top equals straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top comma text   we   get   the   Cartesian   form   as end text end cell row cell negative 9 straight x plus 8 straight y minus straight z equals 11 end cell row cell text The   distance   of   the   point   end text left parenthesis text 9 , end text minus text 8 , end text minus text 10 end text right parenthesis text   from   the   plane end text end cell row cell text = end text open vertical bar fraction numerator negative 9 left parenthesis 9 right parenthesis plus 8 left parenthesis negative 8 right parenthesis minus left parenthesis negative 10 right parenthesis minus 11 over denominator square root of 9 squared plus 8 squared plus 1 squared end root end fraction close vertical bar equals fraction numerator 146 over denominator square root of 146 end fraction equals square root of 146 end cell end table end style

Question 19

Solution 19

Question 20

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   coordinates   of   the   point   where   the   line   end text end cell row cell fraction numerator text x -2 end text over denominator text 3 end text end fraction equals fraction numerator straight y plus 1 over denominator 4 end fraction equals fraction numerator straight z minus 2 over denominator 2 end fraction text   intersects   the   plane   x-y + z -5 = 0 .  end text end cell row cell text Also ,  find   the   angle   between   the   line   and   the   plane. end text end cell end table end style

Solution 20

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   coordinates   of   the   point   where   the   line   end text end cell row cell fraction numerator text x -2 end text over denominator text 3 end text end fraction equals fraction numerator straight y plus 1 over denominator 4 end fraction equals fraction numerator straight z minus 2 over denominator 2 end fraction text = r end text end cell row cell rightwards double arrow straight x equals 3 straight r plus 2 comma straight y equals 4 straight r minus 1 comma straight z equals 2 straight r plus 2 end cell row cell text Substituting   in   the   equation   of   the   plane   x  -  y  +  z  -  5  =  0 ,  end text end cell row cell text we   get   end text left parenthesis 3 straight r plus 2 right parenthesis minus left parenthesis 4 straight r minus 1 right parenthesis plus left parenthesis 2 straight r plus 2 right parenthesis minus 5 equals 0 end cell row cell rightwards double arrow straight r equals 0 end cell row cell rightwards double arrow straight x equals 2 comma straight y equals negative 1 comma straight z equals 2 end cell row cell text Direction   ratios   of   the   line   are   end text 3 comma 4 comma 2 end cell row cell text Direction   ratios   of   a   line   perpendicular   to   the   plane   are   end text 1 comma negative 1 comma 1 end cell row cell sinθ equals fraction numerator 3 cross times 1 plus 4 cross times negative 1 plus 2 cross times 1 over denominator square root of 9 plus 16 plus 4 end root square root of 1 plus 1 plus 1 end root end fraction equals fraction numerator 1 over denominator square root of 87 end fraction end cell row cell straight theta equals sin to the power of negative 1 end exponent fraction numerator 1 over denominator square root of 87 end fraction end cell end table end style

Question 21

Solution 21

Question 22

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   angle   between   the   line   end text fraction numerator text x + 1 end text over denominator text 2 end text end fraction equals straight y over 3 equals fraction numerator straight z minus 3 over denominator 6 end fraction text   end text end cell row cell text and   the   plane   end text 10 straight x plus 2 straight y minus 11 straight z text   end text equals text   end text 3 end cell end table end style

Solution 22

begin mathsize 12px style table attributes columnalign left end attributes row cell text Direction   ratios   of   the   line   end text fraction numerator text x + 1 end text over denominator text 2 end text end fraction equals straight y over 3 equals fraction numerator straight z minus 3 over denominator 6 end fraction text   end text end cell row cell text are   end text left parenthesis 2 comma 3 comma 6 right parenthesis end cell row cell text Direction   ratio   of   a   line   perpendicular   to   the   plane   end text end cell row cell 10 straight x plus 2 straight y minus 11 straight z text   end text equals text   end text 3 text   are   end text 10 comma 2 comma negative 11 end cell row cell text If   end text straight theta text   is   the   angle   between   the   line   and   the   plane ,  then   end text end cell row cell sinθ equals fraction numerator 2 cross times 10 plus 3 cross times 2 plus 6 cross times negative 11 over denominator square root of 2 squared plus 3 squared plus 6 squared end root square root of 10 squared plus 2 squared plus 11 squared end root end fraction equals negative fraction numerator 40 over denominator square root of 49 square root of 225 end fraction equals negative fraction numerator 40 over denominator 7 cross times 15 end fraction equals negative 8 over 21 end cell row cell rightwards double arrow straight theta equals sin to the power of negative 1 end exponent open parentheses negative 8 over 21 close parentheses end cell end table end style

Question 23

Solution 23

Question 24

Solution 24

Question 25

Find the equation of the plane passing through the points

(-1, 2, 0),(2, 2, -1) and parallel to the line

Solution 25

Chapter 29 The plane Exercise Ex. 29.12

Question 1(i)

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the yz-plane.Solution 1(i)

begin mathsize 12px style table attributes columnalign left end attributes row cell text Direction   ratios   of   the   given   line   are end text end cell row cell left parenthesis 5 minus 3 comma 1 minus 4 comma 6 minus 1 right parenthesis equals left parenthesis 2 comma negative 3 comma 5 right parenthesis end cell row cell text Hence ,  equation   of   the   line   is   end text end cell row cell fraction numerator straight x minus 5 over denominator 2 end fraction equals fraction numerator straight y minus 1 over denominator negative 3 end fraction equals fraction numerator straight z minus 6 over denominator 5 end fraction equals straight r end cell row cell rightwards double arrow straight x equals 2 straight r plus 5 comma straight y equals negative 3 straight r plus 1 comma straight z equals 5 straight r plus 6 end cell row cell text For   any   point   on   the   end text yz minus text plane   end text straight x equals 0 end cell row cell rightwards double arrow 2 straight r plus 5 equals 0 rightwards double arrow straight r equals negative 5 over 2 end cell row cell straight y equals negative 3 left parenthesis negative 5 over 2 right parenthesis plus 1 equals 17 over 2 end cell row cell straight z equals 5 left parenthesis negative 5 over 2 right parenthesis plus 6 equals negative 13 over 2 end cell row cell text Hence ,  the   coordinates   of   the   point   are   end text open parentheses 0 comma 17 over 2 comma negative 13 over 2 close parentheses. end cell end table end style

Question 1(ii)

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the zx-plane.Solution 1(ii)

begin mathsize 12px style table attributes columnalign left end attributes row cell text Direction   ratios   of   the   given   line   are end text end cell row cell left parenthesis 5 minus 3 comma 1 minus 4 comma 6 minus 1 right parenthesis equals left parenthesis 2 comma negative 3 comma 5 right parenthesis end cell row cell text Hence ,  equation   of   the   line   is   end text end cell row cell fraction numerator straight x minus 5 over denominator 2 end fraction equals fraction numerator straight y minus 1 over denominator negative 3 end fraction equals fraction numerator straight z minus 6 over denominator 5 end fraction equals straight r end cell row cell rightwards double arrow straight x equals 2 straight r plus 5 comma straight y equals negative 3 straight r plus 1 comma straight z equals 5 straight r plus 6 end cell row cell text For   any   point   on   end text zx minus text plane   end text straight y equals 0 end cell row cell rightwards double arrow negative 3 straight r plus 1 equals 0 rightwards double arrow straight r equals 1 third end cell row cell straight x equals 2 open parentheses 1 third close parentheses plus 5 equals 17 over 3 end cell row cell straight z equals 5 open parentheses 1 third close parentheses plus 6 equals 23 over 3 end cell row cell text Hence ,  the   coordinates   of   the   point   are   end text open parentheses 17 over 3 comma 0 comma 23 over 3 close parentheses. end cell end table end style

Question 2

F i n d space t h e space c o o r d i n a t e s space o f space t h e space p o i n t space w h e r e space t h e space l i n e space t h r o u g h space open parentheses 3 comma negative 4 comma negative 5 close parentheses space a n d
open parentheses 2 comma negative 3 comma 1 close parentheses space c r o s s e s space t h e space p l a n e space 2 x plus y plus z equals 7

Solution 2

L e t space t h e space c o o r d i n a t e s space o f space t h e space p o i n t s space A space a n d space B space b e space
open parentheses 3 comma negative 4 comma negative 5 close parentheses space a n d space open parentheses 2 comma negative 3 comma 1 close parentheses space r e p e c t i v e l y.
E q u a t i o n space o f space t h e space l i n e space j o i n i n g space t h e space p o i n t s space open parentheses x subscript 1 comma y subscript 1 comma z subscript 1 close parentheses space a n d space open parentheses x subscript 2 comma y subscript 2 comma z subscript 2 close parentheses space i s
fraction numerator x minus x subscript 1 over denominator x subscript 2 minus x subscript 1 end fraction equals fraction numerator y minus y subscript 1 over denominator y subscript 2 minus y subscript 1 end fraction equals fraction numerator z minus z subscript 1 over denominator z subscript 2 minus z subscript 1 end fraction equals r comma space w h e r e space r space i s space s o m e space c o n s tan t.
T h u s space e q u a t i o n space o f space A B space i s
fraction numerator x minus 3 over denominator 2 minus 3 end fraction equals fraction numerator y minus open parentheses negative 4 close parentheses over denominator open parentheses negative 3 close parentheses minus open parentheses negative 4 close parentheses end fraction equals fraction numerator z minus open parentheses negative 5 close parentheses over denominator 1 minus open parentheses negative 5 close parentheses end fraction equals r
rightwards double arrow fraction numerator x minus 3 over denominator negative 1 end fraction equals fraction numerator y plus 4 over denominator 1 end fraction equals fraction numerator z plus 5 over denominator 6 end fraction equals r
A n y space p o i n t space o n space t h e space l i n e space A B space i s space o f space t h e space f o r m
minus r plus 3 comma space r minus 4 comma space 6 r minus 5
L e t space P space b e space t h e space p o i n t space o f space i n t e r s e c t i o n space o f space t h e space l i n e space A B space a n d space t h e space p l a n e space 2 x plus y plus z equals 7
T h u s comma space w e space h a v e comma
2 open parentheses negative r plus 3 close parentheses plus space r minus 4 plus 6 r minus 5 equals 7
rightwards double arrow negative 2 r plus 6 plus space r minus 4 plus 6 r minus 5 equals 7
rightwards double arrow 5 r equals 10
rightwards double arrow r equals 2
S u b s t i t u t i n g space t h e space v a l u e space o f space r space i n space minus r plus 3 comma space r minus 4 comma space 6 r minus 5 comma space t h e space c o o r d i n a t e s space o f space P space a r e colon
open parentheses negative 2 plus 3 comma space 2 minus 4 comma space 6 cross times 2 minus 5 close parentheses equals open parentheses 1 comma negative 2 comma 7 close parentheses

Question 3

F i n d space t h e space d i s tan c e space o f space t h e space p o i n t space open parentheses negative 1 comma negative 5 comma negative 10 close parentheses space f r o m space t h e space p o i n t space o f space i n t e r s e c t i o n
o f space t h e space l i n e space r with rightwards arrow on top equals open parentheses 2 i with hat on top minus j with hat on top plus 2 k with overparenthesis on top close parentheses plus lambda open parentheses 3 i with hat on top plus 4 j with hat on top plus 2 k with hat on top close parentheses space a n d space t h e space p l a n e
r with rightwards arrow on top times open parentheses i with hat on top minus j with hat on top plus k with hat on top close parentheses equals 5

Solution 3

Question 4

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   distance   of   the   point   end text left parenthesis text 2 , 12 , 5 end text right parenthesis text   from   the   point   end text end cell row cell text of   intersection   of   the   line   end text stack text r end text with rightwards arrow on top equals 2 straight i with hat on top minus 4 straight j with hat on top plus 2 straight k with hat on top plus straight lambda open parentheses 3 straight i with hat on top plus 4 straight j with hat on top plus 2 straight k with hat on top close parentheses end cell row cell text and   the   plane   end text stack text r end text with rightwards arrow on top. open parentheses straight i with hat on top minus 2 straight j with hat on top plus straight k with hat on top close parentheses equals 0. end cell end table end style

Solution 4

begin mathsize 12px style table attributes columnalign left end attributes row cell text To   find   the   point   of   intersection   of   the   line   end text end cell row cell stack text r end text with rightwards arrow on top equals 2 straight i with hat on top minus 4 straight j with hat on top plus 2 straight k with hat on top plus straight lambda open parentheses 3 straight i with hat on top plus 4 straight j with hat on top plus 2 straight k with hat on top close parentheses text   and   the   plane   end text stack text r end text with rightwards arrow on top. open parentheses straight i with hat on top minus 2 straight j with hat on top plus straight k with hat on top close parentheses equals 0 comma end cell row cell text we   substitute   end text stack text r end text with rightwards arrow on top text   of   line   in   the   plane. end text end cell row cell open square brackets 2 straight i with hat on top minus 4 straight j with hat on top plus 2 straight k with hat on top plus straight lambda open parentheses 3 straight i with hat on top plus 4 straight j with hat on top plus 2 straight k with hat on top close parentheses close square brackets. open parentheses straight i with hat on top minus 2 straight j with hat on top plus straight k with hat on top close parentheses equals 0 end cell row cell rightwards double arrow open square brackets open parentheses 2 plus 3 straight lambda close parentheses straight i with hat on top plus open parentheses negative 4 plus 4 straight lambda close parentheses straight j with hat on top plus open parentheses 2 plus 2 straight lambda close parentheses straight k with hat on top close square brackets. open parentheses straight i with hat on top minus 2 straight j with hat on top plus straight k with hat on top close parentheses equals 0 end cell row cell rightwards double arrow 2 plus 3 straight lambda plus 8 minus 8 straight lambda plus 2 plus 2 straight lambda equals 0 end cell row cell rightwards double arrow 3 straight lambda equals 12 rightwards double arrow straight lambda equals 4 end cell row cell stack text r end text with rightwards arrow on top equals 2 straight i with hat on top minus 4 straight j with hat on top plus 2 straight k with hat on top plus 4 open parentheses 3 straight i with hat on top plus 4 straight j with hat on top plus 2 straight k with hat on top close parentheses equals 14 straight i with hat on top plus 12 straight j with hat on top plus 10 straight k with hat on top end cell row cell text Hence ,  the   distance   of   the   point   2 end text straight i with hat on top plus 12 straight j with hat on top plus 5 straight k with hat on top text   from   end text 14 straight i with hat on top plus 12 straight j with hat on top plus 10 straight k with hat on top text   is end text end cell row cell square root of left parenthesis 14 minus 2 right parenthesis squared plus left parenthesis 12 minus 12 right parenthesis squared plus left parenthesis 10 minus 5 right parenthesis squared end root equals square root of 12 squared plus 5 squared end root equals square root of 169 equals 13 end cell end table end style

Question 5

Find the distance of the point P (-1, -5, -10) from the point of intersection of the line joining the points A (2, -1, 2) and B (5, 3, 4) with the plane x – y + z = 5.Solution 5

begin mathsize 12px style table attributes columnalign left end attributes row cell text Equation   of   the   line   through   the   points   A end text left parenthesis 2 comma negative 1 comma 2 right parenthesis end cell row cell text and   B end text left parenthesis 5 comma 3 comma 4 right parenthesis text   is   end text fraction numerator straight x minus 2 over denominator 5 minus 2 end fraction equals fraction numerator straight y plus 1 over denominator 3 plus 1 end fraction equals fraction numerator straight z minus 2 over denominator 4 minus 2 end fraction equals straight r end cell row cell rightwards double arrow fraction numerator straight x minus 2 over denominator 3 end fraction equals fraction numerator straight y plus 1 over denominator 4 end fraction equals fraction numerator straight z minus 2 over denominator 2 end fraction equals straight r end cell row cell rightwards double arrow straight x equals 3 straight r plus 2 comma straight y equals 4 straight r minus 1 comma straight z equals 2 straight r plus 2 end cell row cell text Substituting   these   in   the   plane   equation   we   get end text end cell row cell left parenthesis 3 straight r plus 2 right parenthesis minus left parenthesis 4 straight r minus 1 right parenthesis plus left parenthesis 2 straight r plus 2 right parenthesis equals 5 end cell row cell rightwards double arrow straight r equals 0 end cell row cell rightwards double arrow straight x equals 2 comma straight y equals negative 1 comma straight z equals 2 end cell row cell text Distance   of   end text left parenthesis 2 comma negative 1 comma 2 right parenthesis text   from   end text left parenthesis negative 1 comma negative 5 comma negative 10 right parenthesis text   is end text end cell row cell equals square root of left parenthesis 2 minus left parenthesis negative 1 right parenthesis right parenthesis squared plus left parenthesis negative 1 minus left parenthesis negative 5 right parenthesis right parenthesis squared plus left parenthesis 2 minus left parenthesis negative 10 right parenthesis right parenthesis squared end root equals square root of 3 squared plus 4 squared plus 12 squared end root end cell row cell equals square root of 169 equals 13 end cell end table end style

Question 6

Find the distance of the point P(3, 4,4) from the point, where the line joining the points A(3, -4, -5) and B (2, -3, 1) intersects the plane 2x + y + z =7.Solution 6

Chapter 29 The plane Exercise Ex. 29.13

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Show space that space the space plane space whose space vector space equation space is space straight r with rightwards arrow on top. left parenthesis straight i with hat on top plus 2 straight j with hat on top minus straight k with hat on top right parenthesis space equals 3 space contains space the space line
whose space vector space equation space is space straight r with rightwards arrow on top equals straight i with hat on top plus straight j with hat on top plus straight lambda left parenthesis 2 straight i with hat on top plus straight j with hat on top plus 4 straight k with hat on top right parenthesis.

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Let space the space equation space of space the space plane space be space straight x over straight a plus straight y over straight b plus straight z over straight c equals 1........ left parenthesis straight i right parenthesis
Plane space is space passing space through space left parenthesis 3 comma 4 comma 2 right parenthesis space and space left parenthesis 7 comma 0 comma 6 right parenthesis
3 over straight a plus 4 over straight b plus 2 over straight c equals 1
7 over straight a plus 0 over straight b plus 6 over straight c equals 1
Required space plane space is space perpendicular space to space 2 straight x minus 5 straight y minus 15 equals 0
2 over straight a plus fraction numerator negative 5 over denominator straight b end fraction plus 0 over straight c equals 0
rightwards double arrow 2 straight b equals 5 straight a
therefore straight b equals 2.5 straight a
3 over straight a plus fraction numerator 4 over denominator 2.5 straight a end fraction plus 2 over straight c equals 1
7 over straight a plus 6 over straight c equals 1
Solving space the space above space 2 space equations comma
straight a equals 3.4 equals 17 over 5 comma space straight b equals space 8.5 equals 17 over 2 space and space straight c equals fraction numerator negative 34 over denominator 6 end fraction equals negative 17 over 3
Substituting space the space values space in space left parenthesis straight i right parenthesis
fraction numerator straight x over denominator 17 over 5 end fraction plus fraction numerator straight y over denominator 17 over 2 end fraction plus fraction numerator straight z over denominator negative 17 over 3 end fraction equals 1
rightwards double arrow fraction numerator 5 straight x over denominator 17 end fraction plus fraction numerator 2 straight y over denominator 17 end fraction minus fraction numerator 3 straight z over denominator 17 end fraction equals 1
rightwards double arrow 5 straight x plus 2 straight y minus 3 straight z equals 17
rightwards double arrow left parenthesis straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top right parenthesis. left parenthesis 5 straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals 17
rightwards double arrow straight r with rightwards arrow on top. left parenthesis 5 straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals 17
Vector space equation space of space the space plane space is space straight r with rightwards arrow on top. left parenthesis 5 straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals 17.
The space line space passes space through space straight B left parenthesis 1 comma 3 comma negative 2 right parenthesis.
5 left parenthesis 1 right parenthesis plus 2 left parenthesis 3 right parenthesis minus 3 left parenthesis negative 2 right parenthesis equals 17
The space point space straight B space lies space on space the space plane.
therefore The space line space straight r with rightwards arrow on top equals straight i with hat on top plus 3 straight j with hat on top minus 2 straight k with hat on top plus straight lambda left parenthesis straight i with hat on top minus straight j with hat on top plus straight k with hat on top right parenthesis space lies space on space the space plane space straight r with rightwards arrow on top. left parenthesis 5 straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis equals 17.

Question 9

Solution 9

Question 10

F i n d space t h e space c o o r d i n a t e s space o f space t h e space p o i n t space w h e r e space t h e space l i n e
fraction numerator x minus 2 over denominator 3 end fraction equals fraction numerator y plus 1 over denominator 4 end fraction equals fraction numerator z minus 2 over denominator 2 end fraction space i n t e r s e c t s space t h e space p l a n e space x minus y plus z minus 5 equals 0. space A l s o comma space f i n d
t h e space a n g l e space b e t w e e n space t h e space l i n e space a n d space t h e space p l a n e.

Solution 10

A n y space p o i n t space o n space t h e space l i n e
fraction numerator x minus 2 over denominator 3 end fraction equals fraction numerator y plus 1 over denominator 4 end fraction equals fraction numerator z minus 2 over denominator 2 end fraction space equals k
i s space o f space t h e space f o r m comma space open parentheses 3 k plus 2 comma space 4 k minus 1 comma space 2 k plus 2 close parentheses.
I f space t h e space p o i n t space P open parentheses 3 k plus 2 comma space 4 k minus 1 comma space 2 k plus 2 close parentheses space l i e s space i n space t h e space p l a n e space x minus y plus z minus 5 equals 0 comma space w e space h a v e comma
open parentheses 3 k plus 2 close parentheses minus open parentheses 4 k minus 1 close parentheses plus open parentheses 2 k plus 2 close parentheses minus 5 equals 0
rightwards double arrow 3 k plus 2 minus 4 k plus 1 plus 2 k plus 2 minus 5 equals 0
rightwards double arrow k equals 0
T h u s comma space t h e space c o o r d i n a t e s space o f space t h e space p o i n t space o f space i n t e r s e c t i o n space o f space t h e space l i n e space a n d
t h e space p l a n e space a r e colon space P open parentheses 3 cross times 0 plus 2 comma space 4 cross times 0 minus 1 comma space 2 cross times 0 plus 2 close parentheses equals P open parentheses 2 comma negative 1 comma 2 close parentheses

L e t space theta space b e space t h e space a n g l e space b e t w e e n space t h e space l i n e space a n d space t h e space p l a n e.
T h u s comma
sin theta equals fraction numerator a l plus b m plus c n over denominator square root of a squared plus b squared plus c squared end root square root of l squared plus m squared plus n squared end root end fraction comma space w h e r e comma space l comma m space a n d space n space a r e space t h e space d i r e c t i o n
r a t i o s space o f space t h e space l i n e space a n d space a comma b space a n d space c space a r e space t h e space d i r e c t i o n space r a t i o s space o f space t h e space n o r m a l
t o space t h e space p l a n e.
H e r e comma space l equals 3 comma space m equals 4 comma space n equals 2 comma space a equals 1 comma space b equals negative 1 comma space a n d space c equals 1
H e n c e comma
sin theta equals fraction numerator 1 cross times 3 plus open parentheses negative 1 close parentheses cross times 4 plus 1 cross times 2 over denominator square root of 1 squared plus open parentheses negative 1 close parentheses squared plus 1 squared end root square root of 3 squared plus 4 squared plus 2 squared end root end fraction
rightwards double arrow sin theta equals fraction numerator 1 over denominator square root of 3 square root of 29 end fraction
rightwards double arrow theta equals sin to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 3 square root of 29 end fraction close parentheses

Question 11

F i n d space t h e space v e c t o r space e q u a t i o n space o f space t h e space p l a n e space p a s sin g space t h r o u g h space t h r e e space p o i n t s
w i t h space p o s i t i o n space v e c t o r s space i with hat on top plus j with hat on top minus 2 k with hat on top comma space 2 i with hat on top minus j with hat on top plus k with hat on top space a n d space i with hat on top plus 2 j with hat on top plus k with hat on top. space A l s o comma space f i n d space t h e space
c o o r d i n a t e s space o f space t h e space p o i n t space o f space i n t e r s e c t i o n space o f space t h i s space p l a n e space a n d space
t h e space l i n e space r with rightwards arrow on top equals 3 i with hat on top minus j with hat on top minus k with hat on top plus lambda open parentheses 2 i with hat on top minus 2 j with hat on top plus k with hat on top close parentheses.

