RD SHARMA SOLUTION CHAPTER-19 Indefinite Integrals I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 19 Indefinite Integrals Ex. 19.1

Question 1

Integrate each of the following functions with respect to x:

(i) ∫ x4dx

(ii) ∫ x5/4dx

(iii) begin mathsize 12px style integral 1 over straight x to the power of 5 dx end style

(iv) begin mathsize 12px style integral 1 over x to the power of 3 divided by 2 end exponent d x end style

(v) ∫ 3xdx

(vi) begin mathsize 12px style integral fraction numerator 1 over denominator square root of x squared end root end fraction d x end style

(vii) ∫ 32log3x dx

(viii) ∫ logxx dx Solution 1

Question 2

Evaluate:

begin mathsize 12px style left parenthesis i right parenthesis space integral square root of fraction numerator 1 plus cos 2 x over denominator 2 end fraction end root d x
left parenthesis i i right parenthesis space integral square root of fraction numerator 1 minus cos 2 x over denominator 2 end fraction d x end root end style

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Chapter 19 Indefinite Integrals Ex. 19.2

Question 1

Evaluate begin mathsize 12px style integral open parentheses 3 cross times square root of x plus 4 square root of x plus 5 close parentheses d x end styleSolution 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

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

Evaluate :integral fraction numerator cos x over denominator 1 plus cos x end fraction d xSolution 42

Question 43

Evalute :

integral fraction numerator 1 minus cos x over denominator 1 plus cos x end fraction d x

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Chapter 19 Indefinite Integrals Ex. 19.3

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

Integrate begin mathsize 12px style integral sin x square root of 1 plus cos 2 x end root space d x end styleSolution 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

Evaluate the following integrals:

begin mathsize 12px style integral fraction numerator 1 over denominator cos squared x left parenthesis 1 minus tan x right parenthesis squared end fraction end style

Solution 19

Chapter 19 – Indefinite Integrals Exercise Ex. 19.4

Question 1

Solution 1

Question 2

Integratebegin mathsize 12px style integral fraction numerator x cubed over denominator x minus 2 end fraction d x end styleSolution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Chapter 19 Indefinite Integrals Ex. 19.5

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

Chapter 19 Indefinite Integral Ex. 19.6

Question 1

Solution 1

Question 2

Evalute the following integral :integral sin cubed left parenthesis 2 x plus 1 right parenthesis d xSolution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Integrate begin mathsize 12px style integral sin x square root of 1 minus cos 2 x end root space d x end styleSolution 8

Chapter 19 Indefinite Integrals Ex. 19.7

Question 1

Integrate ∫ sin4x cos7x dxSolution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Chapter 19 Indefinite Integrals Ex. 19.8

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

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

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 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Chapter 19 – Indefinite Integrals Exercise Ex. 19.9

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

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

begin mathsize 12px style Evaluate colon
integral fraction numerator 4 x plus 3 over denominator square root of 2 x squared plus 3 x plus 1 end root end fraction d x end style

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

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

begin mathsize 12px style Integrate space the space function
fraction numerator e to the power of m tan to the power of negative 1 end exponent x end exponent over denominator 1 plus x squared end fraction end style

Solution 54

L e t space tan to the power of minus 1 end exponent x equals t
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 f u n c t i o n space w i t h space r e s p e c t space t o comma space w comma space w e space h a v e comma
fraction numerator 1 over denominator 1 plus x squared end fraction d x equals d t
rightwards double arrow integral fraction numerator e to the power of m tan to the power of minus 1 end exponent x end exponent over denominator 1 plus x squared end fraction equals integral e to the power of m t end exponent cross times d t
rightwards double arrow integral fraction numerator e to the power of m tan to the power of minus 1 end exponent x end exponent over denominator 1 plus x squared end fraction equals e to the power of m t end exponent over m
R e 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 t space i n space t h e space a b o v e space s o l u t i o n comma space w e space h a v e comma
space space space space space space rightwards double arrow integral fraction numerator e to the power of m tan to the power of minus 1 end exponent x end exponent over denominator 1 plus x squared end fraction equals e to the power of m tan to the power of minus 1 end exponent x end exponent over m plus C

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

Solution 64

Question 65

Solution 65

Question 66

Solution 66

Question 67

Solution 67

Question 68

Solution 68

Question 69

Solution 69

Question 70

Solution 70

Question 71

Solution 71

Question 72

Solution 72

Chapter 19 Indefinite Integrals Exercise Ex. 19.10

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

Chapter 19 – Indefinite Integrals Exercise Ex. 19.11

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

begin mathsize 12px style Let space iota equals integral cos e c to the power of 4 3 x d x. space T h e n
iota equals integral cos e c squared 3 x space cos e c squared 3 d x
equals integral open parentheses 1 plus c o t squared 3 x close parentheses cos e c squared 3 x d x
equals integral open parentheses cos e c squared 3 x space plus space c o t squared 3 x space cos e c squared 3 x close parentheses d x
rightwards double arrow iota equals integral cos e c squared 3 x d x space plus space integral c o t squared 3 x space cos e c squared 3 x d x
Substituting space c o t space 3 x space equals t space a n d space cos e c t squared 3 x d x space equals negative d t space i n space 2 nd space integral comma space we space get
iota equals integral cos e c squared 3 x d x space minus space integral t squared fraction numerator d t over denominator 3 end fraction
equals fraction numerator negative 1 over denominator 3 end fraction c o t 3 x minus t cubed over 9 plus C equals fraction numerator negative 1 over denominator 3 end fraction c o t space 3 x minus 1 over 9 c o t cubed space 3 x plus c
therefore iota equals fraction numerator negative 1 over denominator 3 end fraction c o t space 3 x space minus space fraction numerator c o t cubed 3 x over denominator 9 end fraction plus c
end style

Question 9

Solution 9

begin mathsize 12px style L e t space iota equals integral c o t to the power of n x space cos e c squared x d x. n not equal to negative 1......... open parentheses i close parentheses
L e t space c o t space x equals t. space T h e n
d open parentheses c o t x close parentheses equals d t
rightwards double arrow negative cos e c squared x d x equals d t
rightwards double arrow cos e c squared x d x equals negative d t
P u t t i n g space c o t x space equals t space a n d space c os e c squared x d x equals negative d t space i n space e q u a t i o n space open parentheses i close parentheses comma space w e space g e t
iota equals integral t to the power of n cross times open parentheses negative d t close parentheses
equals negative fraction numerator t to the power of n plus 1 end exponent over denominator n plus 1 end fraction plus c
rightwards double arrow equals negative fraction numerator open parentheses c o t x close parentheses to the power of n plus 1 end exponent over denominator n plus 1 end fraction plus c end style

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Chapter 19 Indefinite Integrals Exercise Ex. 19.12

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Evaluate the following integral:

∫ sin5x cosx dxSolution 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

Chapter 19 Indefinite Integrals Exercise Ex. 19.14

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 19 Indefinite Integrals Exercise Ex. 19.15

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 19 – Indefinite Integrals Exercise Ex. 19.16

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

Chapter 19 Indefinite Integrals Exercise Ex. 19.17

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 19 Indefinite Integrals Exercise Ex. 19.18

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

Chapter 19 Indefinite Integrals Exercise Ex. 19.19

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

Evaluate the following integrals:

Solution 10

Question 11

Solution 11

Question 12

Evaluate the following integrals:

Solution 12

Chapter 19 Indefinite Integrals Exercise Ex. 19.20

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 19 – Indefinite Integrals Exercise Ex. 19.32

Question 3

Solution 3

Question 1

Solution 1

Question 2

Solution 2

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

Chapter 19 Indefinite Integrals Exercise Ex. 19.21

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

Integrate the function

begin mathsize 12px style fraction numerator x minus 1 over denominator square root of x squared plus 1 end root end fraction end style

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Evaluate begin mathsize 12px style integral square root of fraction numerator 1 minus x over denominator 1 plus x end fraction end root d x end styleSolution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Evaluate the following integral

integral fraction numerator 5 x plus 3 over denominator square root of x squared plus 4 x plus 10 end root end fraction d x

Solution 17

Question 18

Evaluate the following integral:

Solution 18

Chapter 19 Indefinite Integrals Exercise Ex. 19.22

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

Chapter 19 Indefinite Integrals Exercise Ex. 19.23

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 19 Indefinite Integrals Exercise Ex. 19.24

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

Chapter 19 Indefinite Integrals Exercise Ex. 19.25

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Find ∫xedxSolution 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

Evaluate the following integral:

Solution 27

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

begin mathsize 12px style Evaluate space the space integrals colon
integral sin to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction close parentheses dx end style

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

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

begin mathsize 12px style Evaluate space the space integrals colon
integral tan to the power of negative 1 end exponent square root of fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction end root dx end style

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Question 61

Solution 61

Chapter 19 Indefinite Integrals Exercise Ex. 19.26

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

begin mathsize 12px style Evaluagte space the space following space integrals colon
integral straight e to the power of straight x open parentheses fraction numerator straight x minus 1 over denominator 2 straight x squared end fraction close parentheses dx end style

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

Evaluate the following integral:

Solution 24

Chapter 19 Indefinite Integrals Exercise Ex. 19.27

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 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Chapter 19 Indefinite Integrals Exercise Ex. 19.28

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

Chapter 19 Indefinite Integrals Exercise Ex. 19.29

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

Evaluate the following integral:

Solution 11

Question 12

Evaluate the following integral:

Solution 12

Chapter 19 Indefinite Integrals Exercise Ex. 19.30

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 15

Solution 15

Question 16

Evaluate the following integral:

Solution 16

Question 17

Solution 17

Question 18

Evaluate the following integral:

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

Evaluate the following integral:

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

Solution 46

Question 47

Evaluate the following integral:

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Evaluate the following integral:

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 59

Solution 59

Question 60

Solution 60

Question 61

Solution 61

Question 62

Solution 62

Question 63

Solution 63

Question 64

Solution 64

Question 65

Solution 65

Chapter 19 – Indefinite Integrals Exercise Ex. 19.31

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

Evaluate the following integral:

Solution 11

Chapter 19 – Indefinite Integrals Exercise MCQ

Question 1

Mark the correct alternative in each of the following:

begin mathsize 12px style integral fraction numerator straight x over denominator 4 plus straight x to the power of 4 end fraction dx space is space equal space to

open parentheses straight a close parentheses space 1 fourth tan to the power of negative 1 end exponent straight x squared space plus space straight C
open parentheses straight b close parentheses space 1 fourth tan to the power of negative 1 end exponent open parentheses straight x squared over 2 close parentheses
open parentheses straight c close parentheses space 1 half tan to the power of negative 1 end exponent open parentheses straight x squared over 2 close parentheses
open parentheses straight d close parentheses space none space of space these end style

Solution 1

begin mathsize 12px style Correct space option colon thin space left parenthesis straight b right parenthesis
straight I equals integral fraction numerator straight x over denominator 4 plus straight x to the power of 4 end fraction dx space
Put space straight x squared equals straight t
rightwards double arrow 2 xdx equals dt
rightwards double arrow xdx equals dt over 2
straight I equals integral fraction numerator begin display style dt over 2 end style over denominator 4 plus straight t squared end fraction
straight I equals 1 half tan to the power of negative 1 end exponent open parentheses straight t over 2 close parentheses plus straight C
straight I equals 1 half tan to the power of negative 1 end exponent open parentheses straight x squared over 2 close parentheses plus straight C
end style

Question 2

Mark the correct alternative in each of the following:

begin mathsize 12px style integral fraction numerator 1 over denominator cos space straight x space plus square root of 3 space sin space straight x end fraction dx space is space equal space to

open parentheses straight a close parentheses space log space tan open parentheses straight pi over 3 plus straight x over 2 close parentheses plus straight C
open parentheses straight b close parentheses space log space tan open parentheses straight x over 2 minus straight pi over 3 close parentheses plus straight C
open parentheses straight c close parentheses space 1 half space log space tan open parentheses straight x over 2 plus straight pi over 3 close parentheses plus straight C
open parentheses straight d close parentheses space none space of space these end style

Solution 2

begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
straight I equals integral fraction numerator 1 over denominator cosx plus square root of 3 sinx end fraction dx
straight I equals 1 half integral fraction numerator 1 over denominator begin display style cosx over 2 end style plus begin display style fraction numerator square root of 3 over denominator 2 end fraction end style sinx end fraction dx
straight I equals 1 half integral fraction numerator 1 over denominator cos open parentheses straight x minus begin display style straight pi over 6 end style close parentheses end fraction dx
straight I equals 1 half integral sec open parentheses straight x minus straight pi over 6 close parentheses dx
straight I equals 1 half ln open vertical bar tan open parentheses straight x over 2 plus straight pi over 3 close parentheses close vertical bar plus straight C end style

Question 3

Mark the correct alternative in each of the following:

begin mathsize 12px style integral straight x space sec space straight x squared space dx space is space equal space to

open parentheses straight a close parentheses space 1 half log open parentheses sec space straight x squared space plus space tan space straight x squared close parentheses plus straight C
open parentheses straight b close parentheses space straight x squared over 2 log open parentheses sec space straight x squared space plus space tan space straight x squared close parentheses plus straight C
open parentheses straight c close parentheses space 2 space log open parentheses sec space straight x squared space plus space tan space straight x squared close parentheses plus straight C

open parentheses straight d close parentheses space none space of space these
end style

Solution 3

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
straight I equals integral xsecx squared dx
Put space straight x squared equals straight t rightwards double arrow straight x equals square root of straight t
rightwards double arrow 2 xdx equals dt
rightwards double arrow xdx equals dt over 2
straight I equals integral sect dt over 2
straight I equals 1 half log open vertical bar sect plus tant close vertical bar plus straight C
straight I equals 1 half log open vertical bar sec straight x squared plus tan straight x squared close vertical bar plus straight C
end style

Question 4

begin mathsize 12px style If space integral fraction numerator 1 over denominator 5 plus 4 space sin space straight x end fraction dx equals straight A space tan to the power of negative 1 end exponent open parentheses straight B space tan straight x over 2 plus 4 over 3 close parentheses plus straight C comma space then

open parentheses straight a close parentheses space straight A equals 2 over 3 comma space straight B equals 5 over 3
open parentheses straight b close parentheses space straight A equals 1 third comma space straight B equals 2 over 3
open parentheses straight c close parentheses space straight A equals negative 2 over 3 comma space straight B equals 5 over 3
open parentheses straight d close parentheses space straight A equals 1 third comma space straight B equals negative 5 over 3 end style

Solution 4

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
integral fraction numerator 1 over denominator 5 plus 4 sinx end fraction dx equals Atan to the power of negative 1 end exponent open parentheses Btan straight x over 2 plus 4 over 3 close parentheses plus straight C
Put space tan straight x over 2 equals straight t rightwards double arrow straight x equals 2 tan to the power of negative 1 end exponent straight t
rightwards double arrow dx equals fraction numerator 2 dt over denominator 1 plus straight t squared end fraction
rightwards double arrow sinx equals fraction numerator 2 tan begin display style straight x over 2 end style over denominator 1 plus tan squared straight x over 2 end fraction equals fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction
integral fraction numerator 1 over denominator 5 plus 4 sinx end fraction dx
equals integral fraction numerator begin display style fraction numerator 2 dt over denominator 1 plus straight t squared end fraction end style over denominator 5 plus 4 cross times fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction end fraction
equals integral fraction numerator 2 dt over denominator 5 straight t squared plus 8 straight t plus 5 end fraction
equals 2 over 5 integral fraction numerator dt over denominator straight t squared plus begin display style 8 over 5 end style straight t plus 1 end fraction
Using space completing space square space method
we space get
straight I equals 2 over 3 tan to the power of negative 1 end exponent open parentheses 5 over 3 tan straight x over 2 plus 4 over 3 close parentheses plus straight C
straight A equals 2 over 3 space and space straight B equals 5 over 3
end style

Question 5

begin mathsize 12px style integral straight x to the power of sin space straight x end exponent open parentheses fraction numerator sin space straight x over denominator straight x end fraction plus cos space straight x. log space straight x close parentheses dx space is space equal space to

open parentheses straight a close parentheses space straight x to the power of sin space straight x end exponent plus straight C

open parentheses straight b close parentheses space straight x to the power of sin space straight x end exponent space cos space straight x plus straight C
open parentheses straight c close parentheses space open parentheses straight x to the power of sin space straight x end exponent close parentheses squared over 2 plus straight C
open parentheses straight d close parentheses space none space of space these end style

Solution 5

begin mathsize 12px style Correct space option colon thin space left parenthesis straight a right parenthesis
straight I equals integral straight x to the power of sinx open parentheses sinx over straight x plus cosxlogx close parentheses dx
Put space straight x to the power of sinx equals straight t
taking space log space on space both space sides comma
logt equals sinxlogx
1 over straight t dt equals sinx over straight x plus cosxlogx
rightwards double arrow straight I equals integral straight t cross times dt over straight t
straight I equals straight t plus straight C
straight I equals straight x to the power of sinx plus straight C
end style

Question 6

begin mathsize 12px style Integration space of space fraction numerator 1 over denominator 1 plus open parentheses log subscript straight e straight x close parentheses squared end fraction with space respect space to space log subscript straight e straight x space is

open parentheses straight a close parentheses space fraction numerator tan to the power of negative 1 end exponent open parentheses log subscript straight e straight x close parentheses over denominator straight x end fraction plus straight C
open parentheses straight b close parentheses space tan to the power of negative 1 end exponent open parentheses log subscript straight e straight x close parentheses space plus space straight C
open parentheses straight c close parentheses space fraction numerator tan to the power of negative 1 end exponent straight x over denominator straight x end fraction plus straight C
open parentheses straight d close parentheses space none space of space these end style

Solution 6

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
integral fraction numerator 1 over denominator 1 plus open parentheses log subscript straight e straight x close parentheses squared end fraction straight d open parentheses log subscript straight e straight x close parentheses
Put space log subscript straight e straight x equals straight t
integral fraction numerator dt over denominator 1 plus straight t squared end fraction equals tan to the power of negative 1 end exponent straight t plus straight C equals tan to the power of negative 1 end exponent open parentheses space log subscript straight e straight x close parentheses plus straight C
end style

Question 7

begin mathsize 12px style If space integral fraction numerator cos space 8 straight x plus 1 over denominator tan space 2 straight x space minus space cot space 2 straight x end fraction dx equals straight a space cos space 8 straight x space plus space straight C comma space then space straight a equals

open parentheses straight a close parentheses space minus 1 over 16
open parentheses straight b close parentheses space 1 over 8
open parentheses straight c close parentheses space 1 over 16
open parentheses straight d close parentheses space minus 1 over 8 end style

Solution 7

begin mathsize 12px style Correct space option colon thin space left parenthesis straight c right parenthesis
integral fraction numerator cos 8 straight x plus 1 over denominator tan 2 straight x minus cot 2 straight x end fraction dx
equals integral fraction numerator 2 cos squared 4 straight x over denominator begin display style fraction numerator sin 2 straight x over denominator cos 2 straight x end fraction end style minus begin display style fraction numerator cos 2 straight x over denominator sin 2 straight x end fraction end style end fraction dx
equals integral fraction numerator 2 cos squared 4 straight x over denominator sin squared 2 straight x minus cos squared 2 straight x end fraction cross times sin 2 xcos 2 xdx
equals integral negative fraction numerator cos squared 4 xsin 4 straight x over denominator cos 4 straight x end fraction dx
equals fraction numerator negative 1 over denominator 2 end fraction integral sin 8 xdx
equals fraction numerator cos 8 straight x over denominator 16 end fraction plus straight C
straight a equals 1 over 16 end style