Solution 11

L e t space A comma space B space a n d space C space b e space t h r e e space p o i n t s space w i t h space p o s i t i o n space v e c t o r s space
i with hat on top plus j with hat on top minus 2 k with hat on top comma space 2 i with hat on top minus j with hat on top plus k with hat on top space a n d space i with hat on top plus 2 j with hat on top plus k with hat on top. space
T h u s comma space stack A B with rightwards arrow on top equals b with rightwards arrow on top minus a with rightwards arrow on top equals open parentheses 2 i with hat on top minus j with hat on top plus k with hat on top close parentheses minus open parentheses i with hat on top plus j with hat on top minus 2 k with hat on top close parentheses equals i with hat on top minus 2 j with hat on top plus 3 k with hat on top
stack A C with rightwards arrow on top equals c with rightwards arrow on top minus a with rightwards arrow on top equals open parentheses i with hat on top plus 2 j with hat on top plus k with hat on top close parentheses minus open parentheses i with hat on top plus j with hat on top minus 2 k with hat on top close parentheses equals j with hat on top plus 3 k with hat on top
N o w space c o n s i d e r space stack A B with rightwards arrow on top cross times stack A C with rightwards arrow on top colon space space
n with rightwards arrow on top equals stack A B with rightwards arrow on top cross times stack A C with rightwards arrow on top equals open vertical bar table row cell table row cell i with hat on top end cell cell j with hat on top end cell cell k with hat on top end cell row 1 cell negative 2 end cell 3 row 0 1 3 end table end cell end table close vertical bar
n with rightwards arrow on top equals i with hat on top open parentheses negative 6 minus 3 close parentheses minus 3 j with hat on top plus k with hat on top equals negative 9 i with hat on top minus 3 j with hat on top plus k with hat on top
S o comma space t h e space e q u a t i o n space o f space t h e space r e q u i r e d space p l a n e space i s space
open parentheses r with rightwards arrow on top minus a with rightwards arrow on top close parentheses times n with rightwards arrow on top equals 0
rightwards double arrow open parentheses r with rightwards arrow on top times n with rightwards arrow on top close parentheses equals open parentheses a with rightwards arrow on top times n with rightwards arrow on top close parentheses
rightwards double arrow open parentheses r with rightwards arrow on top times open parentheses negative 9 i with hat on top minus 3 j with hat on top plus k with hat on top close parentheses close parentheses equals open parentheses i with hat on top plus j with hat on top minus 2 k with hat on top close parentheses times open parentheses negative 9 i with hat on top minus 3 j with hat on top plus k with hat on top close parentheses
rightwards double arrow r with rightwards arrow on top times open parentheses 9 i with hat on top plus 3 j with hat on top minus k with hat on top close parentheses equals 14
A l s o comma space f i n d space t h e space c o o r d i n a t e s space o f space t h e space p o i n t space o f space i n t e r s e c t i o n space o f space t h i s space p l a n e space a n d space
t h e space l i n e space r with rightwards arrow on top equals 3 i with hat on top minus j with hat on top minus k with hat on top plus lambda open parentheses 2 i with hat on top minus 2 j with hat on top plus k with hat on top close parentheses
A n y space p o i n t space o n space t h e space l i n e space r with rightwards arrow on top equals 3 i with hat on top minus j with hat on top minus k with hat on top plus lambda open parentheses 2 i with hat on top minus 2 j with hat on top plus k with hat on top close parentheses space i s space o f space t h e space f o r m comma
open parentheses 3 plus 2 lambda comma space minus 1 minus 2 lambda comma space minus 1 plus lambda close parentheses
I f space t h e space p o i n t space P open parentheses 3 plus 2 lambda comma space minus 1 minus 2 lambda comma space minus 1 plus lambda close parentheses space l i e s space i n space t h e space p l a n e comma space
r with rightwards arrow on top times open parentheses 9 i with hat on top plus 3 j with hat on top minus k with hat on top close parentheses equals 14 comma space w e space h a v e comma
9 open parentheses 3 plus 2 lambda close parentheses minus 3 open parentheses 1 plus 2 lambda close parentheses minus open parentheses negative 1 plus lambda close parentheses equals 14
rightwards double arrow 27 plus 18 lambda minus 3 minus 6 lambda plus 1 minus lambda equals 14
rightwards double arrow 11 lambda equals negative 11
rightwards double arrow lambda equals negative 1
T h u s comma space t h e space r e q u i r e d space p o i n t space o f space i n t e r s e c t i o n space i s space
P open parentheses 3 plus 2 lambda comma space minus 1 minus 2 lambda comma space minus 1 plus lambda close parentheses
rightwards double arrow P open parentheses 3 plus 2 open parentheses negative 1 close parentheses comma space minus 1 minus 2 open parentheses negative 1 close parentheses comma space minus 1 plus open parentheses negative 1 close parentheses close parentheses
rightwards double arrow P open parentheses 1 comma space 1 comma space minus 2 close parentheses

space

Question 12

Solution 12

Question 13

Find the equation of a plane which passes through the point (3, 2, 0) and contains the line

Solution 13

Chapter 29 The plane Exercise Ex. 29.14

Question 1

Find space the space shortest space distance space between space the space lines space fraction numerator straight x minus 2 over denominator negative 1 end fraction equals fraction numerator straight y minus 5 over denominator 2 end fraction equals fraction numerator straight z minus 0 over denominator 3 end fraction space and space
fraction numerator straight x minus 0 over denominator 2 end fraction equals fraction numerator straight y plus 1 over denominator negative 1 end fraction equals fraction numerator straight z minus 1 over denominator 2 end fraction.

Solution 1

Question 2

Find space the space shortest space distance space between space the space lines space fraction numerator straight x plus 1 over denominator 7 end fraction equals fraction numerator straight y plus 1 over denominator negative 6 end fraction equals fraction numerator straight z plus 1 over denominator 1 end fraction space and space
fraction numerator straight x minus 3 over denominator 1 end fraction equals fraction numerator straight y minus 5 over denominator negative 2 end fraction equals fraction numerator straight z minus 7 over denominator 1 end fraction.

Solution 2

straight l subscript 1 colon fraction numerator straight x plus 1 over denominator 7 end fraction equals fraction numerator straight y plus 1 over denominator negative 6 end fraction equals fraction numerator straight z plus 1 over denominator 1 end fraction
straight l subscript 2 colon fraction numerator straight x minus 3 over denominator 1 end fraction equals fraction numerator straight y minus 5 over denominator negative 2 end fraction equals fraction numerator straight z minus 7 over denominator 1 end fraction
Let space the space equation space of space the space plane space containing space straight l subscript 1 space be space straight a left parenthesis straight x plus 1 right parenthesis plus straight b left parenthesis straight y plus 1 right parenthesis plus straight c left parenthesis straight z plus 1 right parenthesis equals 0
Plane space is space parallel space to space straight l subscript 1 colon space 7 straight a minus 6 straight b plus straight c equals 0...... left parenthesis straight i right parenthesis
Plane space is space parallel space to space straight l subscript 2 colon space straight a minus 2 straight b plus straight c equals 0........ left parenthesis ii right parenthesis
Solving space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma
fraction numerator straight a over denominator negative 6 plus 2 end fraction equals fraction numerator straight b over denominator 1 minus 7 end fraction equals fraction numerator straight c over denominator negative 14 plus 6 end fraction
fraction numerator straight a over denominator negative 4 end fraction equals fraction numerator straight b over denominator negative 6 end fraction equals fraction numerator straight c over denominator negative 8 end fraction
therefore Equation space of space the space plane space is space minus 4 left parenthesis straight x plus 1 right parenthesis minus 6 left parenthesis straight y plus 1 right parenthesis minus 8 left parenthesis straight z plus 1 right parenthesis equals 0
4 left parenthesis straight x plus 1 right parenthesis plus 6 left parenthesis straight y plus 1 right parenthesis plus 8 left parenthesis straight z plus 1 right parenthesis equals 0 space is space the space equation space of space the space plane.

Question 3

Find the shortest distance between the lines

Solution 3

Chapter 29 The plane Exercise Ex. 29.15

Question 1

Solution 1

Question 2

Solution 2

Question 3

Hence or otherwise deduce the length of the perpendicular.Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

F i n d space t h e space d i r e c t i o n space cos i n e s space o f space t h e space u n i t space v e c t o r space p e r p e n d i c u l a r space t o space t h e space p l a n e
r with rightwards arrow on top times open parentheses 6 i with hat on top minus 3 j with hat on top minus 2 k with hat on top close parentheses plus 1 equals 0 space p a s sin g space t h r o u g h space t h e space o r i g i n.

Solution 12

G i v e n space e q u a t i o n space o f space t h e space p l a n e space r with rightwards arrow on top times open parentheses 6 i with hat on top minus 3 j with hat on top minus 2 k with hat on top close parentheses plus 1 equals 0
T h u s comma space t h e space d i r e c t i o n space r a t i o s space n o r m a l space t o space t h e space p l a n e space a r e space 6 comma space minus 3 space a n d space minus 2
H e n c e space t h e space d i r e c t i o n space cos i n e s space t o space t h e space n o r m a l space t o space t h e space p l a n e space a r e
fraction numerator 6 over denominator square root of 6 squared plus open parentheses negative 3 close parentheses squared plus open parentheses negative 2 close parentheses squared end root end fraction comma fraction numerator negative 3 over denominator square root of 6 squared plus open parentheses negative 3 close parentheses squared plus open parentheses negative 2 close parentheses squared end root end fraction comma fraction numerator negative 2 over denominator square root of 6 squared plus open parentheses negative 3 close parentheses squared plus open parentheses negative 2 close parentheses squared end root end fraction
equals 6 over 7 comma fraction numerator negative 3 over denominator 7 end fraction comma fraction numerator negative 2 over denominator 7 end fraction
equals fraction numerator negative 6 over denominator 7 end fraction comma 3 over 7 comma 2 over 7
T h e space d i r e c t i o n space cos i n e s space o f space t h e space u n i t space v e c t o r space p e r p e n d i c u l a r space t o space t h e space p l a n e space a r e
s a m e space a s space t h e space d i r e c t i o n space cos i n e s space o f space t h e space n o r m a l space t o space t h e space p l a n e.
T h u s comma space t h e space d i r e c t i o n space cos i n e s space o f space t h e space u n i t space v e c t o r space p e r p e n d i c u l a r space t o space t h e space p l a n e
a r e colon space fraction numerator negative 6 over denominator 7 end fraction comma 3 over 7 comma 2 over 7

Question 13

F i n d space t h e space c o o r d i n a t e s space o f space t h e space f o o t space o f space t h e space p e r p e n d i c u l a r space d r a w n space f r o m space t h e space o r i g i n
t o space t h e space p l a n e space 2 x minus 3 y plus 4 z minus 6 equals 0

Solution 13

C o n s i d e r space t h e space g i v e n space e q u a t i o n space o f space t h e space p l a n e space 2 x minus 3 y plus 4 z minus 6 equals 0
T h e space d i r e c t i o n space r a t i o s space o f space t h e space n o r m a l space t o space t h e space p l a n e space a r e space 2 comma space minus 3 space a n d space 4
T h u s comma space t h e space d i r e c t i o space r a t i o s space o f space t h e space l i n e space p e r p e n d i c u l a r space t o space t h e space p l a n e space a r e space 2 comma space minus 3 space a n d space 4.
T h e space e q u a t i o n space o f space t h e space l i n e space p a s sin g space open parentheses x subscript 1 comma y subscript 1 comma z subscript 1 close parentheses space h a v i n g space d i r e c t i o n space r a t i o s space a comma b space a n d space c space i s
fraction numerator x minus x subscript 1 over denominator a end fraction equals fraction numerator y minus y subscript 1 over denominator b end fraction equals fraction numerator z minus z subscript 1 over denominator c end fraction
T h u s comma space t h e space e q u a t i o n space o f space t h e space l i n e space p a s sin g space t h r o u g h space t h e space o r i g i n space w i t h
d i r e c t i o n space r a t i o s space 2 comma space minus 3 space a n d space 4 space i s
fraction numerator x minus 0 over denominator 2 end fraction equals fraction numerator y minus 0 over denominator negative 3 end fraction equals fraction numerator z minus 0 over denominator 4 end fraction
rightwards double arrow x over 2 equals fraction numerator y over denominator negative 3 end fraction equals z over 4 equals r comma space w h e r e space r space i s space s o m e space c o n s tan t
A n y space p o i n t space o n space t h e space l i n e space i s space o f space t h e space f o r m space 2 r comma space minus 3 r space a n d space 4 r
I f space t h e space p o i n t space P open parentheses 2 r comma space minus 3 r comma space 4 r close parentheses space l i e s space o n space t h e space p l a n e space 2 x minus 3 y plus 4 z minus 6 equals 0 comma
i t space s h o u l d space s a t i s f i e s space t h e space e q u a t i o n comma space space 2 x minus 3 y plus 4 z minus 6 equals 0
T h u s comma space w e space h a v e comma
space 2 open parentheses 2 r close parentheses minus 3 open parentheses negative 3 r close parentheses plus 4 open parentheses 4 r close parentheses minus 6 equals 0
rightwards double arrow 4 r plus 9 r plus 16 r minus 6 equals 0
rightwards double arrow 29 r equals 6
rightwards double arrow r equals 6 over 29
T h u s comma space t h e space c o o r d i n a t e s space o f space t h e space p o i n t space o f space i n t e r s e c t i o n space o f space t h e space
p e r p e n d i c u l a r space f r o m space t h e space o r i g i n space a n d space t h e space p l a n e space a r e colon
P open parentheses 2 cross times 6 over 29 comma space minus 3 cross times 6 over 29 comma space 4 cross times 6 over 29 close parentheses equals P open parentheses 12 over 29 comma space 18 over 29 comma space 24 over 29 close parentheses

Question 14

Find the length and the foot of perpendicular from the point (1, 3/2, 2) to the plane 2x – 2y + 4z + 5 = 0.Solution 14

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RD SHARMA SOLUTION CHAPTER- 28 The Straight Line in Space I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 28 Straight line in space Exercise Ex. 28.1

Question 1

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   vector   and   Cartesian   equations   of   the   line   end text end cell row cell text through   the   point  ( 5 ,  2 , - 4 )  and   which   is   parallel   end text end cell row cell text to   the   vector   3 end text straight i with hat on top plus 2 straight j with hat on top minus 8 straight k with hat on top. end cell end table end style

Solution 1

begin mathsize 12px style table attributes columnalign left end attributes row cell Vector text   end text equation text   end text of text   end text straight a text   end text line end cell row cell is text    end text straight r with rightwards arrow on top equals straight a with rightwards arrow on top plus straight lambda straight b with rightwards arrow on top text   end text end cell row cell The text    end text Cartesian text    end text equation text    end text of text    end text straight a text    end text line text    end text is end cell row cell fraction numerator straight x minus straight x subscript 1 over denominator straight a subscript 1 end fraction equals fraction numerator straight y minus straight y subscript 1 over denominator straight a subscript 2 end fraction equals fraction numerator straight x minus straight x subscript 3 over denominator straight a subscript 3 end fraction end cell row cell Using text    end text the text    end text above text    end text formula comma end cell row cell Vector text    end text equation text    end text of text    end text the text    end text line comma end cell row cell straight r with rightwards arrow on top equals left parenthesis 5 straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top right parenthesis plus straight lambda left parenthesis text 3 end text straight i with hat on top plus 2 straight j with hat on top minus 8 straight k with hat on top right parenthesis end cell row cell The text    end text Cartesian text    end text equation text    end text of text    end text the text    end text line end cell row cell fraction numerator straight x minus 5 over denominator 3 end fraction equals fraction numerator straight y minus 2 over denominator 2 end fraction equals fraction numerator straight z plus 4 over denominator negative 8 end fraction end cell end table end style

Question 2

Find the vector equation of the line passing through the points (-1, 0, 2) and (3, 4, 6).Solution 2

begin mathsize 12px style table attributes columnalign left end attributes row cell The text    end text direction text    end text ratios text    end text of text    end text the text    end text line text    end text are end cell row cell left parenthesis 3 plus 1 comma 4 minus 0 comma 6 minus 2 right parenthesis equals left parenthesis 4 comma 4 comma 4 right parenthesis end cell row cell Since text    end text the text    end text line text    end text passes text    end text through text    end text left parenthesis negative 1 comma 0 comma 2 right parenthesis end cell row cell The text    end text vector text    end text equation text    end text of text    end text the text    end text line comma end cell row cell straight r with rightwards arrow on top equals straight a with rightwards arrow on top plus straight lambda straight b with rightwards arrow on top end cell row cell rightwards double arrow straight r with rightwards arrow on top equals straight a with rightwards arrow on top plus straight lambda straight b with rightwards arrow on top end cell row cell rightwards double arrow straight r with rightwards arrow on top equals left parenthesis negative straight i with rightwards arrow on top plus 0 straight j with rightwards arrow on top plus 2 straight k with rightwards arrow on top right parenthesis plus straight lambda left parenthesis 4 straight i with rightwards arrow on top plus 4 straight j with rightwards arrow on top plus 4 straight k with rightwards arrow on top right parenthesis end cell row cell therefore The text    end text vector text    end text equation text    end text of text    end text the text    end text line comma end cell row cell straight r with rightwards arrow on top equals left parenthesis negative straight i with rightwards arrow on top plus 0 straight j with rightwards arrow on top plus 2 straight k with rightwards arrow on top right parenthesis plus straight lambda left parenthesis 4 straight i with rightwards arrow on top plus 4 straight j with rightwards arrow on top plus 4 straight k with rightwards arrow on top right parenthesis end cell end table end style

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

.

Question 17

Solution 17

Chapter 28 Straight line in space Exercise Ex. 28.2

Question 1

begin mathsize 12px style table attributes columnalign left end attributes row cell text Show   that   the   three   lines   with   direction   cosines   end text fraction numerator text 12 end text over denominator text 13 end text end fraction comma fraction numerator negative 3 over denominator 13 end fraction comma fraction numerator negative 4 over denominator 13 end fraction semicolon end cell row cell 4 over 13 comma 12 over 13 comma 3 over 13 semicolon 3 over 13 comma fraction numerator negative 4 over denominator 13 end fraction comma 12 over 13 text   are   mutually   perpendicular. end text end cell end table end style

Solution 1

begin mathsize 12px style table attributes columnalign left end attributes row cell Let text    end text straight l subscript 1 equals 12 over 13 comma straight m subscript 1 equals negative 3 over 13 comma straight n subscript 1 equals negative 4 over 13 end cell row cell straight l subscript 2 equals 4 over 13 comma straight m subscript 2 equals 12 over 13 comma straight n subscript 1 equals 3 over 13 end cell row cell straight l subscript 3 equals 3 over 13 comma straight m subscript 3 equals negative 4 over 13 comma straight n subscript 3 equals 12 over 13 end cell row blank row cell straight l subscript 1 straight l subscript 2 plus straight m subscript 1 straight m subscript 2 plus straight n subscript 1 straight n subscript 2 end cell row cell equals 12 over 13 cross times 4 over 13 plus left parenthesis negative 3 over 13 right parenthesis cross times 12 over 13 plus left parenthesis negative 4 over 13 right parenthesis cross times 3 over 13 end cell row cell equals fraction numerator 48 minus 36 minus 12 over denominator 169 end fraction equals 0 end cell row cell straight l subscript 2 straight l subscript 3 plus straight m subscript 2 straight m subscript 3 plus straight n subscript 2 straight n subscript 3 end cell row cell equals 4 over 13 cross times 3 over 13 plus 12 over 13 cross times left parenthesis negative 4 over 13 right parenthesis plus 3 over 13 cross times 12 over 13 end cell row cell equals fraction numerator 12 minus 48 plus 36 over denominator 169 end fraction equals 0 end cell row cell straight l subscript 1 straight l subscript 3 plus straight m subscript 1 straight m subscript 3 plus straight n subscript 1 straight n subscript 3 end cell row cell equals 12 over 13 cross times 3 over 13 plus left parenthesis negative 3 over 13 right parenthesis cross times left parenthesis negative 4 over 13 right parenthesis plus left parenthesis negative 4 over 13 right parenthesis cross times 12 over 13 end cell row cell equals fraction numerator 36 plus 12 minus 48 over denominator 169 end fraction equals 0 end cell row cell therefore The text    end text lines text    end text are text    end text mutually text    end text perpendicular. end cell end table end style

Question 2

Show that the line through the points (1, -1, 2) and (3, 4, -2) is perpendicular to the line through points (0, 3, 2) and (3, 5, 6).Solution 2

begin mathsize 12px style table attributes columnalign left end attributes row cell The text    end text direction text   ratios   of    end text straight a text    end text line text    end text passing text    end text through text    end text the text    end text points end cell row cell left parenthesis 1 comma negative 1 comma 2 right parenthesis text    end text and text   end text left parenthesis 3 comma 4 comma negative 2 right parenthesis text    end text are end cell row cell left parenthesis 3 minus 1 comma 4 plus 1 comma negative 2 minus 2 right parenthesis end cell row cell equals left parenthesis 2 comma 5 comma negative 4 right parenthesis end cell row blank row cell The text    end text direction text   ratios    end text of text    end text straight a text    end text line text    end text passing text    end text through text    end text the text    end text points end cell row cell left parenthesis 0 comma 3 comma 2 right parenthesis text    end text and text   end text left parenthesis 3 comma 5 comma 6 right parenthesis text    end text are end cell row cell left parenthesis 3 minus 0 comma 5 minus 3 comma 6 minus 2 right parenthesis end cell row cell equals left parenthesis 3 comma 2 comma 4 right parenthesis end cell row blank row cell Angle text    end text between text    end text the text    end text lines text    end text end cell row cell cosθ equals fraction numerator left parenthesis straight a subscript 1 straight a subscript 2 plus straight b subscript 1 straight b subscript 2 plus straight c subscript 1 straight c subscript 2 right parenthesis over denominator square root of straight a subscript 1 squared plus straight b subscript 1 squared plus straight c subscript 1 squared end root square root of straight a subscript 2 squared plus straight b subscript 2 squared plus straight c subscript 2 squared end root end fraction end cell row cell cosθ equals fraction numerator left square bracket 2 cross times 3 plus 5 cross times 2 plus left parenthesis negative 4 right parenthesis cross times 4 right square bracket over denominator square root of 2 squared plus 5 squared plus left parenthesis negative 4 right parenthesis squared end root square root of 3 squared plus 2 squared plus 4 squared end root end fraction end cell row cell cosθ equals fraction numerator 0 over denominator square root of 2 squared plus 5 squared plus left parenthesis negative 4 right parenthesis squared end root square root of 3 squared plus 2 squared plus 4 squared end root end fraction end cell row cell cosθ equals 0 end cell row cell straight theta equals straight pi over 2 end cell row cell The text    end text lines text    end text are text    end text mutually text    end text perpendicular. end cell end table end style

Question 3

Show that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (-1, -2, 1) and (1, 2, 5).Solution 3

begin mathsize 12px style table attributes columnalign left end attributes row cell The text    end text direction text   ratios    end text of text    end text straight a text    end text line text    end text passing text    end text through text    end text the text    end text points end cell row cell left parenthesis 4 comma 7 comma 8 right parenthesis text    end text and text   end text left parenthesis 2 comma 3 comma 4 right parenthesis text    end text are end cell row cell left parenthesis 4 minus 2 comma 7 minus 3 comma 8 minus 4 right parenthesis end cell row cell equals left parenthesis 2 comma 4 comma 4 right parenthesis end cell row cell The text    end text direction text   ratios    end text of text    end text straight a text    end text line text    end text passing text    end text through text    end text the text    end text points end cell row cell left parenthesis negative 1 comma negative 2 comma 1 right parenthesis text    end text and text   end text left parenthesis 1 comma 2 comma 5 right parenthesis text    end text are end cell row cell left parenthesis negative 1 minus 1 comma negative 2 minus 2 comma 1 minus 5 right parenthesis end cell row cell equals left parenthesis negative 2 comma negative 4 comma negative 4 right parenthesis end cell row blank row cell The text    end text direction text   ratios   are   proportional. end text end cell row cell fraction numerator text 2 end text over denominator negative 2 end fraction equals fraction numerator 4 over denominator negative 4 end fraction equals fraction numerator 4 over denominator negative 4 end fraction end cell row cell Hence comma text   end text the text    end text lines text    end text are text    end text mutually text    end text parallel. end cell end table end style

Question 4

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   Cartesian   equation   of   the   line   which   passes   end text end cell row cell text through   the   point  (- 2 ,  4 , - 5 )  and   parallel   to   the   line   end text end cell row cell text given   by   end text fraction numerator text x + 3 end text over denominator text 3 end text end fraction equals fraction numerator straight y minus 4 over denominator 5 end fraction equals fraction numerator straight z plus 8 over denominator 6 end fraction end cell end table end style

Solution 4

begin mathsize 12px style table attributes columnalign left end attributes row cell The text    end text Cartesian text    end text equation text    end text of text    end text straight a text    end text line text    end text passing text    end text through text    end text left parenthesis straight x subscript 1 comma straight y subscript 1 comma straight z subscript 1 right parenthesis end cell row cell and text    end text with text   end text direction text    end text ratios text   end text left parenthesis straight a subscript 1 comma straight b subscript 1 comma straight c subscript 1 right parenthesis end cell row cell fraction numerator straight x minus straight x subscript 1 over denominator straight a subscript 1 end fraction equals fraction numerator straight y minus straight y subscript 1 over denominator straight b subscript 1 end fraction equals fraction numerator straight z minus straight z subscript 1 over denominator straight c subscript 1 end fraction end cell row cell The text    end text Cartesian text    end text equation text    end text of text    end text straight a text    end text line text    end text passing text    end text through text   end text left parenthesis negative 2 comma 4 comma negative 5 right parenthesis end cell row cell and text    end text parallel text    end text to text    end text the text    end text line text     end text fraction numerator text x + 3 end text over denominator text 3 end text end fraction equals fraction numerator straight y minus 4 over denominator 5 end fraction equals fraction numerator straight z plus 8 over denominator 6 end fraction text     end text is end cell row cell fraction numerator straight x plus 2 over denominator 3 end fraction equals fraction numerator straight y minus 4 over denominator 5 end fraction equals fraction numerator straight z plus 5 over denominator 6 end fraction end cell end table end style

Question 5

Solution 5

Question 6

Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, -1) and (4, 3, -1).Solution 6

begin mathsize 12px style table attributes columnalign left end attributes row cell The text    end text direction text    end text ratios text    end text of text   end text straight a text   end text line text    end text joining text    end text the text    end text origin text    end text to text    end text the text    end text point left parenthesis 2 comma 1 comma 1 right parenthesis end cell row cell are text    end text left parenthesis 2 minus 0 comma 1 minus 0 comma 1 minus 0 right parenthesis equals left parenthesis 2 comma 1 comma 1 right parenthesis end cell row cell The text    end text direction text    end text ratios text    end text of text   end text straight a text   end text line text    end text joining text    end text left parenthesis 3 comma 5 comma negative 1 right parenthesis text    end text and text    end text left parenthesis 4 comma 3 comma negative 1 right parenthesis end cell row cell are text    end text left parenthesis 4 minus 3 comma 3 minus 5 comma negative 1 plus 1 right parenthesis equals left parenthesis 1 comma negative 2 comma 0 right parenthesis end cell row blank row cell Angle text     end text between text    end text the text    end text lines end cell row cell cos text   end text straight theta equals fraction numerator straight a subscript 1 straight a subscript 2 plus straight b subscript 1 straight b subscript 2 plus straight c subscript 1 straight c subscript 2 over denominator square root of straight a subscript 1 squared plus straight b subscript 1 squared plus straight c subscript 1 squared end root square root of straight a subscript 2 squared plus straight b subscript 2 squared plus straight c subscript 2 squared end root end fraction end cell row cell cos text   end text straight theta equals fraction numerator 2 cross times 1 plus 1 cross times left parenthesis negative 2 right parenthesis plus 1 cross times 0 over denominator square root of 2 squared plus 1 squared plus 1 squared end root square root of 1 squared plus left parenthesis negative 2 right parenthesis squared plus 0 squared end root end fraction end cell row cell cos text   end text straight theta equals fraction numerator 0 over denominator square root of 6 square root of 5 end fraction end cell row cell cos text   end text straight theta equals 0 end cell row cell therefore straight theta equals straight pi over 2 end cell row cell The text    end text lines text    end text are text    end text mutually text    end text perpendicular. end cell end table end style