Question 8

begin mathsize 12px style If space integral fraction numerator sin to the power of 8 space straight x space minus space cos to the power of 8 space straight x over denominator 1 minus 2 space sin squared space straight x space cos squared space straight x end fraction dx equals straight a space sin space 2 straight x space plus space straight C comma space then space straight a equals

open parentheses straight a close parentheses space minus 1 divided by 2

open parentheses straight b close parentheses space 1 divided by 2

open parentheses straight c close parentheses space minus 1

open parentheses straight d close parentheses space 1 end style

Solution 8

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
integral fraction numerator sin to the power of 8 straight x minus cos to the power of 8 straight x over denominator 1 minus 2 sin squared xcos squared straight x end fraction dx
integral fraction numerator open parentheses sin to the power of 4 straight x plus cos to the power of 4 straight x close parentheses open parentheses sin to the power of 4 straight x minus cos to the power of 4 straight x close parentheses over denominator 1 minus 2 sin squared xcos squared straight x end fraction dx
integral fraction numerator open parentheses sin to the power of 4 straight x plus cos to the power of 4 straight x close parentheses open parentheses sin squared straight x plus cos squared straight x close parentheses open parentheses sin squared straight x minus cos squared straight x close parentheses over denominator open parentheses sin squared straight x plus cos squared straight x close parentheses squared minus 2 sin squared xcos squared straight x end fraction dx
integral fraction numerator open parentheses sin to the power of 4 straight x plus cos to the power of 4 straight x close parentheses open parentheses sin squared straight x minus cos squared straight x close parentheses over denominator open parentheses sin to the power of 4 straight x plus cos to the power of 4 straight x close parentheses end fraction dx
integral negative cos 2 xdx
fraction numerator negative sin 2 straight x over denominator 2 end fraction plus straight C
rightwards double arrow straight a equals fraction numerator negative 1 over denominator 2 end fraction end style

Question 9

begin mathsize 11px style integral open parentheses straight x minus 1 close parentheses straight e to the power of negative straight x end exponent space dx space is space equal space to

open parentheses straight a close parentheses space minus xe to the power of straight x plus straight C
open parentheses straight b close parentheses space xe to the power of straight x space plus space straight C
open parentheses straight c close parentheses space minus xe to the power of negative straight x end exponent plus straight C
open parentheses straight d close parentheses space xe to the power of negative straight x end exponent plus straight C
end style

Solution 9

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
integral open parentheses straight x minus 1 close parentheses straight e to the power of negative straight x end exponent dx
equals open parentheses straight x minus 1 close parentheses integral straight e to the power of negative straight x end exponent dx minus integral open parentheses open square brackets fraction numerator straight d open parentheses straight x minus 1 close parentheses over denominator dx end fraction close square brackets integral straight e to the power of negative straight x end exponent dx close parentheses dx
equals open parentheses straight x minus 1 close parentheses fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction minus integral fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction dx
equals negative open parentheses straight x minus 1 close parentheses straight e to the power of negative straight x end exponent plus fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction plus straight C
equals negative xe to the power of negative straight x end exponent plus straight e to the power of negative straight x end exponent minus straight e to the power of negative straight x end exponent plus straight C
equals negative xe to the power of negative straight x end exponent plus straight C
end style

Question 10

begin mathsize 11px style If space integral 2 to the power of 1 divided by straight x end exponent over straight x squared dx space equals space straight k space 2 to the power of 1 divided by straight x end exponent plus straight C comma space then space straight k space is space equal space to
open parentheses straight a close parentheses space minus fraction numerator 1 over denominator log subscript straight e 2 end fraction
open parentheses straight b close parentheses space minus log subscript straight e 2
open parentheses straight c close parentheses space minus 1
open parentheses straight d close parentheses space 1 half end style

Solution 10

begin mathsize 12px style Correct space option colon thin space left parenthesis straight a right parenthesis
straight I equals integral 2 to the power of begin display style 1 over straight x end style end exponent over straight x squared dx
Put space 1 over straight x equals straight t
rightwards double arrow fraction numerator negative 1 over denominator straight x squared end fraction dx equals dt rightwards double arrow 1 over straight x squared dx equals negative dt
straight I equals integral 2 to the power of straight t open parentheses negative dt close parentheses
straight I equals fraction numerator negative 2 to the power of straight t over denominator log subscript straight e 2 end fraction plus straight C
rightwards double arrow straight I equals fraction numerator negative 2 to the power of begin display style 1 over straight x end style end exponent over denominator log subscript straight e 2 end fraction plus straight C
straight k equals fraction numerator negative 1 over denominator log subscript straight e 2 end fraction
end style

Question 11

begin mathsize 11px style integral fraction numerator 1 over denominator 1 plus tan space straight x end fraction dx equals

open parentheses straight a close parentheses space log subscript straight e open parentheses straight x plus sin space straight x close parentheses plus straight C
open parentheses straight b close parentheses space log subscript straight e open parentheses sin space straight x plus cos space straight x close parentheses plus straight C
open parentheses straight c close parentheses space 2 space sec squared straight x over 2 plus straight C
open parentheses straight d close parentheses space 1 half open curly brackets straight x space plus space log open parentheses sin space straight x space plus space cos space straight x close parentheses close curly brackets space plus space straight C end style

Solution 11

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
straight I equals integral fraction numerator 1 over denominator 1 plus tanx end fraction dx
equals integral fraction numerator cosx over denominator sinx plus cosx end fraction dx
Numerator space can space be space written space as colon
cosx equals straight A open parentheses sinx plus cosx close parentheses plus straight B fraction numerator straight d open parentheses sinx plus cosx close parentheses over denominator dx end fraction
cosx equals open parentheses straight A minus straight B close parentheses sinx plus open parentheses straight A plus straight B close parentheses cosx
rightwards double arrow straight A minus straight B equals 0 space and space straight A plus straight B equals 1
rightwards double arrow straight A equals 1 half equals straight B
straight I equals integral fraction numerator open square brackets begin display style 1 half end style open parentheses sinx plus cosx close parentheses plus 1 half open parentheses cosx minus sinx close parentheses close square brackets dx over denominator sinx plus cosx end fraction
straight I equals 1 half integral open parentheses 1 plus fraction numerator cosx minus sinx over denominator sinx plus cosx end fraction close parentheses dx
straight I equals 1 half open square brackets 1 plus ln open parentheses sinx plus cosx close parentheses close square brackets plus straight C end style

Question 12

begin mathsize 12px style integral open vertical bar straight x close vertical bar cubed space dx space is space equal space to

open parentheses straight a close parentheses space fraction numerator negative straight x to the power of 4 over denominator 4 end fraction plus straight C
open parentheses straight b close parentheses space open vertical bar straight x close vertical bar to the power of 4 over 4 plus straight C
open parentheses straight c close parentheses space straight x to the power of 4 over 4 plus straight C
open parentheses straight d close parentheses space none space of space these end style

Solution 12

begin mathsize 12px style Correct space option colon thin space left parenthesis straight d right parenthesis
integral open vertical bar straight x close vertical bar cubed dx
If space straight x greater than 0
rightwards double arrow integral straight x cubed dx
equals straight x to the power of 4 over 4 plus straight C
If space straight x less than 0
rightwards double arrow integral negative straight x cubed dx
equals negative straight x to the power of 4 over 4 plus straight C end style

Question 13

begin mathsize 12px style The space value space of space integral fraction numerator cos space square root of straight x over denominator square root of straight x end fraction dx space is
open parentheses straight a close parentheses space 2 space cos square root of straight x plus straight C
open parentheses straight b close parentheses space square root of fraction numerator cos space straight x over denominator straight x end fraction end root plus straight C
open parentheses straight c close parentheses space sin space square root of straight x plus straight C
open parentheses straight d close parentheses space 2 space sin space square root of straight x plus space straight C end style

Solution 13

begin mathsize 12px style Correct space option colon thin space left parenthesis straight d right parenthesis
straight I equals integral fraction numerator cos square root of straight x over denominator square root of straight x end fraction dx
Put comma space
square root of straight x space equals straight t
fraction numerator 1 over denominator 2 square root of straight x end fraction dx equals dt
rightwards double arrow fraction numerator 1 over denominator square root of straight x end fraction dx equals 2 dt
straight I equals integral cost space 2 dt
straight I equals 2 sint plus straight C equals 2 sin square root of straight x plus straight C end style

Question 14

begin mathsize 12px style integral straight e to the power of straight x open parentheses 1 minus cot space straight x space plus space cot squared straight x close parentheses space dx equals

open parentheses straight a close parentheses space straight e to the power of straight x space cot space straight x space plus space straight C
open parentheses straight b close parentheses space minus straight e to the power of straight x space cot space straight x space plus space straight C
open parentheses straight c close parentheses space straight e to the power of straight x space cosec space straight x space plus space straight C
open parentheses straight d close parentheses space minus straight e to the power of straight x space cosec space straight x space plus space straight C end style

Solution 14

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
straight I equals integral straight e to the power of straight x open parentheses 1 minus cotx plus cot squared straight x close parentheses dx
straight I equals integral straight e to the power of straight x open parentheses 1 plus cot squared straight x minus cotx close parentheses dx
straight I equals integral straight e to the power of straight x open parentheses cosec squared straight x minus cotx close parentheses dx
Here comma space straight f left parenthesis straight x right parenthesis equals negative cotx rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis equals cosec squared straight x
straight I equals negative straight e to the power of straight x cotx plus straight C end style

Question 15

Error converting from MathML to accessible text.