Question 7

Find the equation of a line parallel to x-axis and passing through the origin.Solution 7

begin mathsize 12px style table attributes columnalign left end attributes row cell Vector text    end text equation text    end text of text    end text straight a text    end text line text    end text is text    end text end cell row cell straight r with rightwards arrow on top equals straight a with rightwards arrow on top plus straight lambda straight b with rightwards arrow on top end cell row cell The text    end text direction text    end text cosines text    end text of text    end text the text   end text straight x minus axis text    end text are text    end text left parenthesis 1 comma 0 comma 0 right parenthesis. text    end text Equation text    end text of text    end text straight a text    end text line text    end text parallel text   end text end cell row cell text   end text to text    end text the text    end text straight x minus axis text    end text and text    end text passing text    end text through text    end text the text    end text origin text    end text is text    end text end cell row cell straight r with rightwards arrow on top equals left parenthesis 0 straight i with hat on top plus 0 straight j with hat on top plus 0 straight k with hat on top right parenthesis plus straight lambda left parenthesis 1 straight i with hat on top plus 0 straight j with hat on top plus 0 straight k with hat on top right parenthesis end cell row cell straight r with rightwards arrow on top equals straight lambda straight i with hat on top end cell end table end style

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 9(iii)

Solution 9(iii)

Question 9(iv)

Solution 9(iv)

Question 9(v)

Solution 9(v)

Question 9(vi)

find the angle between the following pairs of line :

fraction numerator minus x plus 2 over denominator minus 2 end fraction equals fraction numerator y minus 1 over denominator 7 end fraction equals fraction numerator z plus 3 over denominator minus 3 end fraction space a n d space fraction numerator x plus 2 over denominator minus 1 end fraction equals fraction numerator 2 y minus 8 over denominator 4 end fraction equals fraction numerator z minus 5 over denominator 4 end fraction

Solution 9(vi)

Question 10(i)

Solution 10(i)

Question 10(ii)

Solution 10(ii)

Question 10(iii)

Solution 10(iii)

Question 10(iv)

find the angle between the pairs of lines with directions ratios proposal to a, b, c and b – c, c – a, a – b.Solution 10(iv)

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   direction   cosines   of   the   line   end text fraction numerator x plus 2 over denominator 2 end fraction equals fraction numerator 2 y minus 7 over denominator 6 end fraction equals fraction numerator 5 minus z over denominator 6 end fraction. end cell row cell text Also ,  find   the   vector   equation   of   the   line   through   the   point   end text end cell row cell text A (- 1 ,  2 ,  3 )  and   parallel   to   the   given   line. end text end cell end table end style

Solution 24

begin mathsize 12px style table attributes columnalign left end attributes row cell The text    end text direction text    end text ratios text    end text of text    end text the text    end text line text    end text are end cell row cell fraction numerator straight x plus 2 over denominator 2 end fraction equals fraction numerator 2 straight y minus 7 over denominator 6 end fraction equals fraction numerator 5 minus straight z over denominator 6 end fraction end cell row cell 2 comma 6 comma 6 end cell row blank row cell The text    end text direction text    end text cosines text    end text of text    end text the text    end text line text    end text are end cell row cell straight l equals fraction numerator 2 over denominator square root of 2 squared plus 6 squared plus 6 squared end root end fraction equals fraction numerator 2 over denominator square root of 76 end fraction end cell row cell straight m equals fraction numerator 6 over denominator square root of 2 squared plus 6 squared plus 6 squared end root end fraction equals fraction numerator 6 over denominator square root of 76 end fraction end cell row cell straight n equals fraction numerator 6 over denominator square root of 2 squared plus 6 squared plus 6 squared end root end fraction equals fraction numerator 6 over denominator square root of 76 end fraction end cell row cell left parenthesis fraction numerator 2 over denominator square root of 76 end fraction comma fraction numerator 6 over denominator square root of 76 end fraction comma fraction numerator 6 over denominator square root of 76 end fraction right parenthesis end cell row blank row cell therefore Vector text    end text equation text    end text of text    end text the text    end text line text    end text is end cell row cell straight r with rightwards arrow on top equals straight a with rightwards arrow on top plus straight lambda straight b with rightwards arrow on top end cell row cell straight r with rightwards arrow on top equals left parenthesis negative straight i with rightwards arrow on top plus 2 straight j with rightwards arrow on top plus 3 straight k with rightwards arrow on top right parenthesis plus straight lambda left parenthesis 2 straight i with rightwards arrow on top plus 6 straight j with rightwards arrow on top plus 6 straight k with rightwards arrow on top right parenthesis end cell end table end style

Chapter 28 Straight line in space Exercise Ex. 28.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

Solution 6(iv)

Question 7

begin mathsize 12px style table attributes columnalign left end attributes row cell text Show   that   the   lines   end text stack text r end text with rightwards arrow on top equals 3 straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top plus straight lambda left parenthesis straight i with hat on top plus 2 straight j with hat on top plus 2 straight k with hat on top right parenthesis text   and   end text end cell row cell stack text r end text with rightwards arrow on top equals 5 straight i with hat on top minus 2 straight j with hat on top plus straight mu left parenthesis 3 straight i with hat on top plus 2 straight j with hat on top plus 6 straight k with hat on top right parenthesis text   are   intersecting .  Hence , end text end cell row cell text find   their   point   of   intersection. end text end cell end table end style

Solution 7

begin mathsize 12px style table attributes columnalign left end attributes row cell stack text r end text with rightwards arrow on top equals 3 straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top plus straight lambda left parenthesis straight i with hat on top plus 2 straight j with hat on top plus 2 straight k with hat on top right parenthesis end cell row cell stack text r end text with rightwards arrow on top equals 5 straight i with hat on top minus 2 straight j with hat on top plus straight mu left parenthesis 3 straight i with hat on top plus 2 straight j with hat on top plus 6 straight k with hat on top right parenthesis end cell row cell If text    end text the text    end text lines text    end text intersect text    end text each text    end text other comma text    end text then text    end text the text    end text shortest end cell row cell distance text    end text between text    end text the text    end text lines text    end text should text    end text be text    end text zero. end cell row cell Here comma end cell row cell stack straight a subscript 1 with rightwards arrow on top equals 3 straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top end cell row cell stack straight a subscript 2 with rightwards arrow on top equals 5 straight i with hat on top minus 2 straight j with hat on top end cell row cell stack straight b subscript 1 with rightwards arrow on top equals straight i with hat on top plus 2 straight j with hat on top plus 2 straight k with hat on top end cell row cell stack straight b subscript 2 with rightwards arrow on top equals 3 straight i with hat on top plus 2 straight j with hat on top plus 6 straight k with hat on top end cell row cell left parenthesis stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top right parenthesis equals vertical line table row cell straight i with rightwards arrow on top end cell cell straight j with rightwards arrow on top end cell cell straight k with rightwards arrow on top end cell row 1 2 2 row 3 2 6 end table vertical line end cell row cell equals straight i with rightwards arrow on top left parenthesis 12 minus 4 right parenthesis minus straight j with rightwards arrow on top left parenthesis 6 minus 6 right parenthesis plus straight k with rightwards arrow on top left parenthesis 2 minus 6 right parenthesis end cell row cell equals 8 straight i with rightwards arrow on top minus 0 straight j with rightwards arrow on top minus 4 straight k with rightwards arrow on top end cell row cell left parenthesis stack straight a subscript 2 with rightwards arrow on top minus stack straight a subscript 1 with rightwards arrow on top right parenthesis equals left parenthesis 5 straight i with hat on top minus 2 straight j with hat on top minus 3 straight i with hat on top minus 2 straight j with hat on top plus 4 straight k with hat on top right parenthesis equals left parenthesis 2 straight i with hat on top minus 4 straight j with hat on top plus 4 straight k with hat on top right parenthesis end cell row cell Shortest text    end text Distance comma straight d equals vertical line fraction numerator left parenthesis stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top right parenthesis. left parenthesis stack straight a subscript 2 with rightwards arrow on top minus stack straight a subscript 1 with rightwards arrow on top right parenthesis over denominator vertical line stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top vertical line end fraction vertical line end cell row cell equals vertical line fraction numerator left parenthesis 8 straight i with rightwards arrow on top minus 0 straight j with rightwards arrow on top minus 4 straight k with rightwards arrow on top right parenthesis. left parenthesis 2 straight i with hat on top minus 4 straight j with hat on top plus 4 straight k with hat on top right parenthesis over denominator vertical line 8 straight i with rightwards arrow on top minus 0 straight j with rightwards arrow on top minus 4 straight k with rightwards arrow on top vertical line end fraction vertical line end cell row cell equals vertical line fraction numerator 8 cross times 2 minus 0 cross times 4 plus left parenthesis negative 4 right parenthesis cross times 4 over denominator vertical line 8 straight i with rightwards arrow on top minus 0 straight j with rightwards arrow on top minus 4 straight k with rightwards arrow on top vertical line end fraction vertical line end cell row cell equals vertical line fraction numerator 0 over denominator vertical line 8 straight i with rightwards arrow on top minus 0 straight j with rightwards arrow on top minus 4 straight k with rightwards arrow on top vertical line end fraction vertical line equals 0 end cell row cell Since text    end text the text    end text shortest text    end text distance text    end text is text    end text zero comma text    end text the text    end text lines end cell row cell are text    end text intersect text    end text each text    end text other. end cell end table end style
begin mathsize 12px style table attributes columnalign left end attributes row cell Point text    end text of text    end text intersection text    end text of text    end text the text    end text lines comma end cell row cell stack text r end text with rightwards arrow on top equals 3 straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top plus straight lambda left parenthesis straight i with hat on top plus 2 straight j with hat on top plus 2 straight k with hat on top right parenthesis end cell row cell stack text r end text with rightwards arrow on top equals 5 straight i with hat on top minus 2 straight j with hat on top plus straight mu left parenthesis 3 straight i with hat on top plus 2 straight j with hat on top plus 6 straight k with hat on top right parenthesis end cell row cell Lines text    end text in text    end text the text    end text Cartesian text    end text form comma end cell row cell fraction numerator straight x minus 3 over denominator 1 end fraction equals fraction numerator straight y minus 2 over denominator 2 end fraction equals fraction numerator straight z plus 4 over denominator 2 end fraction equals straight lambda end cell row cell straight x equals straight lambda plus 3 comma straight y equals 2 straight lambda plus 2 comma straight z equals 2 straight lambda minus 4 end cell row cell fraction numerator straight x minus 5 over denominator 3 end fraction equals fraction numerator straight y plus 2 over denominator 2 end fraction equals straight z over 6 equals straight mu end cell row cell straight x equals 3 straight mu plus 5 comma straight y equals 2 straight mu minus 2 comma straight z equals 6 straight mu end cell end table end style
begin mathsize 12px style table attributes columnalign left end attributes row cell From text    end text coordinates text    end text of text    end text straight x comma end cell row cell straight lambda plus 3 equals 3 straight mu plus 5 end cell row cell straight lambda equals 3 straight mu plus 2..... left parenthesis straight i right parenthesis end cell row cell From text    end text coordinates text    end text of text    end text straight y comma end cell row cell 2 straight lambda plus 2 equals 2 straight mu minus 2 end cell row cell straight lambda equals straight mu minus 2....... left parenthesis ii right parenthesis end cell row cell Solving text    end text left parenthesis straight i right parenthesis text    end text and text    end text left parenthesis ii right parenthesis comma end cell row cell straight lambda equals negative 4 comma straight mu equals negative 2 end cell row blank row cell Coordinates text    end text of text    end text the text    end text point text    end text of text    end text intersection comma end cell row cell straight x equals 3 left parenthesis negative 2 right parenthesis plus 5 comma straight y equals 2 left parenthesis negative 2 right parenthesis minus 2 comma straight z equals 6 left parenthesis negative 2 right parenthesis end cell row cell straight x equals negative 1 comma straight y equals negative 6 comma straight z equals negative 12 end cell row cell left parenthesis negative 1 comma negative 6 comma negative 12 right parenthesis end cell end table end style

Chapter 28 Straight line in space Exercise Ex. 28.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

..Question 4

Solution 4

Question 5

Find the foot of perpendicular from the point (2, 3, 4) to the line begin mathsize 12px style fraction numerator 4 minus straight x over denominator 2 end fraction equals straight y over 6 equals fraction numerator 1 minus straight z over denominator 3 end fraction end style. Also find the perpendicular distance from the given point to the line.Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Find the coordinates of the foot of perpendicular drawn from the point A (1, 8, 4) to the line joining the points B (0, -1, 3) and C (2, -3, -1).Solution 14

Chapter 28 Straight line in space Exercise Ex. 28.5

Question 1(i)

Find the shortest distance between the pair of lines whose vector equation are

begin mathsize 12px style space space space space space space space space space space space r with rightwards arrow on top equals 3 i with hat on top space plus space 8 j with hat on top space plus space 3 k with hat on top space plus space lambda open parentheses 3 i with hat on top space minus space j with hat on top space plus space k with hat on top close parentheses
and comma space space r with rightwards arrow on top equals negative 3 i with hat on top space minus space 7 j with hat on top space plus space 6 k with hat on top space plus space mu open parentheses negative 3 i with hat on top space plus space 2 j with hat on top space plus space 4 k with hat on top close parentheses end style

Solution 1(i)

Question 1(ii)

Find the shortest distance between the pair of lines whose vector equations are:

begin mathsize 12px style space space space space space space space space space space space r with rightwards arrow on top equals 3 i with hat on top space plus space 5 j with hat on top space plus space 7 k with hat on top space plus space lambda open parentheses i with hat on top space minus space 2 j with hat on top space plus 7 space k with hat on top close parentheses
and comma space space r with rightwards arrow on top equals negative i with hat on top space minus space j with hat on top space minus space 6 k with hat on top space plus space mu open parentheses 7 i with hat on top space minus space 6 j with hat on top space plus space k with hat on top close parentheses end style

Solution 1(ii)

Question 1(iii)

Find the shortest distance between the pair of lines whose vector equations are:

begin mathsize 12px style space space space space space space space space space space space r with rightwards arrow on top equals open parentheses i with hat on top space plus space 2 j with hat on top space plus space 3 k with hat on top close parentheses space plus space lambda open parentheses 2 i with hat on top space plus 3 j with hat on top space plus space 4 k with hat on top close parentheses
and comma space space r with rightwards arrow on top equals open parentheses 2 i with hat on top space plus space 4 j with hat on top space plus space 5 k with hat on top close parentheses space plus space mu open parentheses 3 i with hat on top space plus space 4 j with hat on top space plus space 5 k with hat on top close parentheses end style

Solution 1(iii)

Question 1(iv)

Find the shortest distance between the lines whose vector equations are:

begin mathsize 12px style r with rightwards arrow on top equals open parentheses 1 minus t close parentheses stack i space with hat on top space plus space open parentheses t space minus space 2 close parentheses j with hat on top space plus space open parentheses 3 space minus space t close parentheses k with hat on top space a n d
r with rightwards arrow on top equals open parentheses s plus 1 close parentheses i with hat on top space plus space open parentheses 2 s minus 1 close parentheses j with hat on top space minus space open parentheses 2 s plus 1 close parentheses k with hat on top end style

Solution 1(iv)

Question 1(v)

Find the shortest distance between the pair of lines whose vector equations are:

begin mathsize 12px style space space space space space space space space space space space r with rightwards arrow on top equals open parentheses lambda space minus space 1 close parentheses i with hat on top space plus space open parentheses lambda space plus space 1 close parentheses j with hat on top space minus space space left parenthesis 1 space plus space lambda right parenthesis k with hat on top
and comma space space r with rightwards arrow on top equals left parenthesis 1 space minus space mu right parenthesis i with hat on top space plus space left parenthesis 2 mu space minus space 1 right parenthesis j with hat on top space plus space space left parenthesis mu space plus space 2 right parenthesis k with hat on top end style

Solution 1(v)

Question 1(vi)

Find the shortest distance between the pair of lines whose vector equations are:

begin mathsize 12px style r with rightwards arrow on top equals open parentheses 2 i with hat on top space minus space j with hat on top space minus space k with hat on top close parentheses space plus space lambda open parentheses 2 i with hat on top space minus space 5 j with hat on top space plus space 2 k with hat on top close parentheses semicolon space r with rightwards arrow on top equals open parentheses i with hat on top space plus space 2 j with hat on top space plus space k with hat on top close parentheses space plus space mu open parentheses i with hat on top space minus space j with hat on top space plus space k with hat on top close parentheses end style

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 2(i)

Find the shortest distance between the pair of lines whose cartesian equations are:

begin mathsize 12px style fraction numerator straight x minus 1 over denominator 2 end fraction equals fraction numerator straight y minus 2 over denominator 3 end fraction equals fraction numerator straight z minus 3 over denominator 4 end fraction space and space fraction numerator straight x minus 2 over denominator 3 end fraction equals fraction numerator straight y minus 3 over denominator 4 end fraction equals fraction numerator straight z minus 5 over denominator 5 end fraction end style.Solution 2(i)

Question 2(ii)

Find the shortest distance between the pair of lines whose cartesian equations are:

begin mathsize 12px style fraction numerator straight x minus 1 over denominator 2 end fraction equals fraction numerator straight y plus 1 over denominator 3 end fraction equals straight z space and space fraction numerator straight x plus 1 over denominator 3 end fraction equals fraction numerator straight y minus 2 over denominator 1 end fraction semicolon space straight z equals 2 end style.Solution 2(ii)

Question 2(iii)

Find the shortest distance between the pair of lines whose cartesian equations are:

begin mathsize 12px style fraction numerator straight x minus 1 over denominator negative 1 end fraction equals fraction numerator straight y plus 2 over denominator 1 end fraction equals fraction numerator straight z minus 3 over denominator negative 2 end fraction and fraction numerator straight x minus 1 over denominator 1 end fraction equals fraction numerator straight y plus 1 over denominator 2 end fraction equals fraction numerator straight z plus 1 over denominator negative 2 end fraction end style.Solution 2(iii)

Question 2(iv)

Find the shortest distance between the pair of lines whose Cartesian equations are:

begin mathsize 12px style fraction numerator x minus 3 over denominator 1 end fraction equals fraction numerator y minus 5 over denominator negative 2 end fraction equals fraction numerator z minus 7 over denominator 1 end fraction semicolon fraction numerator x plus 1 over denominator 7 end fraction equals fraction numerator y plus 1 over denominator negative 6 end fraction equals fraction numerator z plus 1 over denominator 1 end fraction end style.Solution 2(iv)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 5

Solution 5

Question 6

Solution 6

Question 7(i)

Find the shortest distance between the lines

begin mathsize 12px style stack text r end text with rightwards arrow on top equals left parenthesis straight i with rightwards arrow on top plus 2 straight j with rightwards arrow on top plus straight k with rightwards arrow on top right parenthesis plus straight lambda left parenthesis straight i with rightwards arrow on top minus straight j with rightwards arrow on top plus straight k with rightwards arrow on top right parenthesis text   and   end text stack text r end text with rightwards arrow on top equals 2 straight i with rightwards arrow on top minus straight j with rightwards arrow on top minus straight k with rightwards arrow on top plus straight mu left parenthesis 2 straight i with rightwards arrow on top plus straight j with rightwards arrow on top plus 2 straight k with rightwards arrow on top right parenthesis end style

Solution 7(i)

begin mathsize 12px style table attributes columnalign left end attributes row cell Here comma end cell row cell straight a subscript 1 equals straight i with hat on top plus 2 straight j with hat on top plus straight k with hat on top end cell row cell straight b subscript 1 equals straight i with hat on top minus straight j with hat on top plus straight k with hat on top end cell row cell straight a subscript 2 equals 2 straight i with hat on top minus straight j with hat on top minus straight k with hat on top end cell row cell straight b subscript 2 equals 2 straight i with hat on top plus straight j with hat on top plus 2 straight k with hat on top end cell row blank row cell straight a with rightwards arrow on top subscript 2 minus straight a with rightwards arrow on top subscript blank subscript 1 end subscript equals 2 straight i with hat on top minus straight j with hat on top minus straight k with hat on top minus straight i with hat on top minus 2 straight j with hat on top minus straight k with hat on top equals straight i with hat on top minus 3 straight j with hat on top minus 2 straight k with hat on top end cell row cell straight b subscript 1 cross times straight b subscript 2 equals open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 1 cell negative 1 end cell 1 row 2 1 2 end table close vertical bar equals straight i with hat on top left parenthesis negative 2 minus 1 right parenthesis minus straight j with hat on top left parenthesis 2 minus 2 right parenthesis plus straight k with hat on top left parenthesis 1 plus 2 right parenthesis equals negative 3 straight i with hat on top plus 3 straight k with hat on top end cell row blank row cell The text    end text shortest text    end text distance text    end text between text    end text the text    end text two text    end text lines comma end cell row cell straight d equals open vertical bar fraction numerator left parenthesis straight b with rightwards arrow on top subscript 1 cross times straight b with rightwards arrow on top subscript 2 right parenthesis. left parenthesis straight a with rightwards arrow on top subscript 2 minus straight a with rightwards arrow on top subscript 1 right parenthesis over denominator vertical line straight b with rightwards arrow on top subscript 1 cross times straight b with rightwards arrow on top subscript 2 vertical line end fraction close vertical bar end cell row cell straight d equals open vertical bar fraction numerator left parenthesis negative 3 straight i with hat on top plus 3 straight k with hat on top right parenthesis. left parenthesis straight i with hat on top minus 3 straight j with hat on top minus 2 straight k with hat on top right parenthesis over denominator vertical line minus 3 straight i with hat on top minus 3 straight k with hat on top vertical line end fraction close vertical bar equals open vertical bar fraction numerator negative 3 minus 6 over denominator square root of left parenthesis negative 3 right parenthesis squared plus left parenthesis negative 3 right parenthesis squared end root end fraction close vertical bar equals fraction numerator 9 over denominator 3 square root of 2 end fraction end cell row cell The text    end text shortest text    end text distance text    end text between text    end text the text    end text two text    end text lines equals fraction numerator 3 over denominator square root of 2 end fraction units end cell end table end style

Question 7(ii)

Find the shortest distance between the lines

begin mathsize 12px style fraction numerator straight x plus 1 over denominator 7 end fraction equals fraction numerator straight y plus 1 over denominator negative 6 end fraction equals fraction numerator straight z plus 1 over denominator 1 end fraction text   and   end text fraction numerator text x end text minus text 3 end text over denominator text 1 end text end fraction equals fraction numerator straight y minus 5 over denominator negative 2 end fraction equals fraction numerator straight z minus 7 over denominator 1 end fraction end style

Solution 7(ii)

begin mathsize 12px style table attributes columnalign left end attributes row cell Here comma end cell row cell straight a with rightwards arrow on top subscript 1 equals negative straight i with hat on top minus straight j with hat on top minus straight k with hat on top end cell row cell straight a with rightwards arrow on top subscript 2 equals 3 straight i with hat on top plus 5 straight j with hat on top plus 7 straight k with hat on top end cell row cell straight b with rightwards arrow on top subscript 1 equals 7 straight i with hat on top minus 6 straight j with hat on top plus straight k with hat on top end cell row cell straight b with rightwards arrow on top subscript 2 equals straight i with hat on top minus 2 straight j with hat on top plus straight k with hat on top end cell row cell straight b with rightwards arrow on top subscript 1 cross times straight b with rightwards arrow on top subscript 2 equals open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 7 cell negative 6 end cell 1 row 1 cell negative 2 end cell 1 end table close vertical bar end cell row cell equals straight i with hat on top left parenthesis negative 6 plus 2 right parenthesis minus straight j with hat on top left parenthesis 7 minus 1 right parenthesis plus straight k with hat on top left parenthesis negative 14 plus 6 right parenthesis end cell row cell equals negative 4 straight i with hat on top minus 6 straight j with hat on top minus 8 straight k with hat on top end cell row cell straight a with rightwards arrow on top subscript 2 minus straight a with rightwards arrow on top subscript 1 equals straight i with hat on top left parenthesis 3 plus 1 right parenthesis plus straight j with hat on top left parenthesis 5 plus 1 right parenthesis plus straight k with hat on top left parenthesis 7 plus 1 right parenthesis end cell row cell equals 4 straight i with hat on top plus 6 straight j with hat on top plus 8 straight k with hat on top end cell row blank row cell The text    end text shortest text    end text distance text    end text between text    end text two text    end text lines comma end cell row cell straight d equals open vertical bar fraction numerator left parenthesis straight b with rightwards arrow on top subscript 1 cross times straight b with rightwards arrow on top subscript 2 right parenthesis. left parenthesis straight a with rightwards arrow on top subscript 2 minus straight a with rightwards arrow on top subscript 1 right parenthesis over denominator vertical line straight b subscript 1 cross times straight b subscript 2 vertical line end fraction close vertical bar end cell row cell equals open vertical bar fraction numerator left parenthesis negative 4 straight i with hat on top minus 6 straight j with hat on top minus 8 straight k with hat on top right parenthesis. left parenthesis 4 straight i with hat on top plus 6 straight j with hat on top plus 8 straight k with hat on top right parenthesis over denominator square root of left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 6 right parenthesis squared plus left parenthesis negative 8 right parenthesis squared end root end fraction close vertical bar end cell row cell equals open vertical bar fraction numerator negative 16 minus 36 minus 64 over denominator square root of 116 end fraction close vertical bar end cell row cell equals open vertical bar fraction numerator negative 116 over denominator square root of 116 end fraction close vertical bar end cell row cell equals 2 square root of 29 text   end text units end cell end table end style

Question 7(iii)

Find the shortest distance between the lines

begin mathsize 12px style stack text r end text with rightwards arrow on top equals straight i with hat on top plus 2 straight j with hat on top plus 3 straight k with hat on top plus straight lambda left parenthesis straight i with hat on top minus 3 straight j with hat on top plus 2 straight k with hat on top right parenthesis text and   end text stack text r end text with rightwards arrow on top equals 4 straight i with hat on top plus 5 straight j with hat on top plus 6 straight k with hat on top plus straight mu left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus straight k with hat on top right parenthesis end style