Solution 15

begin mathsize 12px style Correct space option colon thin space left parenthesis straight b right parenthesis
straight I equals integral fraction numerator sin to the power of 6 xdx over denominator cos to the power of 8 straight x end fraction
straight I equals integral tan to the power of 6 xsec squared xdx
Put space tanx equals straight t rightwards double arrow sec squared xdx equals dt
straight I equals integral straight t to the power of 6 dt
straight I equals straight t to the power of 7 over 7 plus straight C
straight I equals fraction numerator tan to the power of 7 straight x over denominator 7 end fraction plus straight C end style

Question 16

begin mathsize 12px style integral fraction numerator 1 over denominator 7 plus 5 cos space straight x end fraction dx equals

open parentheses straight a close parentheses space fraction numerator 1 over denominator square root of 6 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 6 end fraction tan straight x over 2 close parentheses plus straight C
open parentheses straight b close parentheses space fraction numerator begin display style 1 end style over denominator begin display style square root of 3 end style end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 1 end style over denominator begin display style square root of 3 end style end fraction tan fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close parentheses plus straight C
open parentheses straight c close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 4 end style end fraction tan to the power of negative 1 end exponent open parentheses tan fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close parentheses plus straight C
open parentheses straight d close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 7 end style end fraction tan to the power of negative 1 end exponent open parentheses tan fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close parentheses plus straight C end style

Solution 16

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
straight I equals integral fraction numerator dx over denominator 7 plus 5 cosx end fraction
put space tan straight x over 2 equals straight t rightwards double arrow dx equals fraction numerator 2 dt over denominator 1 plus straight t squared end fraction
rightwards double arrow cosx equals fraction numerator 1 minus tan squared begin display style straight x over 2 end style over denominator 1 plus tan squared straight x over 2 end fraction equals fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction
straight I equals integral fraction numerator fraction numerator 2 dt over denominator 1 plus straight t squared end fraction over denominator 7 plus 5 cross times fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction end fraction
rightwards double arrow straight I equals integral fraction numerator 1 over denominator straight t squared plus 6 end fraction dt
straight I equals fraction numerator 1 over denominator square root of 6 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator straight t over denominator square root of 6 end fraction close parentheses plus straight C
rightwards double arrow fraction numerator 1 over denominator square root of 6 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator tan begin display style straight x over 2 end style over denominator square root of 6 end fraction close parentheses plus straight C
end style

Question 17

begin mathsize 12px style integral fraction numerator 1 over denominator 1 minus cosx minus sinx end fraction dx equals
open parentheses straight a close parentheses space log space open vertical bar 1 plus cot straight x over 2 close vertical bar plus straight C
open parentheses straight b close parentheses space log space open vertical bar 1 minus tan fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close vertical bar plus straight C
open parentheses straight c close parentheses space log space open vertical bar 1 minus cot fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close vertical bar plus straight C
open parentheses straight d close parentheses space log space open vertical bar 1 plus tan fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close vertical bar plus straight C end style

Solution 17

begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
integral fraction numerator dx over denominator 1 minus cosx minus sinx end fraction
Put space straight t equals space tan straight x over 2 rightwards double arrow dx equals fraction numerator 2 dt over denominator 1 plus straight t squared end fraction
cosx equals fraction numerator 1 minus tan squared straight x over 2 over denominator 1 plus tan squared straight x over 2 end fraction equals fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction
sinx equals fraction numerator 2 tan straight x over 2 over denominator 1 plus tan squared straight x over 2 end fraction equals fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction
put space in space the space question
straight I equals integral fraction numerator begin display style fraction numerator 2 dt over denominator 1 plus straight t squared end fraction end style over denominator 1 minus fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction minus fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction end fraction
straight I equals integral fraction numerator dt over denominator straight t squared minus straight t end fraction
straight I equals integral fraction numerator dt over denominator straight t squared minus straight t plus begin display style 1 fourth end style minus begin display style 1 fourth end style end fraction
rightwards double arrow straight I equals ln open vertical bar 1 minus cot straight x over 2 close vertical bar plus straight C

end style

Question 18

begin mathsize 12px style integral fraction numerator straight x plus 3 over denominator open parentheses straight x plus 4 close parentheses squared end fraction straight e to the power of straight x dx equals
open parentheses straight a close parentheses space fraction numerator straight e to the power of straight x over denominator straight x plus 4 end fraction plus straight C
open parentheses straight b close parentheses space fraction numerator begin display style straight e to the power of straight x end style over denominator begin display style straight x plus 3 end style end fraction plus straight C
open parentheses straight c close parentheses space fraction numerator begin display style 1 end style over denominator begin display style open parentheses straight x plus 4 close parentheses squared end style end fraction plus straight C
open parentheses straight d close parentheses space fraction numerator begin display style straight e to the power of straight x end style over denominator begin display style open parentheses straight x plus 4 close parentheses squared end style end fraction plus straight C end style

Solution 18

begin mathsize 12px style Correct space option colon thin space left parenthesis straight a right parenthesis
straight I equals integral fraction numerator straight x plus 3 over denominator open parentheses straight x plus 4 close parentheses squared end fraction straight e to the power of straight x dx
straight I equals integral open parentheses fraction numerator straight x plus 4 minus 1 over denominator open parentheses straight x plus 4 close parentheses squared end fraction close parentheses straight e to the power of straight x dx
straight I equals integral open parentheses fraction numerator 1 over denominator straight x plus 4 end fraction minus 1 over open parentheses straight x plus 4 close parentheses squared close parentheses straight e to the power of straight x dx
straight f left parenthesis straight x right parenthesis equals fraction numerator 1 over denominator straight x plus 4 end fraction rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis equals negative 1 over open parentheses straight x plus 4 close parentheses squared
rightwards double arrow straight I equals fraction numerator straight e to the power of straight x over denominator straight x plus 4 end fraction plus straight C
end style

Question 19

begin mathsize 12px style integral fraction numerator sin space straight x over denominator 3 space plus space 4 space cos squared straight x end fraction dx
open parentheses straight a close parentheses space log space open parentheses 3 plus 4 cos squared straight x close parentheses plus straight C
open parentheses straight b close parentheses space fraction numerator 1 over denominator 2 square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator cos begin display style space end style begin display style straight x end style over denominator square root of 3 end fraction close parentheses plus straight C
open parentheses straight c close parentheses space minus fraction numerator begin display style 1 end style over denominator begin display style 2 square root of 3 end style end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 2 space cos space straight x end style over denominator square root of 3 end fraction close parentheses plus straight C
open parentheses straight d close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 2 square root of 3 end style end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 2 space cos space straight x end style over denominator square root of 3 end fraction close parentheses plus straight C end style

Solution 19

begin mathsize 12px style Correct space option colon thin space left parenthesis straight c right parenthesis
straight I equals integral fraction numerator sinx over denominator 3 plus 4 cos squared straight x end fraction dx
Put space cosx equals straight t
rightwards double arrow negative sinxdx equals dt
rightwards double arrow sinxdx equals negative dt
straight I equals integral fraction numerator negative dt over denominator 3 plus 4 straight t squared end fraction
straight I equals fraction numerator negative 1 over denominator 2 square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight t over denominator square root of 3 end fraction close parentheses plus straight C
straight I equals fraction numerator negative 1 over denominator 2 square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 2 cosx over denominator square root of 3 end fraction close parentheses plus straight C end style

Question 20

begin mathsize 12px style integral straight e to the power of straight x open parentheses fraction numerator 1 minus sin space straight x over denominator 1 minus cos space straight x end fraction close parentheses dx
open parentheses straight a close parentheses space minus straight e to the power of straight x space tan space straight x over 2 plus straight C
open parentheses straight b close parentheses space minus straight e to the power of straight x space cot space straight x over 2 plus straight C
open parentheses straight c close parentheses space minus 1 half space straight e to the power of straight x space tan space straight x over 2 plus straight C
open parentheses straight a close parentheses space minus 1 half space straight e to the power of straight x space cot space straight x over 2 plus straight C end style

Solution 20

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
straight I equals integral straight e to the power of straight x open parentheses fraction numerator 1 minus sinx over denominator 1 minus cosx end fraction close parentheses dx
straight I equals integral straight e to the power of straight x open parentheses fraction numerator 1 over denominator 1 minus cosx end fraction minus fraction numerator sinx over denominator 1 minus cosx end fraction close parentheses dx
straight I equals integral straight e to the power of straight x open parentheses fraction numerator 1 over denominator 2 sin squared begin display style straight x over 2 end style end fraction minus fraction numerator 2 sin begin display style straight x over 2 end style cos straight x over 2 over denominator 2 sin squared straight x over 2 end fraction close parentheses dx
straight I equals integral straight e to the power of straight x open parentheses fraction numerator cosec squared straight x over 2 over denominator 2 end fraction minus cot straight x over 2 close parentheses dx
straight f left parenthesis straight x right parenthesis equals negative cot straight x over 2 rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis equals fraction numerator cosec squared straight x over 2 over denominator 2 end fraction
straight I equals negative straight e to the power of straight x cot straight x over 2 plus straight C end style

Question 21

begin mathsize 12px style integral 2 over open parentheses straight e to the power of straight x space plus space straight e to the power of negative straight x end exponent close parentheses squared dx
open parentheses straight a close parentheses space fraction numerator negative straight e to the power of negative straight x end exponent over denominator straight e to the power of straight x plus straight e to the power of negative straight x end exponent end fraction plus straight C
open parentheses straight b close parentheses space minus fraction numerator begin display style 1 end style over denominator begin display style straight e to the power of straight x plus straight e to the power of negative straight x end exponent end style end fraction plus straight C
open parentheses straight c close parentheses space fraction numerator begin display style negative 1 end style over denominator begin display style open parentheses straight e to the power of straight x plus space 1 close parentheses squared end style end fraction plus straight C
open parentheses straight d close parentheses space fraction numerator begin display style 1 end style over denominator begin display style straight e to the power of straight x plus straight e to the power of negative straight x end exponent end style end fraction plus straight C end style

Solution 21

Error converting from MathML to accessible text.

Question 22

begin mathsize 12px style integral fraction numerator straight e to the power of straight x open parentheses 1 plus straight x close parentheses over denominator cos squared open parentheses xe to the power of straight x close parentheses end fraction dx equals

open parentheses straight a close parentheses space 2 space log subscript straight e cos open parentheses xe to the power of straight x close parentheses plus straight C
open parentheses straight b close parentheses space sec open parentheses xe to the power of straight x close parentheses plus straight C
open parentheses straight c close parentheses space tan open parentheses xe to the power of straight x close parentheses plus straight C
open parentheses straight d close parentheses space tan open parentheses straight x plus straight e to the power of straight x close parentheses plus straight C end style

Solution 22

begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
straight I equals integral fraction numerator straight e to the power of straight x open parentheses 1 plus straight x close parentheses over denominator cos squared open parentheses xe to the power of straight x close parentheses end fraction dx
Put space xe to the power of straight x equals straight t
rightwards double arrow straight e to the power of straight x open parentheses 1 plus straight x close parentheses dx equals dt
straight I equals integral fraction numerator dt over denominator cos squared straight t end fraction
straight I equals integral sec squared tdt
straight I equals tant plus straight C
straight I equals tan open parentheses xe to the power of straight x close parentheses plus straight C end style

Question 23

begin mathsize 12px style integral fraction numerator sin squared straight x over denominator cos to the power of 4 straight x end fraction dx equals

open parentheses straight a close parentheses space 1 third space tan squared straight x space plus space straight C
open parentheses straight b close parentheses space 1 half space tan squared straight x space plus space straight C
open parentheses straight c close parentheses space 1 third space tan cubed straight x space plus space straight C
open parentheses straight d close parentheses space none space of space these end style