Solution 7(iii)

begin mathsize 12px style table attributes columnalign left end attributes row cell Here comma end cell row cell straight a with rightwards arrow on top subscript 1 equals straight i with hat on top plus 2 straight j with hat on top plus 3 straight k with hat on top comma end cell row cell straight b with rightwards arrow on top subscript 1 equals straight i with hat on top minus 3 straight j with hat on top plus 2 straight k with hat on top end cell row cell straight a with rightwards arrow on top subscript 2 equals 4 straight i with hat on top plus 5 straight j with hat on top plus 6 straight k with hat on top comma end cell row cell straight b with rightwards arrow on top subscript 2 equals 2 straight i with hat on top plus 3 straight j with hat on top plus straight k with hat on top end cell row blank row cell left parenthesis straight a with rightwards arrow on top subscript blank subscript 2 end subscript minus straight a with rightwards arrow on top subscript blank subscript 1 end subscript right parenthesis equals 4 straight i with hat on top plus 5 straight j with hat on top plus 6 straight k with hat on top minus straight i with hat on top minus 2 straight j with hat on top minus 3 straight k with hat on top equals 3 straight i with hat on top plus 3 straight j with hat on top plus 3 straight k with hat on top end cell row cell straight b with rightwards arrow on top subscript 1 cross times straight b with rightwards arrow on top subscript 2 end cell row cell equals left parenthesis straight i with hat on top minus 3 straight j with hat on top plus 2 straight k with hat on top right parenthesis cross times left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus straight k with hat on top right parenthesis end cell row cell equals open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 1 cell negative 3 end cell 2 row 2 3 1 end table close vertical bar end cell row cell equals straight i with hat on top left parenthesis negative 3 minus 6 right parenthesis minus straight j with hat on top left parenthesis 1 minus 4 right parenthesis plus straight k with hat on top left parenthesis 3 plus 6 right parenthesis end cell row cell equals negative 9 straight i with hat on top plus 3 straight j with hat on top plus 9 straight k with hat on top end cell row cell Shortest text    end text distance text    end text between text    end text the text    end text two text    end text lines equals open vertical bar fraction numerator left parenthesis straight a subscript 2 minus straight a subscript 1 right parenthesis. left parenthesis straight b with rightwards arrow on top subscript 1 cross times straight b with rightwards arrow on top subscript 2 right parenthesis over denominator vertical line straight b with rightwards arrow on top subscript 1 cross times straight b with rightwards arrow on top subscript 2 vertical line end fraction close vertical bar end cell row cell equals open vertical bar fraction numerator left parenthesis 3 straight i with hat on top plus 3 straight j with hat on top plus 3 straight k with hat on top right parenthesis. left parenthesis negative 9 straight i with hat on top plus 3 straight j with hat on top plus 9 straight k with hat on top right parenthesis over denominator vertical line minus 9 straight i with hat on top plus 3 straight j with hat on top plus 9 straight k with hat on top vertical line end fraction close vertical bar end cell row cell equals open vertical bar fraction numerator 3 cross times left parenthesis negative 9 right parenthesis plus 3 cross times 3 plus 3 cross times 9 over denominator square root of left parenthesis negative 9 right parenthesis squared plus 3 squared plus 9 squared end root end fraction close vertical bar end cell row cell equals open vertical bar fraction numerator negative 27 plus 9 plus 27 over denominator square root of left parenthesis negative 9 right parenthesis squared plus 3 squared plus 9 squared end root end fraction close vertical bar end cell row cell equals open vertical bar fraction numerator 9 over denominator square root of 171 end fraction close vertical bar equals fraction numerator 3 over denominator square root of 19 end fraction units end cell end table end style

Question 7(iv)

Find the shortest distance between the lines

begin mathsize 12px style stack text r end text with rightwards arrow on top equals 6 straight i with hat on top plus 2 straight j with hat on top plus 2 straight k with hat on top plus straight lambda left parenthesis straight i with hat on top minus 2 straight j with hat on top plus 2 straight k with hat on top right parenthesis text   and   end text stack text r end text with rightwards arrow on top equals negative 4 straight i with hat on top minus straight k with hat on top plus straight mu left parenthesis 3 straight i with hat on top minus 2 straight j with hat on top minus 2 straight k with hat on top right parenthesis end style

Solution 7(iv)

begin mathsize 12px style table attributes columnalign left end attributes row cell Here comma end cell row cell straight a with rightwards arrow on top subscript 1 equals 6 straight i with hat on top plus 2 straight j with hat on top plus 2 straight k with hat on top end cell row cell straight a with rightwards arrow on top subscript 2 equals negative 4 straight i with hat on top minus straight k with hat on top end cell row cell straight b with rightwards arrow on top subscript 1 equals straight i with hat on top minus 2 straight j with hat on top plus 2 straight k with hat on top end cell row cell straight b with rightwards arrow on top subscript 2 equals 3 straight i with hat on top minus 2 straight j with hat on top minus 2 straight k with hat on top end cell row blank row cell straight a with rightwards arrow on top subscript 2 minus straight a with rightwards arrow on top subscript 1 equals negative 4 straight i with hat on top minus straight k with hat on top minus 6 straight i with hat on top minus 2 straight j with hat on top minus 2 straight k with hat on top end cell row cell equals negative 10 straight i with hat on top minus 2 straight j with hat on top minus 3 straight k with hat on top end cell row cell straight b with rightwards arrow on top subscript 1 cross times straight b with rightwards arrow on top subscript 2 equals open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 1 cell negative 2 end cell 2 row 3 cell negative 2 end cell cell negative 2 end cell end table close vertical bar end cell row cell equals straight i with hat on top left parenthesis 4 plus 4 right parenthesis minus straight j with hat on top left parenthesis negative 2 minus 6 right parenthesis plus straight k with hat on top left parenthesis negative 2 plus 6 right parenthesis end cell row cell equals 8 straight i with hat on top plus 8 straight j with hat on top plus 4 straight k with hat on top end cell end table end style
begin mathsize 12px style table attributes columnalign left end attributes row cell Shortest text    end text Distance equals open vertical bar fraction numerator left parenthesis straight a with rightwards arrow on top subscript 2 minus straight a with rightwards arrow on top subscript 1 right parenthesis. left parenthesis straight b with rightwards arrow on top subscript 1 cross times straight b with rightwards arrow on top subscript 2 right parenthesis over denominator vertical line straight b with rightwards arrow on top subscript 1 cross times straight b with rightwards arrow on top subscript 2 vertical line end fraction close vertical bar end cell row cell equals open vertical bar fraction numerator left parenthesis negative 10 straight i with hat on top minus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis. left parenthesis 8 straight i with hat on top plus 8 straight j with hat on top plus 4 straight k with hat on top right parenthesis over denominator vertical line 8 straight i with hat on top minus 8 straight j with hat on top plus 4 straight k with hat on top vertical line end fraction close vertical bar end cell row cell equals open vertical bar fraction numerator left parenthesis negative 10 right parenthesis cross times 8 plus left parenthesis negative 2 right parenthesis cross times 8 plus left parenthesis negative 3 right parenthesis cross times 4 over denominator square root of 8 squared plus left parenthesis negative 8 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared end root end fraction close vertical bar end cell row cell equals open vertical bar fraction numerator negative 80 minus 16 minus 12 over denominator square root of 64 plus 64 plus 16 end root end fraction close vertical bar equals open vertical bar fraction numerator negative 108 over denominator square root of 144 end fraction close vertical bar equals 9 text   units end text end cell end table end style

Question 8

Find the distance between the lines land l2 given by

begin mathsize 12px style table attributes columnalign left end attributes row cell stack text r end text with rightwards arrow on top equals straight i with hat on top plus 2 straight j with hat on top minus 4 straight k with hat on top plus straight lambda left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus 6 straight k with hat on top right parenthesis text   and   end text end cell row cell stack text r end text with rightwards arrow on top equals 3 straight i with hat on top plus 3 straight j with hat on top minus 5 straight k with hat on top plus straight mu left parenthesis 2 straight i with hat on top plus 3 straight j with hat on top plus 6 straight k with hat on top right parenthesis end cell end table end style

Solution 8

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RD SHARMA SOLUTION CHAPTER-27 Direction Cosines and Direction Ratios I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 27 Direction Cosines and Direction Ratios Exercise Ex. 27.1

Question 1

If a line makes angles of 90°, 60° and 30° with the positive direction of x,y and z-axis respectively, find its direction cosines.Solution 1

Let l, m and n be the direction cosines of a line.

l = cos 90° = 0

begin mathsize 12px style text m end text equals text cos   60 end text degree equals 1 half end style
begin mathsize 12px style straight n equals cos text   end text 30 degree equals fraction numerator square root of 3 over denominator 2 end fraction end style
begin mathsize 12px style therefore straight T text he   direction   cosines   of   the   line   are   0 , end text 1 half comma fraction numerator square root of 3 over denominator 2 end fraction. end style

Question 2

If a line has direction ratios 2, -1, -2, determine its direction cosines.Solution 2

begin mathsize 12px style table attributes columnalign left end attributes row cell Let text    end text the text    end text direction text    end text cosines text    end text of text    end text the text    end text line text    end text be text    end text straight l comma straight m comma straight n. end cell row cell Here comma end cell row cell straight a equals 2 comma straight b equals negative 1 comma straight c equals negative 2 text    end text are text    end text the text    end text direction text    end text ratios text    end text of text    end text the text    end text line. end cell row cell straight l equals plus-or-minus fraction numerator straight a over denominator square root of straight a squared plus straight b squared plus straight c squared end root end fraction comma straight m equals plus-or-minus fraction numerator straight b over denominator square root of straight a squared plus straight b squared plus straight c squared end root end fraction comma straight n equals plus-or-minus fraction numerator straight c over denominator square root of straight a squared plus straight b squared plus straight c squared end root end fraction end cell row cell straight l equals fraction numerator 2 over denominator square root of 2 squared plus left parenthesis negative 1 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared end root end fraction comma straight m equals fraction numerator negative 1 over denominator square root of 2 squared plus left parenthesis negative 1 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared end root end fraction comma straight n equals fraction numerator negative 2 over denominator square root of 2 squared plus left parenthesis negative 1 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared end root end fraction end cell row cell straight l equals fraction numerator 2 over denominator square root of 9 end fraction comma straight m equals fraction numerator negative 1 over denominator square root of 9 end fraction comma straight n equals fraction numerator negative 2 over denominator square root of 9 end fraction end cell row cell straight l equals 2 over 3 comma straight m equals negative 1 third comma straight n equals negative 2 over 3 end cell row cell therefore The text    end text direction text    end text ratios text    end text of text    end text the text    end text line text    end text are text    end text 2 over 3 comma negative 1 third comma negative 2 over 3. end cell end table end style

Question 3

Find the direction cosines of the line passing through two points (-2, 4, -5) and (1, 2, 3).Solution 3

begin mathsize 12px style table attributes columnalign left end attributes row cell The text    end text direction text    end text ratios text    end text of text    end text the text    end text line text    end text joining text    end text left parenthesis negative 2 comma 4 comma negative 5 right parenthesis text   end text and text   end text left parenthesis 1 comma text   end text 2 comma text   end text 3 right parenthesis text   are end text comma end cell row cell left parenthesis 1 plus 2 comma 2 minus 4 comma 3 plus 5 right parenthesis equals left parenthesis 3 comma negative 2 comma 8 right parenthesis end cell row cell Here comma straight a equals 3 comma straight b equals negative 2 comma straight c equals 8 end cell row cell Direction text    end text cosines text    end text are end cell row cell fraction numerator 3 over denominator square root of 3 squared plus left parenthesis negative 2 right parenthesis squared plus 8 squared end root end fraction comma fraction numerator negative 2 over denominator square root of 3 squared plus left parenthesis negative 2 right parenthesis squared plus 8 squared end root end fraction comma fraction numerator 8 over denominator square root of 3 squared plus left parenthesis negative 2 right parenthesis squared plus 8 squared end root end fraction end cell row cell equals fraction numerator 3 over denominator square root of 77 end fraction comma fraction numerator negative 2 over denominator square root of 77 end fraction comma fraction numerator 8 over denominator square root of 77 end fraction end cell end table end style

Question 4

Using direction ratios show that the points A (2, 3, -4), (1, -2, 3) and (3, 8, -11) are collinear.Solution 4

begin mathsize 12px style table attributes columnalign left end attributes row cell Here text    end text straight A text   end text left parenthesis 2 comma 3 comma negative text 4 end text right parenthesis comma text   end text straight B text   end text left parenthesis 1 comma negative 2 comma 3 right parenthesis text   end text and text   end text straight C text   end text left parenthesis 3 comma 8 comma negative 11 right parenthesis. end cell row cell Direction text    end text ratios text    end text of text    end text AB equals left parenthesis 1 minus 2 comma negative 2 minus 3 comma 3 plus 4 right parenthesis equals left parenthesis negative 1 comma negative 5 comma 7 right parenthesis end cell row cell Direction text    end text ratios text    end text of text    end text BC equals left parenthesis 3 minus 1 comma 8 plus 2 comma negative 11 minus 3 right parenthesis equals left parenthesis 2 comma 10 comma negative 14 right parenthesis end cell row blank row cell Here comma text    end text the text    end text respective text    end text direction text    end text cosines text    end text of text    end text AB text    end text and text    end text AC comma end cell row cell fraction numerator negative 1 over denominator 2 end fraction equals fraction numerator negative 5 over denominator 10 end fraction equals fraction numerator 7 over denominator negative 14 end fraction text    end text are text    end text proportional. end cell row blank row cell Also comma text   end text straight B text    end text is text    end text the text    end text common text    end text point text    end text between text    end text the text    end text two text    end text lines comma end cell row cell therefore The text    end text points text     end text straight A text   end text left parenthesis 2 comma 3 comma negative text 4 end text right parenthesis comma text   end text straight B text   end text left parenthesis 1 comma negative 2 comma 3 right parenthesis text   end text and text   end text straight C text   end text left parenthesis 3 comma 8 comma negative 11 right parenthesis text    end text are text    end text collinear. text   end text end cell row blank end table end style

Question 5

Find the direction cosines of the sides of the triangle whose vertices are (3, 5, -4), (-1, 1, 2) and (-5, -5, -2)Solution 5

begin mathsize 12px style table attributes columnalign left end attributes row cell straight A left parenthesis 3 comma 5 comma negative 4 right parenthesis comma straight B left parenthesis negative 1 comma 1 comma 2 right parenthesis text    end text and text    end text straight C left parenthesis negative 5 comma negative 5 comma negative 2 right parenthesis end cell row cell The text    end text direction text    end text ratios text    end text of text    end text the text    end text side text    end text AB equals left parenthesis negative 1 minus 3 comma 1 minus 5 comma 2 plus 4 right parenthesis end cell row cell equals left parenthesis negative 4 comma negative 4 comma 6 right parenthesis end cell row cell Direction text    end text cosines text    end text of text    end text AB text    end text will text    end text be end cell row cell fraction numerator negative 4 over denominator square root of left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared plus 6 squared end root end fraction comma fraction numerator negative 4 over denominator square root of left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared plus 6 squared end root end fraction comma fraction numerator 6 over denominator square root of left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared plus 6 squared end root end fraction end cell row cell equals fraction numerator negative 4 over denominator square root of 68 end fraction comma fraction numerator negative 4 over denominator square root of 68 end fraction comma fraction numerator 6 over denominator square root of 68 end fraction end cell row cell equals fraction numerator negative 2 over denominator square root of 17 end fraction comma fraction numerator negative 2 over denominator square root of 17 end fraction comma fraction numerator 3 over denominator square root of 17 end fraction end cell row cell The text    end text direction text    end text ratios text    end text of text    end text the text    end text side text    end text BC equals left parenthesis negative 5 plus 1 comma negative 5 minus 1 comma negative 2 minus 2 right parenthesis end cell row cell equals left parenthesis negative 4 comma negative 6 comma negative 4 right parenthesis end cell row cell Direction text    end text cosines text    end text of text    end text BC text    end text will text    end text be end cell row cell fraction numerator negative 4 over denominator square root of left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 6 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared end root end fraction comma fraction numerator negative 6 over denominator square root of left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 6 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared end root end fraction comma fraction numerator negative 4 over denominator square root of left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 6 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared end root end fraction end cell row cell equals fraction numerator negative 4 over denominator square root of 68 end fraction comma fraction numerator negative 6 over denominator square root of 68 end fraction comma fraction numerator negative 4 over denominator square root of 68 end fraction end cell row cell equals fraction numerator negative 2 over denominator square root of 17 end fraction comma fraction numerator negative 3 over denominator square root of 17 end fraction comma fraction numerator negative 2 over denominator square root of 17 end fraction end cell row cell The text    end text direction text    end text ratios text    end text of text    end text the text    end text side text    end text AC equals left parenthesis negative 5 minus 3 comma negative 5 minus 5 comma negative 2 plus 4 right parenthesis end cell row cell equals left parenthesis negative 8 comma negative 10 comma 2 right parenthesis end cell row cell Direction text    end text cosines text    end text of text    end text AC text    end text will text    end text be end cell row cell fraction numerator negative 8 over denominator square root of left parenthesis negative 8 right parenthesis squared plus left parenthesis negative 10 right parenthesis squared plus 2 squared end root end fraction comma fraction numerator negative 10 over denominator square root of left parenthesis negative 8 right parenthesis squared plus left parenthesis negative 10 right parenthesis squared plus 2 squared end root end fraction comma fraction numerator 2 over denominator square root of left parenthesis negative 8 right parenthesis squared plus left parenthesis negative 10 right parenthesis squared plus 2 squared end root end fraction end cell row cell equals fraction numerator negative 8 over denominator square root of 168 end fraction comma fraction numerator negative 10 over denominator square root of 168 end fraction comma fraction numerator 2 over denominator square root of 168 end fraction end cell row cell equals fraction numerator negative 4 over denominator square root of 42 end fraction comma fraction numerator negative 5 over denominator square root of 42 end fraction comma fraction numerator 1 over denominator square root of 42 end fraction end cell end table end style

Question 6

Solution 6

Question 7

Solution 7

Question 8

Find the acute angle between the lines whose direction ratios are proportional to 2:3:6 and 1:2:2.Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16(i)

Solution 16(i)

Question 16(ii)

Solution 16(ii)

Question 16(iii)

Solution 16(iii)

Question 16(iv)

Find the angle between the lines whose direction cosines are given by equations

2l + 2m – n = 0, mn + ln + lm = 0Solution 16(iv)

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RD SHARMA SOLUTION CHAPTER-26 Scalar Triple Product I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 26 Scalar Triple Product Exercise Ex. 26.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 5(iv)

Solution 5(iv)

Question 6

Solution 6

Question 7

S h o w space t h a t space t h e space p o i n t s space A open parentheses minus 1 comma 4 comma minus 3 close parentheses comma space B open parentheses 3 comma 2 comma minus 5 close parentheses comma space C open parentheses minus 3 comma 8 comma minus 5 close parentheses space a n d space D open parentheses minus 3 comma 2 comma 1 close parentheses
a r e space c o minus p l a n a r.

Solution 7

A B equals p o s i t i o n space v e c t o r space o f space B minus p o s i t i o n space v e c t o r space o f space A
space space space space space equals 4 i with hat on top minus 2 j with hat on top minus 2 k with hat on top
A C equals p o s i t i o n space v e c t o r space o f space C minus p o s i t i o n space v e c t o r space o f space A
space space space space space equals minus 2 i with hat on top plus 4 j with hat on top minus 2 k with hat on top
A D equals p o s i t i o n space v e c t o r space o f space D minus p o s i t i o n space v e c t o r space o f space A
space space space space space equals minus 2 i with hat on top minus 2 j with hat on top plus 4 k with hat on top
T h e space f o u r space p o i n t s space a r e space c o minus p l a n a r space i f space t h e space v e c t o r s space a r e space c o minus p l a n a r.
T h u s comma space open vertical bar table row 4 cell minus 2 end cell cell minus 2 end cell row cell minus 2 end cell 4 cell minus 2 end cell row cell minus 2 end cell cell minus 2 end cell 4 end table close vertical bar equals 4 open square brackets 16 minus 4 close square brackets plus 2 open square brackets minus 8 minus 4 close square brackets minus 2 open square brackets 4 plus 8 close square brackets equals 48 minus 24 minus 24 equals 0
H e n c e space p r o v e d.

Question 8

Show that the four points whose position vectors are begin mathsize 12px style 6 i with hat on top minus 7 j with hat on top comma space 16 i with hat on top space minus space 19 j with hat on top minus 4 k with hat on top comma space 3 i with hat on top space minus space 6 k with hat on top comma space 2 i with hat on top space minus space 5 j with hat on top space plus space 10 k with hat on top end style Coplanar.Solution 8

L e t space O A equals 6 i with hat on top minus 7 j with hat on top comma space O B equals 16 i with hat on top minus 19 j with hat on top minus 4 k with hat on top comma space O C equals 3 i with hat on top minus 6 k with hat on top comma space O D equals 2 i with hat on top minus 5 j with hat on top plus 10 k with hat on top
T h u s comma
A B equals O B minus O A equals 10 i with hat on top minus 12 j with hat on top minus 4 k with hat on top
A C equals O C minus O A equals minus 3 i with hat on top plus 7 j with hat on top minus 6 k with hat on top
A D equals O D minus O A equals minus 4 i with hat on top plus 2 j with hat on top plus 10 k with hat on top
T h e space f o u r space p o i n t s space a r e space c o minus p l a n a r space i f space v e c t o r s space A B comma space A C space a n d space A D space a r e space c o minus p l a n a r.
T h u s comma space w e space h a v e
open vertical bar table row 10 cell minus 12 end cell cell minus 4 end cell row cell minus 3 end cell 7 cell minus 6 end cell row cell minus 4 end cell 2 10 end table close vertical bar equals 10 open parentheses 70 plus 12 close parentheses plus 12 open parentheses minus 30 minus 24 close parentheses minus 4 open parentheses minus 6 plus 28 close parentheses equals 820 minus 648 minus 88

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12(i)

Solution 12(i)

Question 12(ii)

begin mathsize 12px style table attributes columnalign left end attributes row cell text let   end text straight a with rightwards arrow on top text  =  end text straight i with hat on top text + end text straight j with hat on top text + end text straight k with hat on top text ,  end text straight b with rightwards arrow on top text  =  end text straight i with hat on top text   and   end text stack text c end text with rightwards arrow on top text  =  end text straight c subscript 1 straight i with hat on top text + end text straight c subscript 2 straight j with hat on top text + end text straight c subscript 3 straight k with hat on top text .  end text end cell row cell text Then ,  If   end text straight c subscript text 2 end text end subscript text  = - 1   and   end text straight c subscript text 3 end text end subscript equals 1 comma text   show   that   no   value   end text end cell row cell text of   end text straight c subscript text 1 end text end subscript text   can   make   end text straight a with rightwards arrow on top comma text   end text straight b with rightwards arrow on top text   and   end text straight c with rightwards arrow on top text   coplanar. end text end cell end table end style

Solution 12(ii)

begin mathsize 12px style table attributes columnalign left end attributes row cell straight a with rightwards arrow on top text  =  end text straight i with hat on top text + end text straight j with hat on top text + end text straight k with hat on top text ,  end text straight b with rightwards arrow on top text  =  end text straight i with hat on top text   and   end text stack text c end text with rightwards arrow on top text  =  end text straight c subscript 1 straight i with hat on top text + end text straight c subscript 2 straight j with hat on top text + end text straight c subscript 3 straight k with hat on top text. end text end cell row cell straight c subscript text 2 end text end subscript text  = - 1   and   end text straight c subscript text 3 end text end subscript equals 1 comma end cell row cell text If   end text straight a with rightwards arrow on top comma straight b with rightwards arrow on top comma text    and   end text stack text c end text with rightwards arrow on top text   are   coplanar ,  then   their   scalar   triple   product   is   zero. end text end cell row cell left square bracket table row cell straight a with rightwards arrow on top end cell cell straight b with rightwards arrow on top end cell cell stack text c end text with rightwards arrow on top end cell end table right square bracket equals vertical line table row 1 1 1 row 1 0 0 row cell straight c subscript 1 end cell cell straight c subscript 2 end cell cell straight c subscript 3 end cell end table vertical line equals 0 end cell row cell rightwards double arrow negative straight c subscript 3 plus straight c subscript 2 equals 0 end cell row cell rightwards double arrow straight c subscript 2 equals straight c subscript 3 end cell row cell text But   this   is   a   contradiction   as   it   is   given   that   end text straight c subscript text 2 end text end subscript text  = - 1   and   end text straight c subscript text 3 end text end subscript equals 1. end cell row cell text Hence ,  no   value   of   end text straight c subscript 1 text   can   make   the   vectors   coplanar. end text end cell end table end style

Question 13

Solution 13

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RD SHARMA SOLUTION CHAPTER-25 Vector or Cross Product I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 25 Vector or Cross Product Exercise Ex. 25.1

Question 1

Solution 1

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7(i)

Solution 7(i)

Question 7(ii)

Solution 7(ii)

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 8(iv)

Solution 8(iv)

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 9(iii)

Solution 9(iii)

Question 9(iv)

Solution 9(iv)

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27(i)

Solution 27(i)

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Using Vectors, find the area of the triangle with vertices: (i) A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5) (ii) A (1, 2, 3), B(2, -1, 4) and C (4, 5, -1).Solution 34

Question 35

Solution 35

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RD SHARMA SOLUTION CHAPTER-24 Scalar or Dot Product I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 24 Scalar Or Dot Product Exercise Ex. 24.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5 (i)

Solution 5 (i)

Question 5 (ii)

Solution 5 (ii)

Question 5 (iii)

Solution 5 (iii)

Question 5 (iv)

Solution 5 (iv)

Question 5 (v)

Solution 5 (v)

Question 6

Solution 6

Question 7(i)

Solution 7(i)

Question 7(ii)

begin mathsize 12px style table attributes columnalign left end attributes row cell text Dot   products   of   a   vector   with   vectors   end text straight i with hat on top minus straight j with hat on top plus straight k with hat on top comma text   2 end text space straight i with hat on top plus straight j with hat on top minus 3 straight k with hat on top text   end text end cell row cell text and   end text straight i with hat on top plus straight j with hat on top plus straight k with hat on top text   are   respectively   4 , 0   and   2 .  Find   the   vector. end text end cell end table end style