Solution 23

begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
straight I equals integral fraction numerator sin squared straight x over denominator cos to the power of 4 straight x end fraction dx
straight I equals integral tan squared xsec squared xdx
Put space tanx equals straight t rightwards double arrow sec squared xdx equals dt
straight I equals integral straight t squared dt
straight I equals straight t cubed over 3 plus straight C
straight I equals fraction numerator tan cubed straight x over denominator 3 end fraction plus straight C end style

Question 24

begin mathsize 12px style The space primitive space of space the space function space straight f left parenthesis straight x right parenthesis space equals space open parentheses 1 minus 1 over straight x squared close parentheses straight a to the power of straight x plus 1 over straight x end exponent comma space straight a space greater than space 0 space is
open parentheses straight a close parentheses space fraction numerator straight a to the power of straight x plus begin display style 1 over straight x end style end exponent over denominator log subscript straight e straight a end fraction
open parentheses straight b close parentheses space log subscript straight e straight a times straight a to the power of straight x plus 1 over straight x end exponent
open parentheses straight c close parentheses space straight a to the power of straight x plus begin display style 1 over straight x end style end exponent over straight x log subscript straight e straight a
open parentheses straight d close parentheses space straight x fraction numerator straight a to the power of straight x plus begin display style 1 over straight x end style end exponent over denominator log subscript straight e straight a end fraction end style

Solution 24

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
straight f left parenthesis straight x right parenthesis equals open parentheses 1 minus 1 over straight x squared close parentheses straight a to the power of straight x plus 1 over straight x end exponent
rightwards double arrow integral straight f left parenthesis straight x right parenthesis dx equals integral open parentheses 1 minus 1 over straight x squared close parentheses straight a to the power of straight x plus 1 over straight x end exponent dx
Put space straight x plus 1 over straight x equals straight t
rightwards double arrow open parentheses 1 minus 1 over straight x squared close parentheses dx equals dt
straight I equals integral straight a to the power of straight t dt
straight I equals fraction numerator straight a to the power of straight t over denominator log subscript straight e straight a end fraction plus straight C
straight I equals fraction numerator straight a to the power of straight x plus 1 over straight x end exponent over denominator log subscript straight e straight a end fraction plus straight C end style

Question 25

begin mathsize 12px style The space value space of space integral fraction numerator 1 over denominator straight x plus straight x space log space straight x end fraction dx space is
open parentheses straight a close parentheses space 1 space plus space log space straight x
open parentheses straight b close parentheses space straight x space plus space log space straight x
open parentheses straight c close parentheses space straight x space log space open parentheses 1 space plus space log space straight x close parentheses
open parentheses straight d close parentheses space log open parentheses 1 space plus space log space straight x close parentheses end style

Solution 25

begin mathsize 12px style Correct space option colon thin space left parenthesis straight d right parenthesis
straight I equals integral fraction numerator 1 over denominator straight x plus xlogx end fraction dx
straight I equals integral fraction numerator dx over denominator straight x open parentheses 1 plus logx close parentheses end fraction
Put space 1 plus logx equals straight t
rightwards double arrow 1 over straight x dx equals dt
straight I equals integral 1 over straight t dt
rightwards double arrow straight I equals log open vertical bar straight t close vertical bar plus straight C
straight I equals log open parentheses 1 plus logx close parentheses plus straight C
end style

Question 26

begin mathsize 12px style integral square root of fraction numerator straight x over denominator 1 minus straight x end fraction end root dx space is space equal space to
open parentheses straight a close parentheses space sin to the power of negative 1 end exponent square root of straight x plus straight C
open parentheses straight b close parentheses space sin to the power of negative 1 end exponent open curly brackets square root of straight x minus square root of straight x left parenthesis 1 minus straight x right parenthesis end root close curly brackets plus straight C
open parentheses straight c close parentheses space sin to the power of negative 1 end exponent open curly brackets square root of straight x left parenthesis 1 minus straight x right parenthesis end root close curly brackets plus straight C
open parentheses straight d close parentheses space sin to the power of negative 1 end exponent square root of straight x minus square root of straight x left parenthesis 1 minus straight x right parenthesis end root plus straight C end style

Solution 26

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
straight I equals integral square root of fraction numerator straight x over denominator 1 minus straight x end fraction end root dx
straight I equals integral square root of fraction numerator straight x over denominator 1 minus straight x end fraction cross times straight x over straight x end root dx
straight I equals integral fraction numerator xdx over denominator square root of straight x minus straight x squared end root end fraction
Consider comma
straight x equals straight A fraction numerator straight d open parentheses straight x minus straight x squared close parentheses over denominator dx end fraction plus straight B
straight x equals straight A open parentheses 1 minus 2 straight x close parentheses plus straight B
straight x equals negative 2 Ax plus straight A plus straight B
minus 2 straight A equals 1 rightwards double arrow straight A equals fraction numerator negative 1 over denominator 2 end fraction
rightwards double arrow straight A plus straight B equals 0 rightwards double arrow straight B equals 1 half
straight I equals integral fraction numerator fraction numerator negative 1 over denominator 2 end fraction open parentheses 1 minus 2 straight x close parentheses plus 1 half over denominator square root of straight x minus straight x squared end root end fraction dx
straight I equals integral open parentheses fraction numerator negative 1 over denominator 2 end fraction fraction numerator 1 minus 2 straight x over denominator square root of straight x minus straight x squared end root end fraction plus fraction numerator 1 over denominator 2 square root of straight x minus straight x squared end root end fraction close parentheses dx
straight I equals fraction numerator negative 1 over denominator 2 end fraction cross times 2 square root of straight x minus straight x squared end root plus 1 half integral fraction numerator 1 over denominator square root of straight x minus straight x squared end root end fraction dx
Second space term space after space completing space square space method space you space will space get space as
straight I equals negative square root of straight x minus straight x squared end root plus sin to the power of negative 1 end exponent square root of straight x plus straight C end style

Question 27

begin mathsize 12px style integral straight e to the power of straight x open curly brackets straight f open parentheses straight x close parentheses space plus space straight f apostrophe open parentheses straight x close parentheses close curly brackets dx space equals

open parentheses straight a close parentheses space straight e to the power of straight x space straight f open parentheses straight x close parentheses space plus space straight C
open parentheses straight b close parentheses space straight e to the power of straight x space plus space straight f open parentheses straight x close parentheses space plus space straight C
open parentheses straight c close parentheses space 2 straight e to the power of straight x space straight f open parentheses straight x close parentheses space plus space straight C
open parentheses straight d close parentheses space straight e to the power of straight x space minus space straight f open parentheses straight x close parentheses space plus space straight C end style

Solution 27

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
integral straight e to the power of straight x open curly brackets straight f open parentheses straight x close parentheses plus straight f apostrophe open parentheses straight x close parentheses close curly brackets dx equals straight e to the power of straight x straight f open parentheses straight x close parentheses plus straight C
end style

Question 28

begin mathsize 12px style The space value space of space integral fraction numerator sin space straight x space plus space cos space straight x over denominator square root of 1 minus sin space 2 straight x end root end fraction dx space is space equal space to
open parentheses straight a close parentheses space square root of sin space 2 straight x space end root plus space straight C
open parentheses straight b close parentheses space square root of cos 2 straight x end root space plus space straight C
open parentheses straight c close parentheses space plus-or-minus open parentheses sin space straight x space minus space cos space straight x close parentheses space plus space straight C
open parentheses straight d close parentheses space plus-or-minus space log space open parentheses sin space straight x space minus space cos space straight x close parentheses space plus space straight C end style

Solution 28

begin mathsize 12px style Correct space option colon thin space left parenthesis straight d right parenthesis
straight I equals integral fraction numerator sinx plus cosx over denominator square root of 1 minus sin 2 straight x end root end fraction dx
straight I equals integral fraction numerator sinx plus cosx over denominator cosx minus sinx end fraction dx
Put space cosx minus sinx equals straight t rightwards double arrow open parentheses sinx plus cosx close parentheses dx equals plus-or-minus dt
rightwards double arrow open parentheses sinx plus cosx close parentheses dx equals plus-or-minus dt
straight I equals plus-or-minus integral dt over straight t
straight I equals plus-or-minus log open vertical bar straight t close vertical bar plus straight C
straight I equals plus-or-minus log open vertical bar sinx minus cosx close vertical bar plus straight C
end style

Question 29

begin mathsize 12px style If space integral straight x space sin space straight x space dx space equals space minus straight x space cos space straight x space plus space straight alpha comma space then space straight alpha space is space equal space to

open parentheses straight a close parentheses space sin space straight x space plus space straight C
open parentheses straight b close parentheses space cos space straight x space plus space straight C
open parentheses straight c close parentheses space straight C
open parentheses straight d close parentheses space none space fo space these end style

Solution 29

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
integral xsinxdx equals negative xcosx plus straight alpha
straight I equals integral xsinxdx
straight I equals straight x integral sinxdx minus integral open parentheses fraction numerator d straight x over denominator d straight x end fraction integral sinxdx close parentheses dx
straight I equals negative xcosx plus integral cosxdx
straight I equals xcosx plus sinx plus straight C
rightwards double arrow straight alpha equals sinx plus straight C
end style

Question 30

begin mathsize 12px style integral fraction numerator cos space 2 straight x space minus space 1 over denominator cos space 2 straight x space plus space 1 end fraction dx equals

open parentheses straight a close parentheses space tan space straight x space minus space straight x space plus space straight C
open parentheses straight b close parentheses space straight x space plus space tan space straight x space plus space straight C
open parentheses straight c close parentheses space straight x space minus space tan space straight x space plus space straight C
open parentheses straight d close parentheses space minus straight x space minus space cot space straight x space plus space straight C end style

Solution 30

begin mathsize 12px style Correct space option colon space left parenthesis straight c right parenthesis
straight I equals integral fraction numerator cos 2 straight x minus 1 over denominator cos 2 straight x plus 1 end fraction dx
straight I equals negative integral fraction numerator 1 minus cos 2 straight x over denominator 1 plus cos 2 straight x end fraction dx
straight I equals negative integral fraction numerator 2 sin squared straight x over denominator 2 cos squared straight x over 2 end fraction dx
straight I equals negative integral tan squared xdx
straight I equals negative integral open parentheses sec squared straight x minus 1 close parentheses dx
straight I equals negative open parentheses tanx minus straight x close parentheses plus straight C
straight I equals straight x minus tanx plus straight C end style