Solution 7(ii)

begin mathsize 12px style table attributes columnalign left end attributes row cell Let text    end text the text    end text unknown text    end text vector text    end text be text    end text apostrophe straight a with rightwards arrow on top equals straight a subscript 1 straight i with hat on top plus straight b subscript 1 straight j with hat on top plus straight c subscript 1 straight k with hat on top apostrophe end cell row cell straight b with rightwards arrow on top equals straight i with hat on top minus straight j with hat on top plus straight k with hat on top comma text    end text straight c with rightwards arrow on top equals 2 straight i with hat on top plus straight j with hat on top minus 3 straight k with hat on top comma straight d with rightwards arrow on top equals straight i with hat on top plus straight j with hat on top plus straight k with hat on top end cell row cell It text    end text is text    end text given text    end text that text    end text straight a with rightwards arrow on top. straight b with rightwards arrow on top equals 4 end cell row cell straight a subscript 1 minus straight b subscript 1 plus straight c subscript 1 equals 4..... left parenthesis straight i right parenthesis end cell row cell straight a with rightwards arrow on top. straight c with rightwards arrow on top equals 0 end cell row cell 2 straight a subscript 1 plus straight b subscript 1 minus 3 straight c subscript 1 equals 0....... left parenthesis ii right parenthesis end cell row cell straight a with rightwards arrow on top. straight d with rightwards arrow on top equals 2 end cell row cell straight a subscript 1 plus straight b subscript 1 plus straight c subscript 1 equals 2........ left parenthesis iii right parenthesis end cell row blank row cell Solving text    end text left parenthesis straight i right parenthesis comma left parenthesis ii right parenthesis text    end text and text    end text left parenthesis iii right parenthesis comma end cell row cell straight a subscript 1 equals 2 comma straight b subscript 1 equals negative 1 comma straight c subscript 1 equals 1 end cell row blank row cell therefore the text    end text vector text    end text straight a with rightwards arrow on top equals 2 straight i with hat on top minus straight j with hat on top plus straight k with hat on top end cell end table end style

Question 8 (i)

Solution 8 (i)

Question 8 (ii)

Solution 8 (ii)

Question 9

Solution 9

Question 10

Solution 10

Given that begin mathsize 12px style straight a with rightwards arrow on top comma stack straight b comma with rightwards arrow on top straight c with rightwards arrow on top end style are mutually perpendicular, so,

begin mathsize 12px style straight a with rightwards arrow on top straight b with rightwards arrow on top equals straight b with rightwards arrow on top. straight c with rightwards arrow on top equals straight c with rightwards arrow on top. straight a with rightwards arrow on top equals 0
and space straight a with rightwards arrow on top comma straight b with rightwards arrow on top space and space straight c with rightwards arrow on top are space unit space vectors comma space so
open vertical bar straight a with rightwards arrow on top close vertical bar equals open vertical bar straight b with rightwards arrow on top close vertical bar equals open vertical bar straight c with rightwards arrow on top close vertical bar equals 1
Now comma
open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar squared equals open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close parentheses squared
equals open parentheses straight a with rightwards arrow on top close parentheses plus open parentheses straight b with rightwards arrow on top close parentheses squared open parentheses straight c with rightwards arrow on top close parentheses squared plus open parentheses straight c with rightwards arrow on top close parentheses squared plus 2 straight a with rightwards arrow on top straight b with rightwards arrow on top plus 2 straight b with rightwards arrow on top straight c with rightwards arrow on top plus 2 straight b with rightwards arrow on top straight c with rightwards arrow on top plus 2 straight c with rightwards arrow on top straight a with rightwards arrow on top
equals open vertical bar straight a with rightwards arrow on top close vertical bar squared plus open vertical bar straight b with rightwards arrow on top close vertical bar squared plus open vertical bar straight c with rightwards arrow on top close vertical bar squared plus 2 left parenthesis 0 right parenthesis plus 2 left parenthesis 0 right parenthesis plus 2 left parenthesis 0 right parenthesis
equals open parentheses 1 close parentheses squared plus open parentheses 1 close parentheses squared plus left parenthesis 1 right parenthesis squared plus 0
open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar squared equals 1 plus 1 plus 1
open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar squared equals 3
open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar equals square root of 3 end style

Question 11

Solution 11

Question 12

Show that the vector begin mathsize 12px style i with hat on top thin space plus thin space j with hat on top space plus space k with hat on top end style is equally inclined with the coordinate axes.Solution 12

begin mathsize 12px style Let space straight theta space be space the space angle space between space straight i with hat on top plus straight j with hat on top plus straight k with hat on top space and space straight i with hat on top
Then comma
cos space straight theta equals fraction numerator open parentheses space straight i with hat on top plus straight j with hat on top plus straight k with hat on top close parentheses. left parenthesis straight i with hat on top right parenthesis over denominator space open vertical bar straight i with hat on top plus straight j with hat on top plus straight k with hat on top close vertical bar open vertical bar left parenthesis straight i with hat on top right parenthesis close vertical bar end fraction
space space space space space space space space space space equals fraction numerator 1 over denominator begin display style fraction numerator 1 over denominator square root of 3 end fraction end style end fraction
space space space space space space space space space space equals square root of 3
Similarly comma space if space straight alpha space and space straight gamma space are space angles space that space straight i with hat on top plus straight j with hat on top plus straight k with hat on top space make space with space straight j with hat on top space and space straight k with hat on top
Then comma
space space space space space space space space space space space space space cos space straight alpha equals square root of 3
and space space space space space space space cos space straight gamma equals square root of 3
Therefore comma space straight i with hat on top plus straight j with hat on top plus straight k with hat on top space is space equally space inclined space the space three space axes. end style

Question 13

Show that the vectors begin mathsize 12px style straight a with rightwards arrow on top equals 1 over 7 open parentheses 2 straight i with hat on top space plus thin space 3 straight j with hat on top space plus space 6 straight k with hat on top close parentheses comma straight b with rightwards arrow space on top equals 1 over 7 open parentheses 3 straight i with hat on top space minus thin space 6 straight j with hat on top space plus space 2 straight k with hat on top close parentheses comma space straight c with rightwards arrow on top equals 1 over 7 open parentheses 6 straight i with hat on top space plus thin space 2 straight j with hat on top space minus space 3 straight k with hat on top close parentheses end style are mutually perpendicualr unit vectors.Solution 13

begin mathsize 12px style we space have comma
straight a with rightwards harpoon with barb upwards on top equals 1 over 7 open parentheses 2 straight i with overbrace on top plus 3 straight j with overbrace on top plus 6 straight k with overbrace on top close parentheses
straight b with rightwards harpoon with barb upwards on top equals 1 over 7 open parentheses 3 straight i with overbrace on top plus 6 straight j with overbrace on top plus 2 straight k with overbrace on top close parentheses
straight c with rightwards harpoon with barb upwards on top equals 1 over 7 open parentheses 6 straight i with overbrace on top plus 2 straight j with overbrace on top plus 3 straight k with overbrace on top close parentheses
Then comma
straight a with rightwards harpoon with barb upwards on top equals 1 over 7 open parentheses 2 straight i with overbrace on top plus 3 straight j with overbrace on top plus 6 straight k with overbrace on top close parentheses cross times 1 over 7 open parentheses 3 straight i with overbrace on top plus 6 straight j with overbrace on top plus 2 straight k with overbrace on top close parentheses
space space space space space space space space space 1 over 49 open parentheses 6 minus 18 plus 12 close parentheses equals 0
Similary comma
straight b with rightwards harpoon with barb upwards on top. straight c with rightwards harpoon with barb upwards on top equals straight a with rightwards harpoon with barb upwards on top. straight c with rightwards harpoon with barb upwards on top equals straight c
therefore space space straight a with rightwards harpoon with barb upwards on top comma straight b with rightwards harpoon with barb upwards on top comma straight c with rightwards harpoon with barb upwards on top space are space mutually space perpendicular end style

Question 14

begin mathsize 12px style for space any space two space vectors space straight a with rightwards arrow on top and space straight b with rightwards arrow on top comma end style

Solution 14

Question 15

Solution 15

Question 16

begin mathsize 12px style table attributes columnalign left end attributes row cell If space straight p with rightwards arrow on top equals 5 straight i with hat on top plus straight lambda straight j with hat on top minus 3 straight k with hat on top space and space straight q with rightwards arrow on top  = straight i with hat on top plus 3 straight j with hat on top minus 5 straight k with hat on top ,  then space find space the space value space of space straight lambda straight comma end cell row cell so space that space straight p with rightwards arrow on top plus straight q with rightwards arrow on top space and space straight p with rightwards arrow on top minus straight q with rightwards arrow on top space are space perpendicular space vectors. end cell end table end style

Solution 16

begin mathsize 12px style table attributes columnalign left end attributes row cell straight p with rightwards arrow on top equals 5 straight i with hat on top plus straight lambda straight j with hat on top minus 3 straight k with hat on top text   end text and text   end text straight q with rightwards arrow on top text   end text equals straight i with hat on top plus 3 straight j with hat on top minus 5 straight k with hat on top end cell row cell straight p with rightwards arrow on top plus straight q with rightwards arrow on top text   end text end cell row cell equals 5 straight i with hat on top plus straight lambda straight j with hat on top minus 3 straight k with hat on top plus straight i with hat on top plus 3 straight j with hat on top minus 5 straight k with hat on top end cell row cell equals 6 straight i with hat on top plus left parenthesis straight lambda plus 3 right parenthesis straight j with hat on top minus 8 straight k with hat on top end cell row blank row cell straight p with rightwards arrow on top minus straight q with rightwards arrow on top text   end text end cell row cell equals 5 straight i with hat on top plus straight lambda straight j with hat on top minus 3 straight k with hat on top minus straight i with hat on top minus 3 straight j with hat on top plus 5 straight k with hat on top end cell row cell equals 4 straight i with hat on top plus left parenthesis straight lambda minus 3 right parenthesis straight j with hat on top plus 2 straight k with hat on top end cell row blank row cell left parenthesis straight p with rightwards arrow on top plus straight q with rightwards arrow on top right parenthesis. left parenthesis straight p with rightwards arrow on top minus straight q with rightwards arrow on top right parenthesis equals 0 end cell row cell rightwards double arrow left square bracket 6 straight i with hat on top plus left parenthesis straight lambda plus 3 right parenthesis straight j with hat on top minus 8 straight k with hat on top right square bracket. left square bracket 4 straight i with hat on top plus left parenthesis straight lambda minus 3 right parenthesis straight j with hat on top plus 2 straight k with hat on top right square bracket equals 0 end cell row cell rightwards double arrow 24 plus left parenthesis straight lambda squared minus 9 right parenthesis minus 16 equals 0 end cell row cell rightwards double arrow straight lambda squared minus 9 plus 8 equals 0 end cell row cell rightwards double arrow straight lambda squared minus 1 equals 0 end cell row cell therefore straight lambda equals plus-or-minus 1 end cell end table end style

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

If two vector begin mathsize 12px style a with rightwards arrow on top space a n d space b with rightwards arrow on top end style are such that begin mathsize 12px style open vertical bar a with rightwards arrow on top close vertical bar equals 2 comma open vertical bar b with rightwards arrow on top close vertical bar space equals space 1 space a n d space a with rightwards arrow on top b with rightwards arrow on top equals 1 end style, then find the value of (3a – 5b) . (2a + 7b).Solution 29

Question 30(i)

Solution 30(i)

Question 30(ii)

Solution 30(ii)

Question 31(i)

Solution 31(i)

Question 31(ii)

Solution 31(ii)

Question 31(iii)

Solution 31(iii)

Question 32(i)

Solution 32(i)

Question 32(ii)

Solution 32(ii)

Question 32(iii)

Solution 32(iii)

Question 33(i)

Solution 33(i)

Question 33(ii)

Solution 33(ii)

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

begin mathsize 12px style table attributes columnalign left end attributes row cell text If   end text straight a with rightwards arrow on top comma text   end text straight b with rightwards arrow on top text   are   two   vectors   such   that   end text vertical line straight a with rightwards arrow on top plus straight b with rightwards arrow on top vertical line equals vertical line straight b with rightwards arrow on top vertical line comma text   then   prove   end text end cell row cell text that   end text straight a with rightwards arrow on top plus 2 straight b with rightwards arrow on top text   is   perpendicular   to   end text straight a with rightwards arrow on top. end cell end table end style

Solution 49

begin mathsize 12px style table attributes columnalign left end attributes row cell vertical line straight a with rightwards arrow on top plus straight b with rightwards arrow on top vertical line squared equals vertical line straight b with rightwards arrow on top vertical line squared end cell row cell rightwards double arrow left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis. left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis equals straight b with rightwards arrow on top. straight b with rightwards arrow on top end cell row cell rightwards double arrow straight a with rightwards arrow on top. straight a with rightwards arrow on top plus straight a with rightwards arrow on top. straight b with rightwards arrow on top plus straight b with rightwards arrow on top. straight a with rightwards arrow on top plus straight b with rightwards arrow on top. straight b with rightwards arrow on top equals straight b with rightwards arrow on top. straight b with rightwards arrow on top end cell row cell rightwards double arrow straight a with rightwards arrow on top. straight a with rightwards arrow on top plus straight a with rightwards arrow on top. straight b with rightwards arrow on top plus straight a with rightwards arrow on top. straight b with rightwards arrow on top plus straight b with rightwards arrow on top. straight b with rightwards arrow on top equals straight b with rightwards arrow on top. straight b with rightwards arrow on top end cell row cell rightwards double arrow straight a with rightwards arrow on top. straight a with rightwards arrow on top plus 2 straight a with rightwards arrow on top. straight b with rightwards arrow on top equals 0 end cell row cell rightwards double arrow straight a with rightwards arrow on top. left parenthesis straight a with rightwards arrow on top plus 2 straight b with rightwards arrow on top right parenthesis equals 0 end cell row cell therefore straight a with rightwards arrow on top plus 2 straight b with rightwards arrow on top text    end text is text    end text perpendicular text    end text to text   end text straight a with rightwards arrow on top. end cell end table end style

Chapter 24 Scalar Or Dot Product Exercise Ex. 24.2

Question 1

Solution 1

Question 2

Prove that: If the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.Solution 2

Question 3

(Pythagoras’s Theorem) Prove by vector method that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.Solution 3

Question 4

Prove by vector method that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.Solution 4

Question 5

Prove using vectors: The quadrilateral obtained by joining mid-points of adjacent sides of a rectangle is a rhombus.Solution 5

Question 6

Prove that the diagonals of a rhombus are perpendicular bisectors of each other.Solution 6

Question 7

Prove that the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.Solution 7

Question 8

If AD is the median of D ABC, using vectors, prove that

AB2 + AC2 = 2 (AD2 + CD2).Solution 8

Question 9

If the median to the base of a triangle is perpendicular to the base, then triangle is isosceles.Solution 9

Question 10

In a quadrilateral ABCD, prove that AB2 + BC2 + CD2+ DA2 = AC2 + BD2 + 4 PQ2, where P and Q are middle points of diagonals AC and BD. Solution 10

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RD SHARMA SOLUTION CHAPTER-23 Algebra of Vectors I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 23 Algebra of Vectors Exercise Ex. 23.1

Question 1(i)

Represent graphically a dispacement of 40 km, 30° east of north.Solution 1(i)

Question 1(ii)

Represent graphically a displacement of 50 km, south-eastSolution 1(ii)

Here, vector begin mathsize 12px style OP with rightwards arrow on top end style represents the displacement of 50 km, south-east.Question 1(iii)

Represent graphically a displacement of 70 km, 40° north of west.Solution 1(iii)

Here, vector begin mathsize 12px style OP with rightwards arrow on top end style represents the displacement of 70 km, 40° north of west.Question 2

Classify the following measures as scalars and vectors.

(i) 15 kg

(ii) 20 kg weight

(iii) 45°

(iv) 10 metres south-east

(v) 50 m/s2Solution 2

(i) 15 kg is a scalar quantity because it involves only

(ii) 20 kg weight is a vector quantity as it involves both magnitude and direction.

(iii) 45° is a scalar quantity as it involves only magnitude.

(iv) 10 metres south-east is a vector quantity as it involve direction.

(v) 50 m/s2 is a scalar quantity as it involves magnitude of acceleration.Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 23 Algebra of Vectors Exercise Ex. 23.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 23 Algebra of Vectors Exercise Ex. 23.3

Question 1

Find the position vector of a point R which divides the line joining the two points P and Q with position vectors

Solution 1

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

T h e space v e r t i c e s space A comma space B comma space C space o f space t r i a n g l e space A B C space h a v e space r e s p e c t i v e l y space p o s i t i o n space v e c t o r s space
a with rightwards arrow on top comma space b with rightwards arrow on top comma space c with rightwards arrow on top space w i t h space r e s p e c t space t o space a space g i v e n space o r i g i n space O. space S h o w space t h a t space t h e space p o i n t space D space w h e r e space t h e space b i s e c t o r space o f space
angle A space m e e t s space B C space h a s space p o s i t i o n space v e c t o r
d with rightwards arrow on top equals fraction numerator beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator beta plus gamma end fraction comma space w h e r e space beta equals open vertical bar c with rightwards arrow on top minus a with rightwards arrow on top close vertical bar equals gamma equals open vertical bar a with rightwards arrow on top minus b with rightwards arrow on top close vertical bar
H e n c e space d e d u c e space t h a t space t h e space i n c e n t r e space I space h a s space p o s i t i o n space v e c t o r space fraction numerator alpha a with rightwards arrow on top plus beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator alpha plus beta plus gamma end fraction comma space w h e r e
alpha equals open vertical bar b with rightwards arrow on top minus c with rightwards arrow on top close vertical bar

Solution 7

L e t space A B C space b e space a space t r i a n g l e.
L e t space t h e space p o s i t i o n space v e c t o r s space o f space A comma space B space a n d space C space w i t h space r e s p e c t space t o space s o m e space o r i g i n comma space O space b e
a with rightwards arrow on top comma space b with rightwards arrow on top space a n d space c with rightwards arrow on top space r e s p e c t i v e l y.
L e t space D space b e space t h e space p o i n t space o n space B C space w h e r e space t h e space b i s e c t o r space o f space angle A space m e e t s.
L e t space d with rightwards arrow on top space p o s i t i o n space v e c t o r space o f space D space w h i c h space d i v i d e s space B C space i n t e r n a l l y space i n space t h e space r a t i o space beta space a n d space gamma comma
w h e r e space beta equals open vertical bar stack A C with rightwards arrow on top close vertical bar space a n d space gamma equals open vertical bar stack A B with rightwards arrow on top close vertical bar
T h u s comma space beta equals open vertical bar c with rightwards arrow on top minus a with rightwards arrow on top close vertical bar space a n d space gamma equals open vertical bar b with rightwards arrow on top minus a with rightwards arrow on top close vertical bar
T h u s comma space b y space s e c t i o n space f o r m u l a comma space t h e space p o s i t i o n space v e c t o r space o f space D space i s space g i v e n space b y
stack O D with rightwards arrow on top equals fraction numerator beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator beta plus gamma end fraction
L e t space alpha equals open vertical bar b with rightwards arrow on top minus c with rightwards arrow on top close vertical bar
I n c e n t r e space i s space t h e space c o n c u r r e n t space p o i n t space o f space a n g l e space b i s e c t o r s.
T h u s comma space I n c e n t r e space d i v i d e s space t h e space l i n e space A D space i n space t h e space r a t i o space alpha : beta plus gamma
T h u s comma space t h e space p o s i t i o n space v e c t o r space o f space i n c e n t r e space i s
e q u a l space t o space fraction numerator alpha a with rightwards arrow on top plus fraction numerator beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator open parentheses beta plus gamma close parentheses end fraction cross times open parentheses beta plus gamma close parentheses over denominator alpha plus beta plus gamma end fraction equals fraction numerator alpha a with rightwards arrow on top plus beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator alpha plus beta plus gamma end fraction

Chapter 23 – Algebra of Vectors Exercise Ex. 23.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Prove by vector method that the internal bisectors of the angles of a triangle are concurrent.Solution 6

Chapter 23 – Algebra of Vectors Exercise Ex. 23.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Find |AB| in each case.Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

rightwards double arrow b with hat on top equals 1 half i with hat on top plus fraction numerator square root of 3 over denominator 2 end fraction j with hat on top

Question 12

Solution 12

Chapter 23 Algebra of Vectors Exercise Ex. 23.6

Question 1

Solution 1

Question 2

Solution 2

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Question 3

Solution 3

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

F i n d space a space v e c t o r space o f space m a g n i t u d e space o f space 5 space u n i t s space p a r a l l e l space t o space t h e space r e s u l tan t space o f space t h e space v e c t o r s space
a with rightwards arrow on top equals 2 i with hat on top plus 3 j with hat on top minus k with hat on top space a n d space b with rightwards arrow on top equals i with hat on top minus 2 j with hat on top plus k with hat on top

Solution 18

G i v e n space t h a t space
a with rightwards arrow on top equals 2 stack i space with hat on top plus 3 j with hat on top minus k with hat on top
space a n d
b with rightwards arrow on top equals i with hat on top minus 2 j with hat on top plus k with hat on top
T h u s comma space F i n d space a space v e c t o r space o f space m a g n i t u d e space o f space 5 space u n i t s space p a r a l l e l space t o space t h e space r e s u l tan t space o f space t h e space v e c t o r s space
a with rightwards arrow on top plus b with rightwards arrow on top equals 2 i with hat on top plus 3 j with hat on top minus k with hat on top plus space i with hat on top minus 2 j with hat on top plus k with hat on top
rightwards double arrow a with rightwards arrow on top plus b with rightwards arrow on top equals 3 i with hat on top plus j with hat on top
rightwards double arrow open vertical bar a with rightwards arrow on top plus b with rightwards arrow on top close vertical bar equals square root of 9 plus 1 end root equals square root of 10
T h u s comma space t h e space u n i t space v e c t o r space a l o n g space t h e space r e s u l tan t space v e c t o r space a with rightwards arrow on top plus b with rightwards arrow on top space i s space
fraction numerator 3 i with hat on top plus j with hat on top over denominator square root of 10 end fraction
T h e space v e c t o r space o f space m a g n i t u d e space o f space 5 space u n i t s space p a r a l l e l space t o space t h e space r e s u l tan t
v e c t o r equals fraction numerator 3 i with hat on top plus j with hat on top over denominator square root of 10 end fraction cross times 5 equals square root of 5 over 2 end root open parentheses 3 i with hat on top plus j with hat on top close parentheses

Question 19

Solution 19

Chapter 23 Algebra of Vectors Exercise Ex. 23.7

Question 1

Solution 1

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

U sin g space v e c t o r space s h o w space t h a t space t h e space p o i n t s space A open parentheses minus 2 comma 3 comma 5 close parentheses comma space B open parentheses 7 comma 0 comma minus 1 close parentheses space a n d space C open parentheses minus 3 comma minus 2 comma minus 5 close parentheses space
a n d space D open parentheses 3 comma 4 comma 7 close parentheses space a r e space s u c h space t h a t space A B space a n d space C D space i n t e r s e c t space a t space t h e space p o i n t space P open parentheses 1 comma 2 comma 3 close parentheses

Solution 12

W e space h a v e
stack A P with rightwards arrow on top equals P o s i t i o n space v e c t o r space o f space P minus P o s i t i o n space v e c t o r space o f space A
rightwards double arrow stack A P with rightwards arrow on top equals i with hat on top plus 2 j with hat on top plus 3 k with hat on top minus open parentheses minus 2 i with hat on top plus 3 j with hat on top plus 5 k with hat on top close parentheses equals 3 i with hat on top minus j with hat on top minus 2 k with hat on top
stack P B with rightwards arrow on top equals P o s i t i o n space v e c t o r space o f space B minus P o s i t i o n space v e c t o r space o f space P
rightwards double arrow stack P B with rightwards arrow on top equals 7 i with hat on top minus k with hat on top minus open parentheses i with hat on top plus 2 j with hat on top plus 3 k with hat on top close parentheses equals 6 i with hat on top minus 2 j with hat on top minus 4 k with hat on top
C l e a r l y comma space stack P B with rightwards arrow on top equals 2 stack A P with rightwards arrow on top
s o space v e c t o r s space stack A P with rightwards arrow on top space a n d space stack P B with rightwards arrow on top space a r e space c o l l i n e a r.
B u t space P space i s space a space p o i n t space c o m m o n space t o space stack A P with rightwards arrow on top space a n d space stack P B with rightwards arrow on top. space
H e n c e space P comma space A comma space B space a r e space c o l l i n e a r space p o i n t s.
S i m i l a r l y comma space stack C P with rightwards arrow on top equals i with hat on top plus 2 j with hat on top plus 3 k with hat on top minus open parentheses minus 3 i with hat on top minus 2 j with hat on top minus 5 k with hat on top close parentheses equals 4 i with hat on top plus 4 j with hat on top plus 8 k with hat on top
a n d space stack P D with rightwards arrow on top equals 3 i with hat on top plus 4 j with hat on top plus 7 k with hat on top minus open parentheses i with hat on top plus 2 j with hat on top plus 3 k with hat on top close parentheses equals 2 i with hat on top plus 2 j with hat on top plus 4 k with hat on top
S o space v e c t o r s space stack C P with rightwards arrow on top space a n d space stack P D with rightwards arrow on top space a r e space c o l l i n e a r.
B u t space P space i s space a space c o m m o n space p o i n t space t o space stack C P with rightwards arrow on top space a n d space stack C D with rightwards arrow on top.
H e n c e comma space C comma space P comma space D space a r e space c o l l i n e a r space p o i n t s.
T h u s comma space A comma space B comma space C comma space D space a n d space P space a r e space p o i n t s space s u c h space t h a t space A comma space P comma space B space a n d space C comma space P comma space D space
a r e space t w o space s e t s space o f space c o l l i n e a r space p o i n t s. space H e n c e space A B space a n d space C D space i n t e r s e c t space a t space t h e
p o i n t space P

Question 13

Using vectors, find the value of λ such that the points

 (λ, – 10, 3), (1 -1, 3) and (3, 5, 3) are collinear.Solution 13

Chapter 23 Algebra of Vectors Exercise Ex. 23.8

Question 1

Solution 1

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 2 (iii)

Solution 2 (iii)

Question 2 (iv)

Solution 2 (iv)

Question 3 (i)

Solution 3 (i)

Question 3 (ii)

Solution 3 (ii)

Question 4

Solution 4

Question 5 (i)

Solution 5 (i)

Question 5 (ii)

Solution 5 (ii)

Question 6 (i)

Solution 6 (i)

Question 6 (ii)

Solution 6 (ii)

Question 7 (i)

Solution 7 (i)

Question 7 (ii)

Solution 7 (ii)

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 23 Algebra of Vectors Exercise Ex. 23.9

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7 (i)

Solution 7 (i)

Question 7 (ii)

Solution 7 (ii)

Question 7 (iii)

Solution 7 (iii)

Question 8

Solution 8

Question 9

Solution 9

Question 10

If a unit vector begin mathsize 12px style straight a with rightwards arrow on top end style makes and angles begin mathsize 12px style straight pi over 3 space with space straight i with hat on top comma space straight pi over 4 space with space straight j with hat on top end style and an acute angle θ with begin mathsize 12px style straight k with hat on top end style, then find θ and hence, the components of begin mathsize 12px style straight a with rightwards arrow on top end style.Solution 10

Question 11

Solution 11

Question 12

Solution 12

Read More

RD SHARMA SOLUTION CHAPTER – 22 Differential Equations I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 22 Differential Equations Exercise Ex. 22.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

D e t e r m i n e space t h e space o r d e r space a n d space d e g r e e space o f space t h e space f o l l o w i n g space d i f f e r e n t i a l space e q u a t i o n. space S t a t e
a l s o space w h e t h e r space i t space i s space l i n e a r space o r space n o n minus l i n e a r.
square root of 1 plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared end root equals open parentheses c fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 1 third end exponent

Solution 4

C o n s i d e r space t h e space g i v e n space d i f f e r e n t i a l space e q u a t i o n comma space square root of 1 plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared end root equals open parentheses c fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 1 third end exponent
S q u a r i n g space o n space b o t h space t h e space s i d e s comma space w e space h a v e
1 plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared equals open parentheses c fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 2 over 3 end exponent
C u b i n g space o n space b o t h space t h e space s i d e s comma space w e space h a v e
open square brackets 1 plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared close square brackets cubed equals open curly brackets open parentheses c fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 2 over 3 end exponent close curly brackets cubed
rightwards double arrow 1 plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 6 plus 3 open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared plus 3 open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 4 equals c squared open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses squared
rightwards double arrow c squared open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses squared minus open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 6 minus 3 open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 4 minus 3 open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared minus 1 equals 0
T h e space h i g h e s t space o r d e r space d i f f e r e n t i a l space c o e f f i c i e n t space i n space t h i s space
e q u a t i o n space i s space fraction numerator d squared y over denominator d x squared end fraction space a n d space i t s space p o w e r space i s space 2.
T h e r e f o r e comma space t h e space g i v e n space d i f f e r e n t i a l space e q u a t i o n space i s space a space
n o n minus l i n e a r space d i f f e r e n t i a l space e q u a t i o n space o f space s e c o n d space o r d e r space a n d space s e c o n d space d e g r e e.

Question 5

Solution 5

Question 6

D e t e r m i n e space t h e space o r d e r space a n d space d e g r e e space o f space t h e space f o l l o w i n g space d i f f e r e n t i a l space e q u a t i o n. space S t a t e
a l s o space w h e t h e r space i s space l i n e a r space o r space n o n minus l i n e a r.
3 root of fraction numerator d squared y over denominator d x squared end fraction end root equals square root of fraction numerator d y over denominator d x end fraction end root

Solution 6

C o n s i d e r space t h e space g i v e n space d i f f e r e n t i a l space e q u a t i o n comma
3 root of fraction numerator d squared y over denominator d x squared end fraction end root equals square root of fraction numerator d y over denominator d x end fraction end root
C u b i n g space o n space b o t h space t h e space s i d e s space o f space t h e space a b o v e space e q u a t i o n comma space w e space h a v e
fraction numerator d squared y over denominator d x squared end fraction equals open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 3 over 2 end exponent
S q u a r i n g space o n space b o t h space t h e space s i d e s space o f space t h e space a b o v e space e q u a t i o n comma space w e space h a v e
open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses squared equals open square brackets open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 3 over 2 end exponent close square brackets squared
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses squared equals open square brackets open parentheses fraction numerator d y over denominator d x end fraction close parentheses close square brackets cubed
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses squared minus open square brackets open parentheses fraction numerator d y over denominator d x end fraction close parentheses close square brackets cubed equals 0
T h e space h i g h e s t space o r d e r space d i f f e r e n t i a l space c o e f f i c i e n t space i n space t h i s space e q u a t i o n space i s space fraction numerator d squared y over denominator d x squared end fraction
a n d space i t s space p o w e r space i s space 2.
T h e r e f o r e comma space t h e space g i v e n space d i f f e r e n t i a l space e q u a t i o n space i s space a space n o n minus l i n e a r space d i f f e r e n t i a l
e q u a t i o n space o f space s e c o n d space o r d e r space a n d space s e c o n d space d e g r e e.

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Determine the order and degree of the following differential equations. State also whether they are linear or non-linear.

Solution 27

The order of a differential equation is the order of the highest order derivative appearing in the equation.

The degree of a differential equation is the degree of the highest order derivative.

Consider the given differential equation

In the above equation, the order of the highest order derivative is 1.

So the differential equation is of order 1.

In the above differential equation, the power of the highest order derivative is 3.

Hence, it is a differential equation of degree 3.

Since the degree of the above differential equation is 3, more than one, it is a non-linear differential equation.

Chapter 22 – Differential Equations Exercise Ex. 22.2

Question 1

Solution 1

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Form the differential equation having y = (sin-1x)2 + A cos -1 x + B, where A and B are arbitrary constants, as its general solution.Solution 14

Question 15(i)

F o r m space t h e space d i f f e r e n t i a l space e q u a t i o n space o f space t h e space f a m i l y space o f space
c u r v e s space r e p r e s e n t e d space b y space t h e space e q u a t i o n space left parenthesis apostrophe a apostrophe space b e i n g space t h e space p a r a m e t e r right parenthesis.
open parentheses 2 x plus a close parentheses squared plus y squared equals a squared

Solution 15(i)

C o n s i d e r space t h e space g i v e n space e q u a t i o n. comma
open parentheses 2 x plus a close parentheses squared plus y squared equals a squared.... left parenthesis 1 right parenthesis
D i f f e r e n t i a t i n g space t h e space a b o v e space e q u a t i o n space w i t h space r e s p e c t space t o space x comma space w e space h a v e comma
2 open parentheses 2 x plus a close parentheses plus 2 y fraction numerator d y over denominator d x end fraction equals 0
rightwards double arrow open parentheses 2 x plus a close parentheses plus y fraction numerator d y over denominator d x end fraction equals 0
rightwards double arrow 2 x plus a equals minus y fraction numerator d y over denominator d x end fraction
rightwards double arrow a equals minus 2 x minus y fraction numerator d y over denominator d x end fraction
S u b s t i t u t i n g space t h e space v a l u e space o f space a space i n space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e
open parentheses 2 x minus 2 x minus y fraction numerator d y over denominator d x end fraction close parentheses squared plus y squared equals open parentheses minus 2 x minus y fraction numerator d y over denominator d x end fraction close parentheses squared
rightwards double arrow open parentheses y fraction numerator d y over denominator d x end fraction close parentheses squared plus y squared equals open parentheses 4 x squared plus y squared open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared plus 4 x y fraction numerator d y over denominator d x end fraction close parentheses

rightwards double arrow y squared equals 4 x squared plus 4 x y fraction numerator d y over denominator d x end fraction
rightwards double arrow y squared minus 4 x squared minus 4 x y fraction numerator d y over denominator d x end fraction equals 0

Question 15(ii)

Solution 15(ii)

Question 15(iii)

F o r m space t h e space d i f f e r e n t i a l space e q u a t i o n space o f space t h e space f a m i l y space o f space c u r v e s space r e p r e s e n t e d space b y space t h e
e q u a t i o n space left parenthesis apostrophe a apostrophe space b e i n g space t h e space p a r a m e t e r right parenthesis :
open parentheses x minus a close parentheses squared plus 2 y squared equals a squared

Solution 15(iii)

C o n s i d e r space t h e space g i v e n space e q u a t i o n comma
open parentheses x minus a close parentheses squared plus 2 y squared equals a squared.... left parenthesis 1 right parenthesis
D i f f e r e n t i a t i n g space t h e space a b o v e space e q u a t i o n space w i t h space r e s p e c t space t o space x comma space w e space h a v e
2 open parentheses x minus a close parentheses plus 4 y fraction numerator d y over denominator d x end fraction equals 0
rightwards double arrow open parentheses x minus a close parentheses plus 2 y fraction numerator d y over denominator d x end fraction equals 0
rightwards double arrow open parentheses x minus a close parentheses equals minus 2 y fraction numerator d y over denominator d x end fraction
rightwards double arrow a equals x plus 2 y fraction numerator d y over denominator d x end fraction
S u b s t i t u t i n g space t h e space v a l u e space o f space a space i n space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e
open parentheses x minus x plus 2 y fraction numerator d y over denominator d x end fraction close parentheses squared plus 2 y squared equals open parentheses x plus 2 y fraction numerator d y over denominator d x end fraction close parentheses squared
rightwards double arrow 4 y squared open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared plus 2 y squared equals x squared plus 4 y squared open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared plus 4 x y fraction numerator d y over denominator d x end fraction
rightwards double arrow 2 y squared minus x squared equals 4 x y fraction numerator d y over denominator d x end fraction

Question 16(i)

Solution 16(i)

Question 16(ii)

Solution 16(ii)

Question 16(iii)

Solution 16(iii)

Question 16(iv)

Represent the following family of curves by forming the corresponding differential equation (a,b being parameters):

x+ (y – b)2 = 1Solution 16(iv)

begin mathsize 12px style straight x squared plus left parenthesis straight y minus straight b right parenthesis squared equals 1 space space space space space space space space space space space space space space space space space space space space space space space space ____ left parenthesis straight i right parenthesis
Differentiating space it space with space respect space to space straight x comma
2 straight x space plus space 2 left parenthesis straight y minus straight b right parenthesis dy over dx equals 0
straight x space plus space left parenthesis straight y minus straight b right parenthesis dy over dx equals 0
left parenthesis straight y minus straight b right parenthesis dy over dx equals negative straight x
left parenthesis straight y minus straight b right parenthesis equals fraction numerator begin display style fraction numerator negative straight x over denominator dy end fraction end style over denominator dx end fraction
Put space the space value space of space left parenthesis straight y minus straight b right parenthesis space is space equation space left parenthesis straight i right parenthesis
straight x squared open parentheses fraction numerator negative straight x over denominator begin display style dy over dx end style end fraction close parentheses squared equals 1
straight x squared open parentheses dy over dx close parentheses squared plus straight x squared equals open parentheses dy over dx close parentheses squared
straight x squared open curly brackets open parentheses dy over dx close parentheses squared plus 1 close curly brackets equals open parentheses dy over dx close parentheses squared
end style

Question 16(v)

Solution 16(v)

Question 16(vi)

Solution 16(vi)

Question 16(vii)

Solution 16(vii)

Question 16(viii)

Solution 16(viii)

Question 16(ix)

Solution 16(ix)

Question 16(x)

Solution 16(x)

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Chapter 22 – Differential Equations Exercise Ex. 22.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

show that y = ae2x + be-x is a solution of the differential equation begin mathsize 12px style fraction numerator straight d squared straight y over denominator dx squared end fraction minus dy over dx minus 2 straight y equals 0. end styleSolution 3

begin mathsize 12px style straight y equals ae to the power of 2 straight x end exponent plus be to the power of negative straight x end exponent space space space space space space space space space space space space space space space space space space space space space ___ left parenthesis straight i right parenthesis
Differentiating space it space with space respect space to space straight x comma
dy over dx equals 2 ae to the power of 2 straight x end exponent minus be to the power of negative straight x end exponent space space space space space space space space space space space space space ___ left parenthesis ii right parenthesis
Differentiating space it space with space respect space to space straight x comma
fraction numerator straight d squared straight y over denominator dx squared end fraction equals 4 ae to the power of 2 straight x end exponent minus be to the power of negative straight x end exponent space space space space space space space space space space space space space ___ left parenthesis iii right parenthesis
Now comma
fraction numerator straight d squared straight y over denominator dx squared end fraction minus dy over dx minus 2 straight y
equals left parenthesis 4 ae to the power of 2 straight x end exponent plus be to the power of negative straight x end exponent right parenthesis minus left parenthesis 2 ae to the power of 2 straight x end exponent minus be to the power of negative straight x end exponent right parenthesis minus 2 left parenthesis ae to the power of 2 straight x end exponent plus be to the power of negative straight x end exponent right parenthesis
equals 4 ae to the power of 2 straight x end exponent plus be to the power of negative straight x end exponent minus 2 ae to the power of 2 straight x end exponent plus be to the power of negative straight x end exponent minus 2 ae to the power of 2 straight x end exponent minus 2 be to the power of negative straight x end exponent
equals 4 ae to the power of 2 straight x end exponent minus 4 ae to the power of 2 straight x end exponent plus 2 be to the power of negative straight x end exponent minus 2 be to the power of negative straight x end exponent
equals 0
S o comma
fraction numerator d squared y over denominator d x squared end fraction minus fraction numerator d y over denominator d x end fraction minus 2 y equals 0 end style

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Verify that y = begin mathsize 12px style straight a over straight x end style + b is a solution of the differential equation begin mathsize 12px style fraction numerator straight d squared straight y over denominator dx squared end fraction plus 2 over straight x open parentheses dy over dx close parentheses equals 0 end styleSolution 7

begin mathsize 12px style straight y equals straight a over straight x plus straight b space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space ___ left parenthesis straight i right parenthesis
Differentiating space it space with space respect space to space straight x comma
dy over dx equals negative straight a over straight x squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space ___ left parenthesis ii right parenthesis
Differentiating space it space with space respect space to space straight x comma
fraction numerator straight d squared straight y over denominator dx squared end fraction equals fraction numerator 2 straight a over denominator straight x cubed end fraction
equals negative 2 over straight x open parentheses negative straight a over straight x squared close parentheses
fraction numerator straight d squared straight y over denominator dx squared end fraction equals negative 2 over straight x open parentheses dy over dx close parentheses
fraction numerator straight d squared straight y over denominator dx squared end fraction plus 2 over straight x open parentheses dy over dx close parentheses equals 0 end style

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Show that y = ex(A cos x + B sin x) is the solution of the differential equation

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

left parenthesis 1 minus straight x squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight x dy over dx minus straight m squared straight y equals 0
straight y equals straight e to the power of mcos to the power of negative 1 end exponent straight x end exponent
dy over dx equals fraction numerator me to the power of mcos to the power of negative 1 end exponent straight x end exponent over denominator negative square root of 1 minus straight x squared end root end fraction
dy over dx equals fraction numerator negative my over denominator square root of 1 minus straight x squared end root end fraction........ left parenthesis straight i right parenthesis space
fraction numerator straight d squared straight y over denominator dx squared end fraction equals fraction numerator square root of left parenthesis 1 minus straight x squared right parenthesis end root. open parentheses negative straight m begin display style dy over dx end style close parentheses minus left parenthesis negative my right parenthesis begin display style fraction numerator left parenthesis negative 2 straight x right parenthesis over denominator 2 square root of left parenthesis 1 minus straight x squared right parenthesis end root end fraction end style over denominator left parenthesis 1 minus straight x squared right parenthesis end fraction space left square bracket From space left parenthesis straight i right parenthesis right square bracket
fraction numerator straight d squared straight y over denominator dx squared end fraction equals fraction numerator left parenthesis negative straight m right parenthesis open parentheses begin display style negative my end style close parentheses minus straight x begin display style dy over dx end style over denominator left parenthesis 1 minus straight x squared right parenthesis end fraction left square bracket From space left parenthesis straight i right parenthesis right square bracket
left parenthesis 1 minus straight x squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction equals straight m squared straight y minus straight x dy over dx
left parenthesis 1 minus straight x squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction plus straight x dy over dx minus straight m squared straight y equals 0
Hence space Proved

Question 18

Solution 18

Question 19

Solution 19

Question 20

Show that y = e-x + ax + b is solution of the differential equation begin mathsize 12px style straight e to the power of straight x fraction numerator straight d squared straight y over denominator dx squared end fraction equals 1 end styleSolution 20

begin mathsize 12px style straight y equals straight e to the power of negative straight x end exponent plus ax plus straight b
Differentiating space it space with space respect space to space straight x comma
dy over dx equals negative straight e to the power of negative straight x end exponent plus straight a
Differentiating space it space with space respect space to space straight x comma
fraction numerator straight d squared straight y over denominator dx squared end fraction equals straight e to the power of negative straight x end exponent
1 over straight e to the power of negative straight x end exponent fraction numerator straight d squared straight y over denominator dx squared end fraction equals 1
straight e to the power of straight x fraction numerator straight d squared straight y over denominator dx squared end fraction equals 1 end style

Question 21(i)

For the following differential equation verify that the accompanying function is a solution in the mentioned domain (a, b are parameters) begin mathsize 12px style straight x dy over dx equals straight y space space space space space space space space space space space space straight y space equals space ax comma space straight x element of straight R minus left curly bracket 0 right curly bracket end styleSolution 21(i)

begin mathsize 12px style straight y equals ax space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis straight i right parenthesis
Differentiating space it space with space respect space to space straight x comma
dy over dx equals straight a
equals ax over straight x space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because straight x element of straight R minus left curly bracket 0 right curly bracket right square bracket
dy over dx equals straight y over straight x space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket Using space equation space left parenthesis straight i right parenthesis right square bracket
straight x dy over dx equals straight y
So comma space straight y space equals space ax space is space the space solution space of space the space given space equation. end style

Question 21(ii)

Solution 21(ii)

Question 21(iii)

Solution 21(iii)

Question 21(iv)

Solution 21(iv)

Question 21(v)

Solution 21(v)

Chapter 22 – Differential Equations Exercise Ex. 22.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

For the following initial value problem verify that the accompanying function is a solution:

begin mathsize 12px style fraction numerator straight d squared straight y over denominator dx squared end fraction plus straight y equals 0 comma space straight y open parentheses 0 close parentheses space equals space 0 comma space straight y apostrophe open parentheses 0 close parentheses equals 1 space space space space space space space space space space space straight y equals sinx end style

Solution 3

begin mathsize 12px style Here comma space straight y space equals space sin space straight x space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis straight i right parenthesis
Differentiating space it space with space respect space to space straight x comma
dy over dx equals cos space straight x space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis ii right parenthesis
Again space differentiating space it space with space respect space to space straight x comma
fraction numerator straight d squared straight y over denominator dx squared end fraction equals negative sin space straight x space
fraction numerator straight d squared straight y over denominator dx squared end fraction equals negative straight y
fraction numerator straight d squared straight y over denominator dx squared end fraction plus straight y equals 0
So comma space straight y equals sin space straight x space is space straight a space solution space of space the space equation.
Put space space space space space straight x equals 0 space in space equation space left parenthesis straight i right parenthesis comma
rightwards double arrow space space space space space straight y space equals space sin space 0
rightwards double arrow space space space space space straight y equals 0
rightwards double arrow space space space space space straight y left parenthesis 0 right parenthesis space equals 0
Put space straight x space equals 0 space in space equation space left parenthesis ii right parenthesis
straight y apostrophe space equals cos space 0
straight y apostrophe space equals 1
rightwards double arrow straight y apostrophe left parenthesis 0 right parenthesis space equals space 1 end style

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Chapter 22 – Differential Equations Exercise Ex. 22.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

w h e r e space x not equal to open parentheses 2 n plus 1 close parentheses pi comma space n element of Z

Question 6

Solution 6

Question 7

Solution 7

w h e r e space x element of R

Question 8

Solution 8

fraction numerator d y over denominator d x end fraction equals log x
rightwards double arrow d y equals log x cross times d x
rightwards double arrow integral d y equals integral log x d x
rightwards double arrow y equals log x cross times integral 1 d x minus integral open parentheses 1 over x integral 1 d x close parentheses d x plus C space space space space space open square brackets U sin g space i n t e g r a t i o n space b y space p a r t s close square brackets
rightwards double arrow y equals x log x minus integral d x plus C
rightwards double arrow y equals x log x minus x plus C
rightwards double arrow y equals x open parentheses log x minus 1 close parentheses plus C comma space w h e r e space x element of open parentheses 0 comma infinity close parentheses

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solve the following differential equation:

(sin x + cos x)dy + (cos x – sin x) dx = 0Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solve the following differential equation:

begin mathsize 12px style cos space x fraction numerator d y over denominator d x end fraction minus cos space 2 x space equals space cos space 3 x end style

Solution 15

begin mathsize 12px style cos space straight x space dy over dx minus cos space 2 space straight x space equals space cos space 3 space straight x
cos space straight x space dy over dx minus cos space 3 space straight x space equals space cos space 2 space straight x
dy over dx equals fraction numerator 4 cos cubed space straight x minus 3 cos space straight x space plus 2 space cos squared straight x space minus 1 over denominator cos space straight x end fraction
dy over dx equals fraction numerator 4 cos cubed straight x over denominator cos space straight x end fraction minus fraction numerator 3 cos space straight x over denominator cos space straight x end fraction plus fraction numerator 2 cos squared straight x over denominator cos space straight x end fraction minus fraction numerator 1 over denominator cos space straight x end fraction
dy over dx equals 4 cos squared straight x minus 3 plus 2 space cos space straight x minus space sec space straight x
dy over dx equals 4 open parentheses fraction numerator cos space 2 straight x plus 1 over denominator 2 end fraction close parentheses minus 32 space cos space straight x space minus space sec space straight x
dy equals left parenthesis 2 space cos space 2 space straight x space plus space 2 minus 3 plus 2 space cos space straight x space minus sec space straight x right parenthesis dx
integral dy equals integral left parenthesis 2 cos space 2 straight x space minus 1 plus 2 space cos space straight x minus space sec space straight x right parenthesis dx
straight y space equals space sin space 2 straight x space minus straight x plus 2 space sin space straight x space minus log open vertical bar sec space straight x space plus space tan space straight x close vertical bar plus straight c end style

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solve the following differential equation

begin mathsize 12px style open parentheses 1 plus straight x squared close parentheses dy over dx minus straight x equals 2 space tan to the power of negative 1 end exponent straight x end style

Solution 18

begin mathsize 12px style left parenthesis 1 plus straight x squared right parenthesis dy over dx minus straight x equals 2 tan to the power of negative 1 end exponent straight x
left parenthesis 1 plus straight x squared right parenthesis dy over dx equals 2 tan to the power of negative 1 end exponent straight x plus straight x
dy equals open parentheses fraction numerator 2 tan to the power of negative 1 end exponent straight x plus straight x over denominator 1 plus straight x squared end fraction close parentheses dx
integral dy equals integral open parentheses fraction numerator 2 tan to the power of negative 1 end exponent straight x plus straight x over denominator 1 plus straight x squared end fraction close parentheses dx
straight y equals integral left parenthesis 2 straight t plus tant right parenthesis dt space space space space space space space space space space space space space space space space left square bracket tan to the power of negative 1 end exponent straight x equals straight t right square bracket
equals 1 half log open vertical bar 1 plus straight x squared close vertical bar plus left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis squared plus straight c end style

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

solve the following differential equation

begin mathsize 12px style straight x open parentheses straight x squared minus 1 close parentheses dy over dx equals 1 comma space straight y left parenthesis 2 right parenthesis space equals space 0 end style

Solution 26

begin mathsize 12px style straight x left parenthesis straight x squared minus 1 right parenthesis dy over dx equals 1 comma straight y left parenthesis 2 right parenthesis equals 0
dy over dx equals fraction numerator 1 over denominator straight x left parenthesis straight x squared minus 1 right parenthesis end fraction
dy equals fraction numerator 1 over denominator straight x left parenthesis straight x squared minus 1 right parenthesis end fraction dx
integral dy equals integral open parentheses fraction numerator 1 over denominator straight x open parentheses straight x squared minus 1 close parentheses end fraction close parentheses dx
straight y equals 1 half integral fraction numerator 1 over denominator straight x minus 1 end fraction dx minus integral 1 over straight x dx plus 1 half integral fraction numerator 1 over denominator straight x plus 1 end fraction dx
equals 1 half log open vertical bar straight x minus 1 close vertical bar minus log open vertical bar straight x close vertical bar plus 1 half log open vertical bar straight x plus 1 close vertical bar plus straight c
Putting space straight x equals 2 comma space straight y equals 0 comma space we space have
straight y equals 1 half log open vertical bar straight x minus 1 close vertical bar minus log open vertical bar straight x close vertical bar plus 1 half log open vertical bar straight x plus 1 close vertical bar plus straight c space
0 equals 1 half log open vertical bar 2 minus 1 close vertical bar minus log open vertical bar 2 close vertical bar plus 1 half log open vertical bar 2 plus 1 close vertical bar plus straight c space
straight c equals log open vertical bar 2 close vertical bar minus 1 half log open vertical bar 3 close vertical bar
Putting space the space value space ofc comma space we space have
straight y equals 1 half log open vertical bar straight x minus 1 close vertical bar minus log open vertical bar straight x close vertical bar plus 1 half log open vertical bar straight x plus 1 close vertical bar plus straight c
equals log 4 over 3 open parentheses fraction numerator straight x squared minus 1 over denominator straight x squared end fraction close parentheses end style

Chapter 22 – Differential Equations Exercise Ex. 22.6

Question 1

Solve the following differential equation:

begin mathsize 12px style dv over dx plus fraction numerator 1 plus straight y squared over denominator straight y end fraction equals 0 end style

Solution 1

Question 2

Solve the following differential equation:

begin mathsize 12px style dy over dx equals fraction numerator 1 plus straight y squared over denominator straight y cubed end fraction end style

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Chapter 22 – Differential Equations Exercise Ex. 22.7

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

begin mathsize 12px style cos space straight x space cos space straight y space dy over dx equals negative sin space straight x space sin space straight y
fraction numerator cos space straight y over denominator sin space straight y end fraction dy space equals space minus fraction numerator sin space straight x over denominator cos space straight x end fraction dx
integral cot space ydy space equals negative integral tan space xdx
log space sin space straight y space equals space log space cos space straight x space plus space log space straight c
sin space straight y space equals space straight c space cos space straight x end style