Question 31

begin mathsize 12px style integral fraction numerator cos space 2 straight x space minus space cos space 2 straight theta over denominator cos space straight x space minus space cos space straight theta end fraction dx space is space equal space to

open parentheses straight a close parentheses space 2 open parentheses sin space straight x space plus space straight x space cos space straight theta close parentheses space plus space straight C
open parentheses straight b close parentheses space 2 open parentheses sin space straight x space minus space straight x space cos space straight theta close parentheses space plus space straight C
open parentheses straight c close parentheses space 2 open parentheses sin space straight x space plus space 2 straight x space cos space straight theta close parentheses space plus space straight C
open parentheses straight d close parentheses space 2 open parentheses sin space straight x space minus space 2 straight x space cos space straight theta close parentheses space plus space straight C end style

Solution 31

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
straight I equals integral fraction numerator cos 2 straight x minus cos 2 straight theta over denominator cosx minus cosθ end fraction dx
straight I equals integral fraction numerator 2 cos squared straight x minus 1 minus open parentheses 2 cos squared straight theta minus 1 close parentheses over denominator cosx minus cosθ end fraction dx
straight I equals integral fraction numerator 2 open parentheses cos squared straight x minus cos squared straight theta close parentheses over denominator cosx minus cosθ end fraction dx
straight I equals 2 integral open parentheses cosx plus cosθ close parentheses dx
straight I equals 2 open parentheses sinx plus xcosθ close parentheses plus straight C end style

Question 32

begin mathsize 12px style integral straight x to the power of 9 over open parentheses 4 straight x squared plus 1 close parentheses to the power of 6 dx space is space equal space to

open parentheses straight a close parentheses space fraction numerator 1 over denominator 5 straight x end fraction open parentheses 4 plus 1 over straight x squared close parentheses to the power of negative 5 end exponent plus straight C
open parentheses straight b close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 5 end style end fraction open parentheses 4 plus fraction numerator begin display style 1 end style over denominator begin display style straight x squared end style end fraction close parentheses to the power of negative 5 end exponent plus straight C
open parentheses straight c close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 10 straight x end style end fraction open parentheses fraction numerator begin display style 1 end style over denominator begin display style straight x squared end style end fraction plus 4 close parentheses to the power of negative 5 end exponent plus straight C
open parentheses straight d close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 10 end style end fraction open parentheses fraction numerator begin display style 1 end style over denominator begin display style straight x squared end style end fraction plus 4 close parentheses to the power of negative 5 end exponent plus straight C
end style

Solution 32

begin mathsize 12px style Correct space option colon left parenthesis straight d right parenthesis
straight I equals integral straight x to the power of 9 over open parentheses 4 straight x squared plus 1 close parentheses to the power of 6 dx
straight I equals integral fraction numerator straight x to the power of 9 over denominator straight x to the power of 12 open parentheses 4 plus begin display style 1 over straight x squared end style close parentheses to the power of 6 end fraction dx
straight I equals integral fraction numerator dx over denominator straight x cubed open parentheses 4 plus 1 over straight x squared close parentheses to the power of 6 end fraction
Put space open parentheses 4 plus 1 over straight x squared close parentheses equals straight t rightwards double arrow fraction numerator negative 2 over denominator straight x cubed end fraction dx equals dt
straight I equals 1 half integral negative straight t to the power of negative 6 end exponent dt
straight I equals 1 half open parentheses fraction numerator negative straight t to the power of negative 5 end exponent over denominator negative 5 end fraction close parentheses plus straight C
straight I equals 1 over 10 1 over straight t to the power of 5 plus straight C
straight I equals 1 over 10 open parentheses 4 plus 1 over straight x squared close parentheses to the power of negative 5 end exponent plus straight C end style

Question 33

begin mathsize 12px style integral fraction numerator straight x cubed over denominator square root of 1 plus straight x squared end root end fraction dx equals straight a open parentheses 1 plus straight x squared close parentheses to the power of 3 divided by 2 end exponent space plus space straight b square root of 1 plus straight x squared end root plus straight C comma space then

open parentheses straight a close parentheses space straight a space equals space 1 third comma space straight b space equals space 1
open parentheses straight b close parentheses space straight a space equals space minus fraction numerator begin display style 1 end style over denominator begin display style 3 end style end fraction comma space straight b space equals space 1
open parentheses straight c close parentheses space straight a space equals negative fraction numerator begin display style 1 end style over denominator begin display style 3 end style end fraction comma space straight b space equals space minus 1
open parentheses straight d close parentheses space straight a space equals space fraction numerator begin display style 1 end style over denominator begin display style 3 end style end fraction comma space straight b space equals space minus 1 end style

Solution 33

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
straight I equals integral fraction numerator straight x cubed over denominator square root of 1 plus straight x squared end root end fraction dx
1 plus straight x squared equals straight t rightwards double arrow 2 xdx equals dt
rightwards double arrow xdx equals dt over 2
straight I equals integral fraction numerator straight x squared over denominator square root of 1 plus straight x squared end root end fraction xdx
straight I equals integral fraction numerator straight t minus 1 over denominator square root of straight t end fraction dt over 2
straight I equals 1 half integral open parentheses square root of straight t minus straight t to the power of fraction numerator negative 1 over denominator 2 end fraction end exponent close parentheses dt
straight I equals 1 half open parentheses 2 over 3 straight t to the power of 3 over 2 end exponent minus 2 square root of straight t close parentheses plus straight C
straight I equals 1 third open parentheses 1 plus straight x squared close parentheses to the power of 3 over 2 end exponent minus square root of 1 plus straight x squared end root plus straight C
straight a equals 1 third comma space straight b equals negative 1 end style

Question 34

begin mathsize 12px style integral fraction numerator straight x cubed over denominator straight x plus 1 end fraction dx space is space equal space to

open parentheses straight a close parentheses space straight x space plus space straight x squared over 2 plus straight x cubed over 3 minus space log space open vertical bar 1 minus straight x close vertical bar plus straight C
open parentheses straight b close parentheses space straight x space plus space fraction numerator begin display style straight x squared end style over denominator begin display style 2 end style end fraction minus fraction numerator begin display style straight x cubed end style over denominator begin display style 3 end style end fraction minus space log space open vertical bar 1 minus straight x close vertical bar plus straight C
open parentheses straight c close parentheses space straight x space minus space fraction numerator begin display style straight x squared end style over denominator begin display style 2 end style end fraction minus fraction numerator begin display style straight x cubed end style over denominator begin display style 3 end style end fraction minus space log space open vertical bar 1 plus straight x close vertical bar plus straight C
open parentheses straight d close parentheses space straight x space minus space fraction numerator begin display style straight x squared end style over denominator begin display style 2 end style end fraction plus fraction numerator begin display style straight x cubed end style over denominator begin display style 3 end style end fraction minus space log space open vertical bar 1 plus straight x close vertical bar plus straight C
end style

Solution 34

begin mathsize 12px style Correct space option colon left parenthesis straight d right parenthesis
straight I equals integral fraction numerator straight x cubed over denominator straight x plus 1 end fraction dx
straight I equals integral fraction numerator straight x cubed plus 1 minus 1 over denominator straight x plus 1 end fraction dx
straight I equals integral fraction numerator open parentheses straight x plus 1 close parentheses open parentheses straight x squared minus straight x plus 1 close parentheses minus 1 over denominator straight x plus 1 end fraction dx
straight I equals integral open parentheses straight x squared minus straight x plus 1 minus fraction numerator 1 over denominator straight x plus 1 end fraction close parentheses dx
straight I equals straight x cubed over 3 minus straight x squared over 2 plus straight x minus log open vertical bar straight x plus 1 close vertical bar plus straight C end style

Question 35

begin mathsize 12px style If space integral fraction numerator 1 over denominator open parentheses straight x plus 2 close parentheses open parentheses straight x squared plus 1 close parentheses end fraction dx space equals space straight a space log space open vertical bar 1 plus straight x squared close vertical bar space plus space straight b space tan to the power of negative 1 end exponent straight x space plus space 1 fifth log open vertical bar straight x space plus space 2 close vertical bar space plus space straight C comma space then

open parentheses straight a close parentheses space straight a space equals space minus 1 over 10 comma space straight b space equals space minus 2 over 5
open parentheses straight b close parentheses space straight a space equals space 1 over 10 comma space straight b space equals space minus 2 over 5
open parentheses straight c close parentheses space straight a space equals space minus 1 over 10 comma space straight b space equals space 2 over 5
open parentheses straight d close parentheses space straight a space equals space 1 over 10 comma space straight b space equals space 2 over 5 end style