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

begin mathsize 12px style left parenthesis straight y space plus space xy right parenthesis dx space plus open parentheses straight x minus xy squared close parentheses dy space equals space 0
straight y left parenthesis 1 plus straight x right parenthesis dx equals open parentheses xy squared minus straight x close parentheses dy
straight y left parenthesis 1 plus straight x right parenthesis dx equals straight x open parentheses straight y squared minus 1 close parentheses dy
fraction numerator open parentheses straight y squared minus 1 close parentheses dy over denominator straight y end fraction equals fraction numerator 1 plus straight x over denominator straight x end fraction dx
integral open parentheses straight y minus 1 over straight y close parentheses dy equals integral open parentheses 1 over straight x plus 1 close parentheses dx
straight y squared over 2 minus log open vertical bar straight y close vertical bar equals log open vertical bar straight x close vertical bar plus straight x plus straight c subscript 1
straight y squared over 2 minus straight x minus log open vertical bar straight y close vertical bar minus log open vertical bar straight x close vertical bar equals straight c subscript 1
log open vertical bar straight x close vertical bar plus straight x plus log open vertical bar straight y close vertical bar minus straight y squared over 2 equals straight c end style

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solve the following differential equation:

begin mathsize 12px style fraction numerator d y over denominator d x end fraction equals e to the power of x plus y end exponent space plus space e to the power of negative x plus y end exponent end style

Solution 35

begin mathsize 12px style dy over dx equals straight e to the power of straight x plus straight y end exponent plus straight e to the power of negative straight x plus straight y end exponent
equals straight e to the power of straight x space xe to the power of straight y plus straight e to the power of negative straight x end exponent space xe to the power of straight y
dy over dx equals straight e to the power of straight y open parentheses straight e to the power of straight x plus straight e to the power of negative straight x end exponent close parentheses
dy over straight e to the power of straight y equals open parentheses straight e to the power of straight x plus straight e to the power of negative straight x end exponent close parentheses dx
integral straight e to the power of negative straight y end exponent dy equals integral open parentheses straight e to the power of straight x plus straight e to the power of negative straight x end exponent close parentheses dx
minus straight e to the power of negative straight y end exponent equals straight e to the power of straight x minus straight e to the power of negative straight x end exponent plus straight c
straight e to the power of negative straight x end exponent minus straight e to the power of negative straight y end exponent equals straight e to the power of straight x plus straight c end style

Question 36

Solve the following differential equation:

begin mathsize 12px style fraction numerator d y over denominator d x end fraction equals open parentheses cos squared x space minus space sin squared x close parentheses cos squared space y end style

Solution 36

begin mathsize 12px style dy over dx equals open parentheses cos squared straight x minus sin squared straight x close parentheses cos squared straight y
fraction numerator dy over denominator cos squared straight y end fraction equals open parentheses cos squared straight x minus sin squared straight x close parentheses dx
integral sec squared ydy equals integral cos 2 xdx
tan space straight y equals fraction numerator sin 2 straight x over denominator 2 end fraction plus straight c end style

Question 37(i)

Solution 37(i)

Question 37(ii)

Solve the following differential equation:

Solution 37(ii)

Question 38(i)

Solution 38(i)

Question 38(ii)

Solution 38(ii)

Question 38(iii)

yex/y dx = (xex/y + y2) dy, y ¹ 0Solution 38(iii)

Question 38(iv)

(1 + y2) tan-1 x dx + 2y (1 + x2)dy = 0Solution 38(iv)

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

begin mathsize 12px style dy over dx equals 2 straight e to the power of straight x straight y cubed comma straight y left parenthesis 0 right parenthesis equals 1 half
integral dy over straight y cubed equals integral 2 straight e to the power of straight x dx
minus fraction numerator 1 over denominator 2 straight y squared end fraction equals 2 straight e to the power of straight x plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative left parenthesis straight i right parenthesis
Put space straight x space equals space 0 comma space straight y space equals space 1 half
minus 4 over 2 equals 2 straight e to the power of 0 plus straight c
minus 2 equals 2 plus straight c
straight c equals negative 4
Put space straight c equals negative 4 space in space equation space left parenthesis straight i right parenthesis
minus fraction numerator 1 over denominator 2 straight y squared end fraction equals 2 straight e to the power of straight x minus 4
minus 1 equals 4 straight e to the power of straight x straight y squared minus 8 straight y squared
minus 1 equals negative straight y squared left parenthesis 8 minus 4 straight e to the power of straight x right parenthesis
straight y squared left parenthesis 8 minus 4 straight e to the power of straight x right parenthesis equals 1 end style

Question 43

Solution 43

begin mathsize 12px style dr over dt equals negative rt comma space straight r left parenthesis 0 right parenthesis space equals space straight r subscript 0
integral dr over straight r equals negative integral tdt
log open vertical bar straight r close vertical bar equals negative straight t squared over 2 plus straight c space space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative negative left parenthesis straight i right parenthesis
Put space straight t space equals space 0 comma space straight r space equals space straight r subscript 0 space inequation space left parenthesis straight i right parenthesis comma
log open vertical bar straight r subscript 0 close vertical bar equals 0 plus straight c
log open vertical bar straight r subscript 0 close vertical bar equals straight c
Now comma
log open vertical bar straight r close vertical bar equals negative straight t squared over 2 plus log open vertical bar straight r subscript 0 close vertical bar
straight r over straight r subscript 0 equals straight e to the power of negative straight t squared over 2 end exponent
straight r equals straight r subscript 0 straight e to the power of negative straight t squared over 2 end exponent end style

Question 44

Solution 44

Question 45(i)

Solution 45(i)

Question 45(ii)

Solution 45(ii)

Question 45(iii)

Solution 45(iii)

begin mathsize 12px style dy over dx equals 2 straight e to the power of 2 straight x end exponent straight y squared comma space straight y space left parenthesis 0 right parenthesis equals negative 1
integral dy over straight y squared equals integral 2 straight e to the power of 2 straight x end exponent dx
minus 1 over straight y equals fraction numerator 2 straight e to the power of 2 straight x end exponent over denominator 2 end fraction plus straight c
minus 1 over straight y equals straight e to the power of 2 straight x end exponent plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative left parenthesis straight i right parenthesis
Put space straight y space equals space minus 1 comma space straight x space equals space 0
1 equals straight e to the power of 0 plus straight c
1 equals 1 plus straight c
straight c equals 0
Put space straight c equals 0 space in space equation space left parenthesis straight i right parenthesis comma
minus 1 over straight y equals straight e to the power of 2 straight x end exponent
straight y equals negative straight e to the power of negative 2 straight x end exponent end style

Question 45(iv)

Solution 45(iv)

begin mathsize 12px style cos space straight y dy over dx equals straight e to the power of straight x comma space straight y space left parenthesis 0 right parenthesis equals straight pi over 2
integral cos space ydy equals integral straight e to the power of straight x dx
sin space straight y equals straight e to the power of straight x plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative left parenthesis straight i right parenthesis
Put space straight x space equals 0 comma space straight y equals straight pi over 2
sin open parentheses straight pi over 2 close parentheses equals straight e to the power of 0 plus straight c
1 equals 1 plus straight c
straight c equals 0
Put space straight c equals 0 space in space equation space left parenthesis straight i right parenthesis comma
sin space straight y space equals space straight e to the power of straight x
straight y equals sin to the power of negative 1 end exponent open parentheses straight e to the power of straight x close parentheses end style

Question 45(v)

Solution 45(v)

Question 45(vi)

Solve the following initial value problem

begin mathsize 12px style dy over dx end style=1 + x2 + y2 + x2y2, y(0) = 1Solution 45(vi)

begin mathsize 12px style dy over dx equals 1 plus straight x squared plus straight y squared plus straight x squared straight y squared comma space straight y left parenthesis 0 right parenthesis equals 1
equals left parenthesis 1 plus straight x squared right parenthesis left parenthesis 1 plus straight y squared right parenthesis
integral fraction numerator dy over denominator 1 plus straight y squared end fraction equals integral left parenthesis 1 plus straight x squared right parenthesis dx
tan to the power of negative 1 end exponent straight y equals straight x plus straight x cubed over 3 plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative left parenthesis straight i right parenthesis
Put space space straight x equals 0 comma space straight y equals 1
tan to the power of negative 1 end exponent straight y equals straight x plus straight x cubed over 3 plus straight c
straight c equals straight pi over 4
Put space straight c equals straight pi over 4 space in space equation space left parenthesis straight i right parenthesis
tan to the power of negative 1 end exponent straight y equals straight x plus straight x cubed over 3 plus straight pi over 4 end style

Question 45(vii)

Solve the following initial value problem

begin mathsize 12px style xy dy over dx equals open parentheses straight x plus 2 close parentheses open parentheses straight y plus 2 close parentheses comma space straight y open parentheses 1 close parentheses space equals space minus 1 end style

Solution 45(vii)

begin mathsize 12px style xy dy over dx equals open parentheses straight x plus 2 close parentheses open parentheses straight y plus 2 close parentheses comma space straight y open parentheses 1 close parentheses equals negative 1
fraction numerator ydy over denominator open parentheses straight y plus 2 close parentheses end fraction equals fraction numerator open parentheses straight x plus 2 close parentheses over denominator straight x end fraction dx
integral open parentheses 1 minus fraction numerator 2 over denominator straight y plus 2 end fraction close parentheses dy equals integral open parentheses 1 plus 2 over straight x close parentheses dx
straight y minus straight x minus 2 log left parenthesis straight y plus 2 right parenthesis minus 2 logx equals straight c
Put space straight x equals 1 comma space straight y equals negative 1
minus 1 minus 1 minus 2 log left parenthesis negative 1 plus 2 right parenthesis minus 2 log 1 equals straight c
rightwards double arrow negative 2 equals straight c
Thus comma space we space have
straight y minus straight x minus 2 log left parenthesis straight y plus 2 right parenthesis minus 2 logx equals negative 2
end style

Question 45(viii)

Solution 45(viii)

Question 45(ix)

Solution 45(ix)

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Find the particular solution of ebegin mathsize 12px style dy over dx end style= x + 1, given that y = 3 when x = 0.Solution 49

begin mathsize 12px style dy over straight e to the power of dx equals straight x plus 1
dy over dx equals log left parenthesis straight x plus 1 right parenthesis comma space straight y equals 3 space at space straight x equals 0
integral dy equals integral log left parenthesis straight x plus 1 right parenthesis dx
straight y equals log open vertical bar straight x plus 1 close vertical bar straight x integral 1 cross times dx minus integral open parentheses fraction numerator 1 over denominator straight x plus 1 end fraction cross times integral 1 dx close parentheses dx plus straight c
Using space in space tegration space by space parts
straight y equals straight x space log open vertical bar straight x space plus 1 close vertical bar minus integral fraction numerator straight x over denominator straight x plus 1 end fraction dx plus straight c
straight y equals straight x space log open vertical bar straight x plus 1 close vertical bar minus open parentheses integral open parentheses 1 minus fraction numerator 1 over denominator straight x plus 1 end fraction close parentheses dx close parentheses plus straight c
equals space straight x space log space open vertical bar straight x plus 1 close vertical bar minus open parentheses straight x minus log open vertical bar straight x plus 1 close vertical bar close parentheses plus straight c
straight y equals xlog open vertical bar straight x plus 1 close vertical bar minus straight x plus log open vertical bar straight x plus 1 close vertical bar plus straight c
straight y equals open parentheses straight x plus 1 close parentheses log open vertical bar straight x plus 1 close vertical bar minus straight x plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative left parenthesis straight i right parenthesis
Put space straight y space equals 3 space and space straight x space equals 0
3 equals 0 minus 0 plus straight c
straight c equals 3
Put space straight c space equals space 3 space in space equation space left parenthesis straight i right parenthesis comma
straight y equals left parenthesis straight x plus 1 right parenthesis log open vertical bar straight x plus 1 close vertical bar minus straight x plus 3 end style

Question 50

Solution 50

begin mathsize 12px style cos space ydy space plus space cos space straight x space sin space ydx equals 0
cos space ydy equals negative cos space straight x space sin space ydx
fraction numerator cos space straight y over denominator sin space straight y end fraction dy equals negative cos space xdx
integral cot space ydy equals negative integral cos space xdx
log open vertical bar sin space straight y close vertical bar equals negative sin space straight x space plus straight c space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative left parenthesis straight i right parenthesis
put space straight y space equals space straight pi over 2 and space straight x space equals straight pi over 2
log space open vertical bar sin straight pi over 2 close vertical bar equals negative sin straight pi over 2 plus straight c
0 equals negative 1 plus straight c
straight c equals 1
Put space straight c space equals 1 space in space equation space left parenthesis 1 right parenthesis comma
log open vertical bar sin space straight y close vertical bar equals 1 minus sin space straight x
log open vertical bar sin space straight y close vertical bar plus sin space straight x equals 1 end style

Question 51

Solution 51

begin mathsize 12px style dy over dx equals negative 4 xy squared comma space straight y equals 1 space when space straight x space equals 0
integral dy over straight y squared equals negative 4 integral xdx
minus 1 over straight y equals negative 4 straight x squared over 2 plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus negative negative negative left parenthesis straight i right parenthesis
Put space straight y space equals 1 space and space straight x space equals 0
minus 1 equals 0 plus straight c
straight c equals negative 1
Put space straight c equals negative 1 space in space equation space left parenthesis ii right parenthesis comma
minus 1 over straight y equals negative 2 straight x squared minus 1
1 over straight y equals 2 straight x squared plus 1
straight y equals fraction numerator 1 over denominator 2 straight x squared plus 1 end fraction end style

Question 52

Find the equation of a curve passing through the point (0,0) and whose differential equation is begin mathsize 12px style dy over dx equals straight e to the power of straight x space sin space straight x. end styleSolution 52

Question 53

Solution 53

Question 54

The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after after t seconds.Solution 54

begin mathsize 12px style Let space the space rate space of space change space of space volume space of space the space balloon space be space straight k left parenthesis where space straight k space is space straight a space constant right parenthesis
rightwards double arrow dv over dt equals straight k
rightwards double arrow straight d over dt open parentheses 4 over 3 πr cubed close parentheses equals straight k space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets Volume space of space sphere equals 4 over 3 πr cubed close square brackets
rightwards double arrow 4 over 3 straight pi times 3 straight r squared times dr over dt equals straight k
rightwards double arrow 4 πr squared dr equals straight k space dt
Intrgrating space both space sides comma space we space get colon
4 straight pi integral straight r squared dr equals straight k integral dt
rightwards double arrow 4 straight pi times straight r squared over 3 equals kt plus straight c
rightwards double arrow 4 πr cubed equals 3 open parentheses kt plus straight c close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space........ left parenthesis straight i right parenthesis
Now comma space at space straight t space equals space 0 comma space straight r space equals space 3 colon
4 straight pi cross times 3 cubed equals 3 left parenthesis straight k cross times 0 plus straight c right parenthesis
108 straight pi equals 3 straight c
straight c equals 36 straight pi end style
begin mathsize 12px style At space straight t space equals 3 comma space straight r equals 6
4 straight pi cross times 6 cubed equals 3 left parenthesis straight k cross times 3 plus straight c right parenthesis
864 straight pi equals 3 left parenthesis 3 straight k plus 36 straight pi right parenthesis
3 straight k equals negative 288 straight pi minus 36 straight pi equals 252 straight pi
straight k equals 84 straight pi
Substituting space the space values space of space straight k space and space straight C space in space equation space left parenthesis 1 right parenthesis comma space we space get colon
4 πr cubed equals 3 open square brackets 84 πt plus 36 straight pi close square brackets
rightwards double arrow 4 πr cubed equals 4 straight pi left parenthesis 63 straight t plus 27 right parenthesis
rightwards double arrow straight r cubed equals 63 straight t plus 27
rightwards double arrow straight r equals left parenthesis 63 straight t plus 27 right parenthesis to the power of 1 third end exponent space Thus comma space the space radius space of space the space balloon space after space straight t space seconds space is space left parenthesis 63 straight t plus 27 right parenthesis to the power of 1 third end exponent. end style

Question 55

in a bank,principal increases continuously at the rate of r% per  year. Find The value of r if Rs 100 doubles itself in 10 years (loge 2 = 0.6931).Solution 55

Let p, t and represent the principal, time, and rate of interest respectively.

It is given that the principal increases continuously at the rate of r% per year.

begin mathsize 12px style rightwards double arrow dp over dt equals open parentheses straight r over 100 close parentheses straight p
rightwards double arrow dp over straight p equals open parentheses straight r over 100 close parentheses dt end style

Integrating both side, we get:

begin mathsize 12px style integral dp over straight p equals straight r over 100 integral dt
rightwards double arrow log space straight p space equals space rt over 100 plus straight k
rightwards double arrow straight p equals straight e to the power of rt over 100 plus straight k end exponent space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space...... left parenthesis straight i right parenthesis
It space is space given space that space when space straight t space equals 0 comma space straight p space 100.
rightwards double arrow 100 space equals straight e to the power of straight k space space space space space space space space space space space space space space space space space space space space space..... left parenthesis 2 right parenthesis
Now comma space it space calligraphic l equals 10 comma space then space straight p space equals 2 space cross times 100 space equals 200.
200 equals straight e to the power of calligraphic l over 10 plus straight k end exponent
rightwards double arrow 200 space equals straight e to the power of calligraphic l over 10 end exponent. straight e to the power of straight k
rightwards double arrow 200 space equals straight e to the power of calligraphic l over 10 end exponent.100 space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis From left parenthesis 2 right parenthesis right parenthesis
rightwards double arrow straight e to the power of calligraphic l over 10 end exponent equals 2
rightwards double arrow straight r over 10 log subscript straight e 2
rightwards double arrow straight r over 10 equals 0.6931
rightwards double arrow straight r space equals 6.931
Hence comma space the space value space of space straight r space is space 6.93 percent sign end style

Question 56

Solution 56

Question 57

Solution 57

..Question 58

Solution 58

Chapter 22 – Differential Equations Exercise Ex. 22.8

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

2 v minus v fraction numerator d v over denominator d x end fraction equals fraction numerator d v over denominator d x end fraction
rightwards double arrow 2 v equals v fraction numerator d v over denominator d x end fraction plus fraction numerator d v over denominator d x end fraction
rightwards double arrow 2 v equals open parentheses v plus 1 close parentheses fraction numerator d v over denominator d x end fraction
rightwards double arrow fraction numerator open parentheses v plus 1 close parentheses over denominator v end fraction d v equals 2 d x

Question 10

Solution 10

Question 11

Solve the following differential equation.

Solution 11

Chapter 22 – Differential Equations Exercise Ex. 22.9

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solve the following differential equation:

begin mathsize 12px style straight x dy over dx equals straight x plus straight y end style

Solution 4

Question 5

Solve the following differential equation:

begin mathsize 12px style open parentheses straight x squared minus straight y squared close parentheses dx minus 2 xydy space equals 0 end style

Solution 5

Question 6

Solve the following initial value problem

begin mathsize 12px style dy over dx equals fraction numerator straight x plus straight y over denominator straight x minus straight y end fraction end style

Solution 6

begin mathsize 12px style dy over dx equals fraction numerator straight x plus straight y over denominator straight x minus straight y end fraction
Here space it space is space straight a space homogeneous space equation
put space straight y equals vx
And
dy over dx equals straight v plus straight x dv over dx
So comma
straight v plus straight x dv over dx equals fraction numerator 1 plus straight v over denominator 1 minus straight v end fraction
straight x dv over dx equals fraction numerator 1 plus straight v over denominator 1 minus straight v end fraction minus straight v
straight x dv over dx equals fraction numerator 1 plus straight v squared over denominator 1 minus straight v end fraction
fraction numerator 1 minus straight v over denominator 1 plus straight v squared end fraction dv equals dx over straight x
integral fraction numerator 1 minus straight v over denominator 1 plus straight v squared end fraction dv equals integral dx over straight x
integral fraction numerator 1 over denominator 1 plus straight v squared end fraction dv minus 1 half integral fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv equals integral dx over straight x
tan to the power of negative 1 end exponent straight v minus 1 half log left parenthesis 1 plus straight v squared right parenthesis equals log space straight x space plus straight c
tan to the power of negative 1 end exponent straight y over straight x equals 1 half log open parentheses straight x squared plus straight y squared close parentheses plus straight c end style

Question 7

Solution 7

Question 8

S o l v e space t h e space d i f f e r e n t i a l space e q u a t i o n space x squared fraction numerator d y over denominator d x end fraction equals x squared minus 2 y squared plus x y

Solution 8

C o n s i d e r space t h e space g i v e n space d i f f e r e n t i a l space e q u a t i o n
space x squared fraction numerator d y over denominator d x end fraction equals x squared minus 2 y squared plus x y
rightwards double arrow fraction numerator d y over denominator d x end fraction equals fraction numerator x squared minus 2 y squared plus x y over denominator x squared end fraction
T h i s space i s space a space h o m o g e n e o u s space d i f f e r e n t i a l space e q u a t i o n.
S u b s t i t u t i n g space y equals v x space a n d space fraction numerator d y over denominator d x end fraction equals v plus x fraction numerator d v over denominator d x end fraction comma space w e space h a v e
v plus x fraction numerator d v over denominator d x end fraction equals fraction numerator x squared minus 2 v squared cross times x squared plus x cross times v cross times x over denominator x squared end fraction
rightwards double arrow v plus x fraction numerator d v over denominator d x end fraction equals 1 minus 2 v squared plus v
rightwards double arrow x fraction numerator d v over denominator d x end fraction equals 1 minus 2 v squared
rightwards double arrow fraction numerator d v over denominator 1 minus 2 v squared end fraction equals fraction numerator d x over denominator x end fraction
rightwards double arrow fraction numerator d v over denominator v squared minus begin display style 1 half end style end fraction equals minus 2 fraction numerator d x over denominator x end fraction
rightwards double arrow integral fraction numerator d v over denominator open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses squared minus v squared end fraction equals 2 integral fraction numerator d x over denominator x end fraction
rightwards double arrow integral fraction numerator d v over denominator open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses squared minus v squared end fraction equals 2 integral fraction numerator d x over denominator x end fraction
rightwards double arrow fraction numerator square root of 2 over denominator 2 end fraction log open parentheses fraction numerator fraction numerator 1 over denominator square root of 2 end fraction plus v over denominator fraction numerator 1 over denominator square root of 2 end fraction minus v end fraction close parentheses equals 2 log x plus log C
rightwards double arrow fraction numerator 1 over denominator square root of 2 end fraction log open parentheses fraction numerator fraction numerator 1 over denominator square root of 2 end fraction plus begin display style y over x end style over denominator fraction numerator 1 over denominator square root of 2 end fraction minus y over x end fraction close parentheses equals 2 log x plus log C
rightwards double arrow fraction numerator 1 over denominator square root of 2 end fraction log open parentheses fraction numerator x plus y square root of 2 over denominator x minus y square root of 2 end fraction close parentheses equals 2 log x plus log C
rightwards double arrow fraction numerator 1 over denominator square root of 2 end fraction log open parentheses fraction numerator x plus y square root of 2 over denominator x minus y square root of 2 end fraction close parentheses equals log x squared plus log C
rightwards double arrow log open parentheses fraction numerator x plus y square root of 2 over denominator x minus y square root of 2 end fraction close parentheses to the power of fraction numerator 1 over denominator square root of 2 end fraction end exponent equals log C x squared
rightwards double arrow open parentheses fraction numerator x plus y square root of 2 over denominator x minus y square root of 2 end fraction close parentheses to the power of fraction numerator 1 over denominator square root of 2 end fraction end exponent equals C x squared
rightwards double arrow open parentheses fraction numerator x plus y square root of 2 over denominator x minus y square root of 2 end fraction close parentheses equals open parentheses C x squared close parentheses to the power of square root of 2 end exponent

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solve the following initial value problem

begin mathsize 12px style dy over dx equals straight y over straight x plus sin open parentheses straight y over straight x close parentheses end style

Solution 19

begin mathsize 12px style dy over dx equals straight y over straight x plus sin open parentheses straight y over straight x close parentheses
Here space it space is space straight a space homogeneous space equation
Put space space space space space space space space space straight y space equals space vx
And
dy over dx equals straight v plus straight x dv over dx
So comma
straight v plus straight x dv over dx equals straight v plus sin space straight v
straight x dv over dx equals sin space straight v
cosecvdv space equals dx over straight x
integral cosecvdv equals integral dx over straight x
log space tan space straight v over 2 equals log space straight x plus log space straight c
tan straight v over 2 equals Cx
tan fraction numerator straight y over denominator 2 straight x end fraction equals Cx end style

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solve the following initial value poblem

begin mathsize 12px style open parentheses 1 plus straight e to the power of straight x over straight y end exponent close parentheses dx plus straight e to the power of straight x over straight y end exponent open parentheses 1 minus straight x over straight y close parentheses dy equals 0 end style

Solution 25

begin mathsize 12px style open parentheses 1 plus straight e to the power of straight x over straight y end exponent close parentheses dx plus straight e to the power of straight x over straight y end exponent open parentheses 1 minus straight x over straight y close parentheses dy equals 0
Here space it space is space straight a space homogeneous space equation
Put space space space space space space space space space space space space space space straight x equals vy
And
dx over dy equals straight v plus straight y dv over dy
So comma
dx over dy equals negative fraction numerator straight e to the power of begin display style straight x over straight y end style end exponent open parentheses 1 minus begin display style straight x over straight y end style close parentheses over denominator open parentheses 1 plus straight e to the power of begin display style straight x over straight y end style end exponent close parentheses end fraction
straight v plus straight y dv over dy equals negative fraction numerator straight e to the power of begin display style vy over straight y end style end exponent open parentheses 1 minus begin display style vy over straight y end style close parentheses over denominator blank end fraction
equals negative straight e to the power of straight v fraction numerator left parenthesis 1 minus straight v right parenthesis over denominator left parenthesis 1 plus straight e to the power of straight v right parenthesis end fraction
straight y dv over dy equals negative fraction numerator straight e to the power of straight v left parenthesis 1 minus straight v right parenthesis over denominator open parentheses 1 plus straight e to the power of straight v close parentheses end fraction
equals fraction numerator negative straight e to the power of straight v left parenthesis 1 minus straight v right parenthesis minus straight v left parenthesis 1 plus straight e to the power of straight v right parenthesis over denominator left parenthesis 1 plus straight e to the power of straight v right parenthesis end fraction
fraction numerator left parenthesis 1 plus straight e to the power of straight v right parenthesis over denominator negative straight e to the power of straight v open parentheses 1 minus straight v close parentheses minus straight v open parentheses 1 plus straight e to the power of straight v close parentheses end fraction dv equals dy over straight y
straight x plus ye to the power of straight x divided by straight y end exponent equals straight c end style

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solve the following differential equation:

begin mathsize 12px style ydx plus open curly brackets xlog space open parentheses straight y over straight x close parentheses dy minus 2 xdy equals 0 close curly brackets end style

Solution 35

Question 36(i)

Solution 36(i)

Question 36(ii)

Solution 36(ii)

Question 36(iii)

Solve the following initial value problem

begin mathsize 12px style dy over dx minus straight y over straight x plus cosec straight y over straight x equals 0 comma space straight y left parenthesis 1 right parenthesis equals 0 end style

Solution 36(iii)

begin mathsize 12px style dy over dx minus straight y over straight x plus cosec straight y over straight x equals 0 comma space straight y left parenthesis 1 right parenthesis equals 0
Here space it space is space straight a space homogeneous space equation
Put space space space space space space space space space straight y space equals vx
And
dy over dx equals straight v plus straight x dv over dx
So comma
straight v plus straight x dv over dx equals vx over straight x minus cosec vx over straight x
straight x dv over dx equals straight v minus cosecv minus straight v
equals negative cosecv
dv over cosecv equals negative dx over straight x
sin space vdv equals negative dx over straight x
minus cos space straight v equals negative log open vertical bar straight x close vertical bar plus straight c
minus cos straight y over straight x equals negative log open vertical bar straight x close vertical bar plus straight c
Now space putting space straight y equals 0 comma space straight x equals 1 comma space we space have
straight c equals negative 1
Now
minus cos straight y over straight x plus 1 equals negative log open vertical bar straight x close vertical bar
log open vertical bar straight x close vertical bar equals cos straight y over straight x minus 1 end style

Question 36(iv)

Solution 36(iv)

Question 36(v)

Solution 36(v)

Question 36(vi)

Solution 36(vi)

Question 36(vii)

Solution 36(vii)

Question 36(viii)

Solution 36(viii)

Question 36(ix)

Solve the following initial value problem

Error converting from MathML to accessible text.