Solution 35

begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
straight I equals integral fraction numerator 1 over denominator open parentheses straight x plus 2 close parentheses open parentheses straight x squared plus 1 close parentheses end fraction dx
Consider comma
fraction numerator 1 over denominator open parentheses straight x plus 2 close parentheses open parentheses straight x squared plus 1 close parentheses end fraction equals fraction numerator straight A over denominator straight x plus 2 end fraction plus fraction numerator Bx plus straight C over denominator open parentheses straight x squared plus 1 close parentheses end fraction
1 equals straight A open parentheses straight x squared plus 1 close parentheses plus open parentheses Bx plus straight C close parentheses open parentheses straight x plus 2 close parentheses
Comaring space coefficeints space and space solving space it space simultaneously space we space get
straight A equals 1 fifth comma space straight B equals negative 1 fifth comma space straight C equals 2 over 5
straight I equals integral open parentheses fraction numerator 1 over denominator 5 straight x plus 2 end fraction plus fraction numerator begin display style fraction numerator negative 1 over denominator 5 end fraction end style straight x plus begin display style 2 over 5 end style over denominator straight x squared plus 1 end fraction close parentheses dx
Integrating space we space get space as
1 fifth log open vertical bar straight x plus 2 close vertical bar minus 1 over 10 log open vertical bar straight x squared plus 1 close vertical bar plus 2 over 5 tan to the power of negative 1 end exponent straight x plus straight C
rightwards double arrow straight a equals negative 1 over 10 comma space straight b equals 2 over 5 end style
begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
straight I equals integral fraction numerator 1 over denominator open parentheses straight x plus 2 close parentheses open parentheses straight x squared plus 1 close parentheses end fraction dx
Consider comma
fraction numerator 1 over denominator open parentheses straight x plus 2 close parentheses open parentheses straight x squared plus 1 close parentheses end fraction equals fraction numerator straight A over denominator straight x plus 2 end fraction plus fraction numerator Bx plus straight C over denominator open parentheses straight x squared plus 1 close parentheses end fraction
1 equals straight A open parentheses straight x squared plus 1 close parentheses plus open parentheses Bx plus straight C close parentheses open parentheses straight x plus 2 close parentheses
Comaring space coefficeints space and space solving space it space simultaneously space we space get
straight A equals 1 fifth comma space straight B equals negative 1 fifth comma space straight C equals 2 over 5
straight I equals integral open parentheses fraction numerator 1 over denominator 5 straight x plus 2 end fraction plus fraction numerator begin display style fraction numerator negative 1 over denominator 5 end fraction end style straight x plus begin display style 2 over 5 end style over denominator straight x squared plus 1 end fraction close parentheses dx
Integrating space we space get space as
1 fifth log open vertical bar straight x plus 2 close vertical bar minus 1 over 10 log open vertical bar straight x squared plus 1 close vertical bar plus 2 over 5 tan to the power of negative 1 end exponent straight x plus straight C
rightwards double arrow straight a equals negative 1 over 10 comma space straight b equals 2 over 5 end style

Chapter 19 – Indefinite Integrals Exercise Ex. 19VSAQ

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

Write the value of ∫ e2x2 + In x dxSolution 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

Solution 46

Question 47

Solution 47

Question 48

Solution 48

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

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Chapter – 6 Physical and Chemical Changes and Salts | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter 6 Physical and Chemical Changes

Q.1.Classify the changes involved in the following processes as physical or chemical changes:
(a) Photosynthesis
(b) Dissolving sugar in water
(c) Burning of coal
(d) Melting of wax
(e) Beating aluminium to make aluminium foil
(f) Digestion of food

Ans.(a) Chemical change (b) Physical change
(c) Chemical change (d) Physical change
(e) Physical change (/) Chemical change

Q.2.(a) Cutting a log of wood into pieces is a chemical change. (True/ False)
(b) Formation of manure from leaves is a physical change. (True/ False)


(c)Iron pipes coated with zinc do not get rusted easily. (True/ False)
(d)Iron and rust are the same substances. (True/ False)
(e)Condensation of steam is not a chemical change. (True/ False)
Ans. (a)False
Correct statement: Cutting a log of wood into pieces is an irreversible physical change.
(b)False
Correct statement: Formation of manure from leaves is a chemical change.
(c) True
(d)False
Correct statement: Iron and rust are two different chemical substances.
(e)True

Q.3.Fill in the blanks in the following statements:
(a) When carbon dioxide is passed through lime water, it turns milky due to the formation of .
(b) The chemical name of baking soda is .
(c) Two methods by which rusting of iron can be prevented are __________ and __________
(d) Changes in which only ____________ properties of a substance change are called physical changes.
(e) Changes in which new substances are formed are called _____________ changes..
Ans. (a)calcium carbonate
(b) sodium hydrogen carbonate
(c) painting or greasing, galvanisation
(d) physical
(e) chemical

Q.4. When baking soda is mixed with lemon juice, bubbles are formed with the evolution of a gas. What type of change is it? Explain.
Ans. The reaction between baking soda and lemon juice can be given as below:
Lemon juice + Baking soda ————-> C02 (bubbles) + Other substances
(Citric acid)                                    (Sodium hydrogen carbonate)  (Carbon dioxide)

It is a chemical change

Q.5. When a candle burns, both physical and chemical changes take place. Identify these changes. Give another example of a familiar process in which both the chemical and physical changes take place.
Ans. When a candle burns, both physical and chemical changes occur:


(i) Physical change: melting of wax, vapourisation of melted wax.
(ii) Chemical change: Burning of vapours of wax to give carbon dioxide, heat and light.
LPG is another example in which physical change occurs when LPG comes out of cylinder and is converted from liquid to gaseous state and a chemical change occurs when gas burns in air.

Q.6. How would you show that setting of a curd is a chemical change?
Ans. We can say that setting of curd is a chemical change because we can not get the original substance, i.e., milk back and a new substance is formed with different taste, smell and other chemical properties

Q.7. Explain why burning of wood and cutting it into small pieces are considered as two different types of changes. ~
Ans. Burning of wood is a chemical change because in burning new substances are formed as
Wood + Oxygen ———–> Charcoal + Carbon dioxide + Heat + Light
But cutting it into small pieces is physical change because no new substance is formed. We can only reduce the size of wood.

Q.8. Describe how crystals of copper sulphate are prepared.
Ans. Take a cupful of water in a beaker and add a few drops of dilute sulphuric acid. Heat the water. When it starts boiling, add copper sulphate powder slowly. Continue to add copper sulphate powder till no more powder can be dissolved. .During this process continuously stir the solution. Filter the solution. Leave it for cooling. Look it after some time, you can see the crystals of copper sulphate
NCERT Solutions Class 7 Science Chapter 6 Physical and Chemical Changes Q8

Q.9. Explain how painting of an iron gate prevents it from rusting?
Ans. It is known that for rusting the presence of oxygen and moisture is essential. Painting prevents the iron gate from coming in contact with oxygen and moisture.

Q.10. Explain why rusting of iron objects is faster in coastal areas than in deserts.
Ans. As content of moisture in the air in coastal areas is higher than in the air in deserts. So, the process of rusting is faster in coastal areas.

Q.11. The gas we use in the kitchen is called liquified petroleum gas (LPG). In the cylinder it exists as a liquid. When it comes out from the cylinder it becomes a gas (Change- A) then it bums (Change-B). The following statements pertain to these changes. Choose the correct one.
(i) Process-A is a chemical change.
(ii) Process-B is a chemical change.
(iii) Both processes A and B are chemical changes.
(iv) None of these processes is a chemical change.
Ans. (ii) Process-B is a chemical change.

Q.12.Anaerobic bacteria digest animal waste and produce biogas (Change-A). The biogas is then burnt as fuel (Change-B). The following statements pertain to these changes. Choose the correct one.
(i) Process-A is a chemical change.
(ii) Process-B is a chemical change.
(iii) Both processes A and B are chemical changes.
(iv) None of these processes is a chemical change.
Ans.(iii) Both processes A and B are chemical change

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Chapter – 5 Acids Bases and Salts | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter 5 Acids Bases and Salts

Q.2.Ammonia is found in many household products, such as window cleaners. It turns red litmus blue. What is its nature?
Ans.Ammonia has basic nature.

Q.3.Name the source from which litmus solution is obtained. What is the use of this solution?
Ans.Litmus solution is extracted from lichens. It is used to determine whether the given solution is acidic or basic.

Q.4.Is the distilled water acidic/basic/neutral? How would you verify it?
Ans.Distilled water will be neutral. We can verify it by showing that neither blue nor red litmus paper changes its colour when dipped in it.

Q.5.Describe the process of neutralisation with the help of an example.
Ans.The reaction between an acid and a base is known as neutralisation. Salt and water are produced in this process with the evolution of heat.
Antacids like milk of magnesia (magnesium hydroxide), baking soda, etc. which contain a base are used for reducing acidity in stomach when excessive acid released by glands.

Q.6.Mark ‘T’ if the statement is true and ‘F’ if it is false:
(i) Nitric acid turns red litmus blue. (T/F)
(ii) Sodium hydroxide turns blue litmus red. {T/F)
(iii) Sodium hydroxide and hydrochloric acid neutralise each other and form salt and water. (T/F)
(id) Indicator is a substance which shows different colours in acidic and basic solutions. . (T/F)
(v) Tooth decay is caused by the presence of a base. (T/F)
Ans.(1) F (ii) F (iii) T (iv) T (V) F

Q.7. Dorji has a few bottles of soft drink in his restaurant. But, unfortunately, these are
not labelled. He has to serve the drinks on the demand of customers. One customer
wants acidic drink, another wants basic and third one wants neutral drink. How
will Dorji decide which drink is to be served to whom?
Ans.Dorji can decide with the help of litmus paper:
(i) The drink which would turn a red litmus blue would be basic.
(ii) If the drink turns a blue litmus to red would be acidic.
(iii) The drink which would not affect both red and blue litmus would be neutral.

Q.8.Explain why:
(a) An antacid tablet is taken when you suffer from acidity.
(b) Calamine solution is applied on the skin when an ant bites.
(c) Factory waste is neutralised before disposing it into the water bodies.
Ans.(a) We take an antacid such as milk of magnesia to neutralises the excessive acid released in stomach.
(b) Ant injects an acidic liquid (Formic acid) into the skin on biting which causes inflammation, to the skin. The effect of the acid can be neutralised by rubbing. Calamine solution which contains zinc carbonate which is very weak base and causes no harm to the skin.
(c) The wastes of factories contain acids. If acids are disposed off in the water body, the acids will harm the organisms. So factory wastes are neutralised by adding basic substances.

Q.9. Three liquids are given to you. One is hydrochloric acid, another is sodium hydroxide and third is a sugar solution. How will you identify them? You have only turmeric indicator.
Ans.Name of the substances Effect on turmeric indicator
1. Hydrochloric acid Yellow to blue
2. Sodium hydroxide Yellow to red
3. Sugar solution No change

Q.10. Blue litmus paper is dipped in a solution. It remains blue. What is the nature of the solution? Explain.
Ans. (i) It can be identified on the basis of the following observations : Bases change the colour of litmus paper to blue. As the colour of blue litmus paper is not affected, the solution must be basic.


(ii) If the solution is neutral, even then colour of litmus will not change.