Solution 36(ix)

Error converting from MathML to accessible text.

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Chapter 22 – Differential Equations Exercise Ex. 22.10

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solve the following differential equation:

begin mathsize 12px style open parentheses sin space straight x close parentheses dy over dx plus ycos space straight x equals 2 space sin squared space straight x space cos space straight x end style

Solution 30

begin mathsize 12px style Here comma space open parentheses sin space straight x close parentheses dy over dx plus space straight y space cos space straight x space equals 2 space sin squared straight x space cos space straight x
dy over dx plus straight y space cot space straight x space equals space 2 space sin space straight x space cos space straight x
It space is space straight a space linear space differential space equation. space Comparing space it space with comma
dy over dx plus py equals straight Q
straight p equals cot space straight x comma space straight Q space equals 2 sin space straight x space cos space straight x
straight I. straight F. space space equals straight e to the power of integral pdx end exponent
space space space space space space space space space equals straight e to the power of integral cotxdx end exponent
space space space space space space space space space equals straight e to the power of log space sin space straight x end exponent
space space space space space space space space space equals space sin space straight x
solution space of space the space equation space is space given space by comma
straight y space cross times open parentheses straight I. straight F close parentheses equals integral straight Q cross times left parenthesis straight I. straight F right parenthesis dx plus straight c
straight y left parenthesis sin space straight x right parenthesis equals integral 2 sin space straight x space cos space straight x left parenthesis sin space straight x right parenthesis space dx space plus straight c
ysin space straight x equals space left parenthesis 2 divided by 3 right parenthesis sin to the power of logical and 3 straight x plus straight C
end style

Question 31

Solve the following differential equation:

begin mathsize 12px style open parentheses straight x squared minus 1 close parentheses dy over dx plus 2 left parenthesis straight x plus 2 right parenthesis straight y equals 2 left parenthesis straight x plus 1 right parenthesis end style

Solution 31

begin mathsize 12px style Here comma space open parentheses straight x squared minus 1 close parentheses dy over dx plus 2 open parentheses straight x plus 2 close parentheses straight y equals 2 left parenthesis straight x plus 1 right parenthesis
dy over dx plus fraction numerator 2 left parenthesis straight x plus 2 right parenthesis over denominator open parentheses straight x squared minus 1 close parentheses end fraction straight y equals fraction numerator 2 left parenthesis straight x plus 1 right parenthesis over denominator open parentheses straight x squared minus 1 close parentheses end fraction
It space is space straight a space linear space differential space equation. space Comparing space it space with comma
dy over dx plus py equals straight Q
straight P equals fraction numerator 2 left parenthesis straight x plus 2 right parenthesis over denominator straight x squared minus 1 end fraction comma straight Q fraction numerator 2 left parenthesis straight x plus 1 right parenthesis over denominator left parenthesis straight x squared minus 1 right parenthesis end fraction
straight I. straight F. space equals straight e to the power of integral pdx end exponent
space space space space space space space space space space equals straight e to the power of 2 integral fraction numerator left parenthesis straight x plus 2 right parenthesis over denominator open parentheses straight x squared minus 1 close parentheses end fraction dx end exponent
space space space space space space space space space space equals straight e to the power of 2 integral fraction numerator 2 straight x over denominator straight x squared minus 1 end fraction dx plus 4 integral fraction numerator 1 over denominator straight x squared minus 1 end fraction dx end exponent
space space space space space space space space space space equals straight e to the power of log open vertical bar straight x squared minus 1 close vertical bar plus 4 straight x 1 half log open vertical bar fraction numerator straight x minus 1 over denominator straight x plus 1 end fraction close vertical bar end exponent
space space space space space space space space space space equals straight e to the power of log open vertical bar straight x squared minus 1 close vertical bar plus log open vertical bar fraction numerator straight x minus 1 over denominator straight x plus 1 end fraction close vertical bar squared end exponent
space space space space space space space space space space equals straight e to the power of log fraction numerator open parentheses straight x squared minus 1 close parentheses left parenthesis straight x minus 1 right parenthesis squared over denominator left parenthesis straight x plus 1 right parenthesis squared end fraction end exponent
straight I. straight F space space space space space space equals fraction numerator open parentheses straight x plus 1 close parentheses left parenthesis straight x minus 1 right parenthesis left parenthesis straight x minus 1 right parenthesis squared over denominator left parenthesis straight x plus 1 right parenthesis squared end fraction
space space space space space space space space space space space space equals fraction numerator left parenthesis straight x minus 1 right parenthesis cubed over denominator left parenthesis straight x plus 1 right parenthesis end fraction
Solution space of space the space equation space is space given space by comma
4 cross times left parenthesis straight I. straight F right parenthesis equals integral straight Q cross times left parenthesis straight I. straight F right parenthesis dx plus straight c
fraction numerator straight y left parenthesis straight x minus 1 right parenthesis cubed over denominator left parenthesis straight x plus 1 right parenthesis end fraction equals 2 straight x squared over 2 minus 6 straight x minus 8 space log open vertical bar straight x plus 1 close vertical bar plus straight c
fraction numerator straight y left parenthesis straight x minus 1 right parenthesis cubed over denominator left parenthesis straight x plus 1 right parenthesis end fraction equals straight x squared minus 6 straight x plus 8 log open vertical bar straight x plus 1 close vertical bar plus straight c
straight y equals fraction numerator straight x plus 1 over denominator left parenthesis straight x minus 1 right parenthesis cubed end fraction open square brackets straight x squared minus 6 straight x minus 8 space log open vertical bar straight x plus 1 close vertical bar plus straight c close square brackets end style

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36(i)

Solution 36(i)

begin mathsize 12px style Here comma space dy over dx plus 3 straight y equals straight e to the power of mx
It space is space straight a space linear space differential space equation. space Comparing space it space with comma
dy over dx plus py equals straight Q
straight P equals 3 comma space straight Q equals straight e to the power of mx
straight I. straight F. equals straight e to the power of integral pdx end exponent
equals straight e to the power of integral 3 dx end exponent
equals straight e to the power of 3 straight x end exponent
Solution space of space the space equation space is space given space by comma
straight y cross times open parentheses straight I. straight F close parentheses equals integral space straight Q space cross times space open parentheses straight I. straight F. close parentheses dx space plus space straight c
straight y open parentheses straight e to the power of 3 straight x end exponent close parentheses space equals space integral straight e to the power of mx space straight e to the power of 3 straight x end exponent space dx space plus space straight c
equals space integral space straight e to the power of open parentheses straight m plus 3 close parentheses straight x end exponent space dx space plus space straight c
straight y open parentheses straight e to the power of 3 straight x end exponent close parentheses equals fraction numerator straight e to the power of open parentheses straight m plus 3 close parentheses straight x end exponent space dx over denominator open parentheses straight m plus 3 close parentheses end fraction plus straight c end style

Question 36(ii)

Solution 36(ii)

Question 36(iii)

Solution 36(iii)

Question 36(iv)

Solution 36(iv)

Question 36(v)

Solution 36(v)

Question 36(vi)

Solution 36(vi)

Question 36(vii)

Solution 36(vii)

Question 36(viii)

Solution 36(viii)

Question 36(ix)

Solution 36(ix)

Question 36(x)

Solution 36(x)

begin mathsize 12px style Here comma space e to the power of negative y space end exponent s e c squared space y d y equals d x space plus space x d y
e to the power of negative y space end exponent s e c squared space y equals fraction numerator d x over denominator d y end fraction plus x
fraction numerator d x over denominator d y end fraction plus x equals e to the power of negative y end exponent space s e c squared space y
It space is space straight a space linear space differential space equation. space Comparing space it space with comma
fraction numerator d x over denominator d y end fraction plus p x equals Q
P space equals space 1 comma space Q equals e to the power of negative y end exponent space s e c squared y
I. F. space equals space e to the power of integral p d y end exponent
space space space space space space space space equals space e to the power of integral d y end exponent
space space space space space space space space equals space e to the power of y
Solution space of space the space equation space is space given space by comma
x space cross times space open parentheses I. F close parentheses equals integral space Q space cross times space open parentheses I. F. close parentheses d y equals c
x open parentheses e to the power of y close parentheses equals integral e to the power of negative y end exponent space s e c squared y open parentheses e to the power of y close parentheses d y plus c
x e to the power of y equals t a n space y space plus c
x equals open parentheses t a n space y space plus space c close parentheses e to the power of negative y end exponent end style

Question 36(xi)

Solution 36(xi)

Question 36(xii)

Solution 36(xii)

Question 37(i)

Solution 37(i)

Question 37(ii)

Solution 37(ii)

Question 37(iii)

Solution 37(iii)

Question 37(iv)

Solution 37(iv)

Question 37(v)

Solve the following initial value problem:

begin mathsize 12px style open parentheses 1 plus straight y squared close parentheses dx space plus space open parentheses straight x minus straight e minus tan to the power of negative 1 end exponent straight t close parentheses dy equals 0 comma space straight y open parentheses 0 close parentheses equals 0 end style

Solution 37(v)

Question 37(vi)

Solution 37(vi)

Question 37(vii)

Solution 37(vii)

Question 37(viii)

Solve the following initial value problem

begin mathsize 12px style dy over dx plus ycotx equals 4 straight x space cosecx comma space straight y open parentheses straight pi over 2 close parentheses equals 0 end style

Solution 37(viii)

begin mathsize 12px style fraction numerator d y over denominator d x end fraction plus y space c o t space x space equals space 4 x space cos e c space x comma space y open parentheses straight pi over 2 close parentheses equals 0
It space is space straight a space linear space differential space equation. space Comparing space it space with comma
fraction numerator d y over denominator d x end fraction plus p y equals Q
p equals c o t x comma space Q equals 4 x space cos e c x
I. F.
equals e to the power of integral p d x end exponent space
equals e to the power of integral c o t space x d x end exponent
equals e to the power of log space sin space x end exponent
equals sin space x
Solution space of space the space equation space is space given space by comma
y cross times open parentheses I. F close parentheses equals integral Q cross times open parentheses I. F close parentheses d x plus c
y open parentheses sin space x close parentheses space equals space integral 4 x space cos e c space x space cross times open parentheses sin space x close parentheses d x space plus space c
space space space space space space space space space space space space space space equals space integral 4 x d x space plus space c
y space sin space x space equals 4 x squared over 2 plus c
space space space space space space space space space space space space space equals 2 x squared space plus thin space c
Put space y space equals space 0 comma space x equals straight pi over 2
space space space space space space space 0 equals straight pi squared over 2 plus c
space space space space space space space c equals negative straight pi squared over 2
Now comma space
y space sin space x space equals space 2 x squared minus straight pi squared over 2 end style

Question 37(ix)

Solution 37(ix)

Question 37(x)

Solution 37(x)

Question 37(xi)

Solution 37(xi)

Question 37(xii)

dy = cos x (2 – y cosec x) dxSolution 37(xii)

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solve the differential equation

begin mathsize 12px style open parentheses straight y plus 3 straight x squared close parentheses dx over dy equals straight x end style

Solution 40

begin mathsize 12px style open parentheses y plus 3 x squared close parentheses fraction numerator d x over denominator d y end fraction equals x
fraction numerator d x over denominator d y end fraction equals fraction numerator x over denominator y plus 3 x squared end fraction
fraction numerator d y over denominator d x end fraction equals fraction numerator y plus 3 x squared over denominator x end fraction
fraction numerator d y over denominator d x end fraction minus y over x equals 3 x
It space is space straight a space linear space differential space equation. space Comparing space it space with comma
fraction numerator d y over denominator d x end fraction plus p y equals Q
p equals 1 over x comma Q equals 3 x
I. F.
equals e to the power of integral p d z end exponent
equals e to the power of negative integral 1 over x d z end exponent
equals e to the power of negative l o g z end exponent
equals 1 over x
Solution space of space the space equation space is space given space by comma
y cross times open parentheses I. F close parentheses equals integral Q cross times open parentheses I. F close parentheses d x plus c
y open parentheses 1 over x close parentheses equals integral 3 x cross times open parentheses 1 over x close parentheses d x plus c
y over x equals 3 x plus c end style

Question 41

Solution 41

Chapter 22 – Differential Equations Exercise Ex. 22.11

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

I n space a space c u l t u r e comma space t h e space b a c t e r i a space c o u n t space i s space 100000. space T h e space n u m b e r space i s space i n c r e a s e d space b y space 10 %
i n space 2 space h o u r s. space I n space h o w space m a n y space h o u r s space w i l l space t h e space c o u n t space r e a c h space 200000 comma space i f space t h e space r a t e space o f space g r o w t h
o f space b a c t e r i a space i s space p r o p o r t i o n a l space t o space t h e space n u m b e r space p r e s e n t ?

Solution 4

L e t space C space b e space t h e space c o u n t space o f space b a c t e r i a space a t space a n y space t i m e space t.
I t space i s space g i v e n space t h a t
fraction numerator d C over denominator d t end fraction infinity C
rightwards double arrow fraction numerator d C over denominator d t end fraction equals lambda C comma space w h e r e space lambda space i s space a space c o n s tan t space o f space p r o p o r t i o n a l i t y
rightwards double arrow fraction numerator d C over denominator C end fraction equals lambda d t
rightwards double arrow integral fraction numerator d C over denominator C end fraction equals lambda integral d t
rightwards double arrow log C equals lambda t plus log K.... left parenthesis 1 right parenthesis
I n i t i a l l y comma space a t space t equals 0 comma space C equals 100000
T h u s comma space w e space h a v e comma
log 100000 equals lambda cross times 0 plus log K.... left parenthesis 2 right parenthesis
rightwards double arrow log 100000 equals log K.... left parenthesis 3 right parenthesis
A t space t equals 2 comma space C equals 100000 plus 100000 cross times 10 over 100 equals 110000
T h u s comma space f r o m space left parenthesis 1 right parenthesis comma space w e space h a v e comma
log 110000 equals lambda cross times 2 plus log K.... left parenthesis 4 right parenthesis
S u b t r a c t i n g space e q u a t i o n space left parenthesis 2 right parenthesis space f r o m space left parenthesis 4 right parenthesis comma space w e space h a v e comma
log 110000 minus log 100000 equals 2 lambda
rightwards double arrow log 11 cross times 10000 minus log 10 cross times 10000 equals 2 lambda
rightwards double arrow log fraction numerator 11 cross times 10000 over denominator 10 cross times 10000 end fraction equals 2 lambda
rightwards double arrow log 11 over 10 equals 2 lambda
rightwards double arrow lambda equals 1 half log 11 over 10.... left parenthesis 5 right parenthesis
W e space n e e d space t o space f i n d space t h e space t i m e space apostrophe t apostrophe space i n space w h i c h space t h e space c o u n t space r e a c h e s space 200000.
S u b s t i t u t i n g space t h e space v a l u e s space o f space lambda space a n d space K space f r o m space e q u a t i o n s space left parenthesis 3 right parenthesis space a n d space left parenthesis 5 right parenthesis space i n space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e
log 200000 equals 1 half log 11 over 10 t plus log 100000
rightwards double arrow 1 half log 11 over 10 t equals log 200000 minus log 100000
rightwards double arrow 1 half log 11 over 10 t equals log 200000 over 100000
rightwards double arrow 1 half log 11 over 10 t equals log 2
rightwards double arrow t equals fraction numerator 2 log 2 over denominator log 11 over 10 end fraction space h o u r s

Question 5

Solution 5

Question 6

Solution 6

Question 7

T h e space p o p u l a t i o n space o f space a space c i t y space i n c r e s e s space a t space a space r a t e space p r o p o r t i o n a l space t o space t h e space n u m b e r space o f space
i n h a b i tan t s space p r e s e n t space a t space a n y space t i m e space t. space I f space t h e space p o p u l a t i o n space o f space t h e space c i t y space w a s space 200000 space i n
1990 space a n d space 250000 space i n space 2000 comma space w h a t space w i l l space b e space t h e space p o p u l a t i o n space i n space 2010 ?

Solution 7


L e t space P space b e space t h e space p o p u l a t i o n space o f space t h e space c i t y space a t space a n y space t i m e space t.
I t space i s space g i v e n space t h a t
fraction numerator d P over denominator d t end fraction infinity P
rightwards double arrow fraction numerator d P over denominator d t end fraction equals lambda P comma space w h e r e space lambda space i s space a space c o n s tan t space o f space p r o p o r t i o n a l i t y
rightwards double arrow fraction numerator d P over denominator P end fraction equals lambda d t
rightwards double arrow integral fraction numerator d P over denominator P end fraction equals lambda integral d t
rightwards double arrow log P equals lambda t plus log K.... left parenthesis 1 right parenthesis
I n i t i a l l y comma space a t space t equals 1990 comma space P equals 200000
T h u s comma space w e space h a v e comma
log 200000 equals lambda cross times 1990 plus log K.... left parenthesis 2 right parenthesis
A t space t equals 2000 comma space P equals 250000
T h u s comma space f r o m space left parenthesis 1 right parenthesis comma space w e space h a v e comma
log 250000 equals lambda cross times 2000 plus log K.... left parenthesis 3 right parenthesis
S u b t r a c t i n g space e q u a t i o n space left parenthesis 2 right parenthesis space f r o m space left parenthesis 3 right parenthesis comma space w e space h a v e comma
log 250000 minus log 200000 equals 10 lambda
rightwards double arrow log 4 over 5 equals 10 lambda
rightwards double arrow lambda equals 1 over 10 log 4 over 5.... left parenthesis 4 right parenthesis
S u b s t i t u t i n g space t h e space v a l u e space o f space lambda space f r o m space e q u a t i o n space left parenthesis 4 right parenthesis space i n space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e
log 200000 equals 1990 cross times 1 over 10 log 4 over 5 plus log K
rightwards double arrow log K equals log 200000 minus 199 cross times log 4 over 5.... left parenthesis 5 right parenthesis space space
S u b s t i t u t i n g space t h e space v a l u e space o f space lambda comma space log K space a n d space t equals 2010 space i n space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e
log P equals open curly brackets 1 over 10 log 4 over 5 close curly brackets 2010 plus log 200000 minus 199 cross times log 4 over 5
rightwards double arrow log P equals log open curly brackets 4 over 5 close curly brackets to the power of 201 plus log open parentheses 200000 cross times open parentheses 5 over 4 close parentheses to the power of 199 close parentheses
rightwards double arrow P equals open curly brackets 4 over 5 close curly brackets to the power of 201 cross times 200000 cross times open parentheses 5 over 4 close parentheses to the power of 199
rightwards double arrow P equals open parentheses 5 over 4 close parentheses squared cross times 200000 equals 25 over 16 cross times 200000 equals 312500

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Chapter 22 – Differential Equations Exercise Ex. 22RE

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

F i n d space t h e space d i f f e r e n t i a l space e q u a t i o n space c o r r e s p o n d i n g space t o space y equals a e to the power of 2 x end exponent plus b e to the power of minus 3 x end exponent plus c e to the power of x comma space w h e r e
a comma space b comma space c space a r e space a r b i t r a r y space c o n s tan t s.

Solution 15

C o n s i d e r space t h e space g i v e n space e q u a t i o n space y equals a e to the power of 2 x end exponent plus b e to the power of minus 3 x end exponent plus c e to the power of x
D i f f e r e n t i a t i n g space t h e space a b o v e space e q u a t i o n space w i t h space r e s p e c t space t o space x comma space w e space h a v e comma
fraction numerator d y over denominator d x end fraction equals 2 a e to the power of 2 x end exponent minus 3 b e to the power of minus 3 x end exponent plus c e to the power of x.... left parenthesis 1 right parenthesis
rightwards double arrow 7 fraction numerator d y over denominator d x end fraction equals 14 a e to the power of 2 x end exponent minus 21 b e to the power of minus 3 x end exponent plus 7 c e to the power of x.... left parenthesis 2 right parenthesis
D i f f e r e n t i a t i n g space e q u a t i o n space left parenthesis 1 right parenthesis space w i t h space r e s p e c t space t o space x comma space w e space h a v e comma
fraction numerator d squared y over denominator d x squared end fraction equals 4 a e to the power of 2 x end exponent plus 9 b e to the power of minus 3 x end exponent plus c e to the power of x.... left parenthesis 3 right parenthesis
A g a i n space d i f f e r e n t i a t i n g space t h e space a b o v e space e q u a t i o n space w i t h space r e s p e c t space t o space x comma space w e space h a v e comma
fraction numerator d cubed y over denominator d x cubed end fraction equals 8 a e to the power of 2 x end exponent minus 27 b e to the power of minus 3 x end exponent plus c e to the power of x.... left parenthesis 4 right parenthesis
N o w space c o n s i d e r space t h e space f o l l o w i n g space e x p r e s s i o n
fraction numerator d cubed y over denominator d x cubed end fraction minus 7 fraction numerator d y over denominator d x end fraction plus 6 y
equals 8 a e to the power of 2 x end exponent minus 27 b e to the power of minus 3 x end exponent plus c e to the power of x minus 14 a e to the power of 2 x end exponent plus 21 b e to the power of minus 3 x end exponent minus 7 c e to the power of x plus 6 open parentheses a e to the power of 2 x end exponent plus b e to the power of minus 3 x end exponent plus c e to the power of x close parentheses
equals 8 a e to the power of 2 x end exponent minus 27 b e to the power of minus 3 x end exponent plus c e to the power of x minus 14 a e to the power of 2 x end exponent plus 21 b e to the power of minus 3 x end exponent minus 7 c e to the power of x plus 6 a e to the power of 2 x end exponent plus 6 b e to the power of minus 3 x end exponent plus 6 c e to the power of x
equals 0
T h u s comma space t h e space r e q u i r e d space d i f f e r e n t i a l space e q u a t i o n space c o r r e s p o n d i n g space t o space
space y equals a e to the power of 2 x end exponent plus b e to the power of minus 3 x end exponent plus c e to the power of x space i s
fraction numerator d cubed y over denominator d x cubed end fraction minus 7 fraction numerator d y over denominator d x end fraction plus 6 y equals 0

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solve the following differential equation:

begin mathsize 12px style y space s e c squared space x space plus space open parentheses y plus 7 close parentheses space tan space x space fraction numerator d y over denominator d x end fraction equals 0 end style

Solution 46

Question 47

Solution 47

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Question 61

Solution 61

Question 62

Solution 62

Question 63

Solution 63

Question 64(i)

Solution 64(i)

Question 64(ii)

Solution 64(ii)

Question 64(iii)

Solution 64(iii)

Question 64(iv)

Solution 64(iv)

Question 64(v)

Solution 64(v)

Question 64(vi)

Solution 64(vi)

Question 65(i)

Solution 65(i)

Question 65(ii)

Solution 65(ii)

Question 65(iii)

Solution 65(iii)

Question 66(i)

Solution 66(i)

Question 66(ii)

Solution 66(ii)

Question 66(iii)

Solution 66(iii)

Question 66(iv)

Solution 66(iv)

Question 66(v)

Solution 66(v)

Question 66(vi)

Solution 66(vi)

Question 66(vii)

Solution 66(vii)

Question 66(viii)

Solution 66(viii)

Question 66(ix)

Solution 66(ix)

Question 66(x)

Solution 66(x)

Question 66(xi)

Solution 66(xi)

Question 66(xii)

Solution 66(xii)

Question 66(xiii)

Solution 66(xiii)

Question 66(xiv)

Solution 66(xiv)

Question 66(xv)

Solution 66(xv)

Question 67(i)

Solution 67(i)

Question 67(ii)

Solution 67(ii)

Question 67(iii)

Solution 67(iii)

Question 68

Solution 68

Question 69

Solution 69

Question 70

Solution 70

Question 71

Solution 71

Question 72

Solution 72

Question 73

Solution 73

Question 74

Solution 74

Question 75

Solution 75

Question 76

Solution 76

Question 77

Solution 77

Question 78

Solution 78

Question 79

Solution 79

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