Q. 11. Consider the following statements:
(a) Both acids and bases change colour of all indicators.
(b) If an indicator gives a colour change with an acid, it does not give a change with a base.
(c) If an indicator changes colour with a base, it does not change colour with an acid.
(d) Change of colour in an acid and a base depends on the type of the indicator. Which of these statements are correct?
(i) All four (ii) (a) and (d) (iii) (b) and (c) (iv) only (d)
Ans. (ii) (a) and (d)

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RD SHARMA SOLUTION CHAPTER- 33 Binomial Distribution I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 33 Binomial Distribution Exercise Ex. 33.1

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

Required Probability =4547 over 8192Question 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

Also, find the mean and variance of this distribution.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

                                = 0.0256Question 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

Solution 49

Question 50

Find the probability that in 10 throws of a fair die a score which is a multiple of 3 will be obtained in at least 8 of the throws.Solution 50

Question 51

A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.Solution 51

Question 52

The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?Solution 52

Question 53

A factory produces bulbs. The probability that one bulb is defective is   and they are packed in boxes of 10. From a single box, find the probability that

i. none of the bulbs is defective.

ii. exactly two bulls are defective.

iii. more than 8 bulbs work properly.Solution 53

Note: Answer given in the book is incorrect.

Chapter 33 Binomial Distribution Exercise Ex. 33.2

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

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

From a lot of 15 bulbs which include 5 defective, sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence, find the mean of the distribution.Solution 22

Out of 15 bulbs 5 are defective.

begin mathsize 12px style table attributes columnalign left end attributes row cell text Hence ,  the   probability   that   the   drawn   bulb   is   defective   is end text end cell row cell text P end text left parenthesis text Defective end text right parenthesis equals 5 over 15 equals 1 third end cell row cell text P end text left parenthesis text Not   defective end text right parenthesis equals 10 over 15 equals 2 over 3 end cell row cell text Let   X   denote   the   number   of   defective   bulbs   out   of   4. end text end cell row cell text Then ,  X   follows   binomial   distribution   with   end text end cell row cell straight n equals 4 comma text   end text straight p equals 1 third text   and   end text straight q equals 2 over 3 text   such   that end text end cell row cell straight P left parenthesis straight X equals straight r right parenthesis equals straight C presuperscript 4 subscript straight r open parentheses 1 third close parentheses to the power of straight r open parentheses 2 over 3 close parentheses to the power of 4 minus straight r end exponent semicolon straight r equals 0 comma 1 comma 2 comma 3 comma 4 end cell row cell text Mean end text equals sum from straight r equals 0 to 4 of rP left parenthesis straight r right parenthesis equals 1 cross times straight C presuperscript 4 subscript 1 open parentheses 1 third close parentheses open parentheses 2 over 3 close parentheses cubed plus 2 cross times straight C presuperscript 4 subscript 2 open parentheses 1 third close parentheses squared open parentheses 2 over 3 close parentheses squared end cell row cell plus 3 cross times straight C presuperscript 4 subscript 3 open parentheses 1 third close parentheses cubed open parentheses 2 over 3 close parentheses plus 4 cross times straight C presuperscript 4 subscript 4 open parentheses 1 third close parentheses to the power of 4 open parentheses 2 over 3 close parentheses to the power of 0 end cell row cell equals 32 over 81 plus 48 over 81 plus 24 over 81 plus 4 over 81 equals 108 over 81 equals 4 over 3 end cell end table end style

Question 23

A die is thrown three times. Let X be’ the number of twos seen’. Find the expectation of X.Solution 23

Question 24

A die is thrown twice. A ‘success’ is getting an even number on a toss. Find the variance of number of successes.Solution 24

Question 25

Three cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability of the number spades. Hence, find the mean of the distribution.Solution 25

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Chapter – 4 Heat | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter 4 Heat

Q.1. State similarities and differences between the laboratory thermometer and the clinical thermometer.
AnsSimilarities:
(i) Both thermometers consist of long narrow uniform glass tubes.
(ii) Both have a bulb at one end.
(iii) Both contain mercury in bulb.
(iv) Both use Celsius scale on the glass tube.
Differences:
(i) A clinical thermometer reads temperature 35°C to 45°C while the range of laboratory thermometer is -10°C to 110°C.
(ii) Clinical thermometer has a kink near the bulb while there is no kink in the laboratory thermometer.
Due to kink mercury does not fall down on its own in clinical thermometer.

Q.2. Give two examples each of conductors and insulators of heat.
Ans. Conductors—aluminium, iron Insulators—plastic, wood.

Q.3.Fill in the blanks
The hotness of an object is detetmined by its ____________ .
(b) Temperature of boiling water cannot be measured by a ____________ thermometer.
(c) Temperature is measured in degree ____________ .


(d) No medium is required for transfer of heat by the process of ____________.
(e) A cold steel spoon is dipped in a cup of hot milk. It transfers heat to its other end by the process of ____________
(f) Clothes of ___________ colours absorb heat better than clothes of light colours.

Ans. (a) temperature (b) clinical (c) Celsius (d) radiation (e) conduction (f) dark

NCERT Solutions for Class 7 Science Chapter 4 Heat Q4

Q.5. Discuss why wearing more layers of clothing during winter keeps us warmer than wearing just one thick piece of clothing?
Ans.More layers of clothing keep us warm in winters as they have a lot of space between them. This space gets filled up with air. Air is a bad conductor, it does not allow the body heat to escape out.

Q.6. Look at figure 4.6. Mark where the heat is being transferred by conduction, by convection and by radiation
Ans.
NCERT Solutions for Class 7 Science Chapter 4 Heat Q6

Q.7. In places of hot climate it is advised that the outer walls of houses be painted white. Explain.
Ans.In places of hot climate it is advised that the outer wall of houses be painted white because white colour reflects heat and the houses do not heat up too much

Q.8. One litre of water at 30°C is mixed with one litre of water at 50°C. The temperature of the mixture will be:
(a) 80°C (b) More than 50°C but less than 80°C
(c) 20°C (d) Between 30°C and 50°C
Ans.(d) Between 30°C and 50°C.

Q.9. An iron ball at 40°C is dropped in a mug containing water at 40°C. The heat will:
(a) flow from iron ball to water.
(b) not flow from iron ball to water or from water to iron ball.
(c) flow from water to iron ball.
(d) increase the temperature of both.
Ans. (b) not flow from iron ball to water or from water to iron ball

Q.10. A wooden spoon is dipped in a cup of ice-cream. Its other end:
(a) becomes cold by the process of conduction
(b) becomes cold by the process of convection
(c) becomes cold by the process of radiation
(d) does not become cold
Ans.(d) does not become cold.

 

Q.11.Stainless steel pans are usually provided with copper bottoms. The reason for this could be that:
(a) copper bottom makes the pan more durable
(b) such pans appear colourful
(c) copper is a better conductor of heat than the stainless steel
(d) copper is easier to clean than the stainless steel
Ans.(c) copper is better conductor of heat than the stainless steel

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RD SHARMA SOLUTION CHAPTER- 32 Mean and Variance of a Random Variable I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 32 Mean and variance of a random variable Exercise Ex. 32.1

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

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

Four balls are to be drawn without replacement from a box containing 8 red and 4 white balls. If X denotes the number of red balls drawn, find the probability distribution of X.Solution 28

Question 29

The probability distribution of a random variable X is given below:

(i) Determine the value of k

(ii) Determine P (X  2) and P b(X > 2)

(iii) Find P (X  2) + P(X > 2)Solution 29

Chapter 32 Mean and variance of a random variable Exercise Ex. 32.2

Question 1(i)

Find the mean and standard deviation of each of the following probability distributions:

xi : 2 3 4

pi : 2.2 0.5 0.3Solution 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 1(viii)

Solution 1(viii)

Question 1(ix)

Find the mean and standard deviation of each of the following probability distributions:

Solution 1(ix)

Question 2

A discrete random variable X has the probability distribution given below:

X : 0.5 1 1.5 2

P(X) : k k2 2k2 k

(i) Find the value of k.

(ii) Determine the mean of the distribution.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

Three cards are drawn at random (without replacement) from a well shuffled pack of 52 cards. Find the probability distribution of number of red cards. Hence find the mean of the distribution.Solution 17

begin mathsize 12px style table attributes columnalign left end attributes row cell Let text    end text apostrophe straight X apostrophe text    end text be text    end text the text    end text random text    end text variable text    end text which text    end text can text    end text assume end cell row cell values text    end text from text    end text 0 text    end text to text    end text 3. end cell row cell straight P left parenthesis straight X equals 0 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 3 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 2600 over 22100 equals 2 over 17 end cell row cell straight P left parenthesis straight X equals 1 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 1 cross times to the power of 26 straight C subscript 2 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 8450 over 22100 equals 13 over 34 end cell row cell straight P left parenthesis straight X equals 2 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 2 cross times to the power of 26 straight C subscript 1 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 8450 over 22100 equals 13 over 34 end cell row cell straight P left parenthesis straight X equals 3 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 3 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 2600 over 22100 equals 2 over 17 end cell row cell Probability text    end text distribution text    end text of text    end text straight X colon end cell row cell table row cell straight X equals straight x subscript straight i end cell 0 1 2 3 row cell straight p left parenthesis straight X equals straight x subscript straight i right parenthesis end cell cell 2 over 17 end cell cell 13 over 34 end cell cell 13 over 34 end cell cell 2 over 17 end cell end table end cell row blank row cell Mean equals sum from straight i equals 0 to 3 of left parenthesis straight x subscript straight i cross times straight p subscript straight i right parenthesis end cell row cell equals straight x subscript 0 straight p subscript 0 plus straight x subscript 1 straight p subscript 1 plus straight x subscript 2 straight p subscript 2 plus straight x subscript 3 straight p subscript 3 end cell row cell equals 0 cross times 2 over 17 plus 1 cross times 13 over 34 plus 2 cross times 13 over 34 plus 3 cross times 2 over 17 end cell row cell equals fraction numerator 13 plus 26 plus 12 over denominator 34 end fraction end cell row cell equals 51 over 34 end cell row cell equals 3 over 2 end cell row cell equals 1.5 end cell end table end style

Question 18

An urn contains 5 are 2 black balls. Two balls are randomly drawn, without replacement. Let X represent the number of black balls drawn. What are the possible values of X? Is X a random variable? If yes, find the mean and variance of X.Solution 18

Question 19

Two numbers are selected at random (without replacement) from positive integers 2,3,4,5, 6 and 7. Let X denote the larger of the two number obtained. Find the mean and variance of the probability distribution of X.Solution 19

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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

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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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