Chapter 24 Scalar Or Dot Product Exercise Ex. 24.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5 (i)

Solution 5 (i)

Question 5 (ii)

Solution 5 (ii)

Question 5 (iii)

Solution 5 (iii)

Question 5 (iv)

Solution 5 (iv)

Question 5 (v)

Solution 5 (v)

Question 6

Solution 6

Question 7(i)

Solution 7(i)

Question 7(ii)

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

Solution 7(ii)

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

Question 8 (i)

Solution 8 (i)

Question 8 (ii)

Solution 8 (ii)

Question 9

Solution 9

Question 10

Solution 10

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

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

Question 11

Solution 11

Question 12

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

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

Question 13

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

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

Question 14

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

Solution 14

Question 15

Solution 15

Question 16

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

Solution 16

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

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

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

Question 30(i)

Solution 30(i)

Question 30(ii)

Solution 30(ii)

Question 31(i)

Solution 31(i)

Question 31(ii)

Solution 31(ii)

Question 31(iii)

Solution 31(iii)

Question 32(i)

Solution 32(i)

Question 32(ii)

Solution 32(ii)

Question 32(iii)

Solution 32(iii)

Question 33(i)

Solution 33(i)

Question 33(ii)

Solution 33(ii)

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

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

Solution 49

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

Chapter 24 Scalar Or Dot Product Exercise Ex. 24.2

Question 1

Solution 1

Question 2

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

Question 3

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

Question 4

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

Question 5

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

Question 6

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

Question 7

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

Question 8

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

AB2 + AC2 = 2 (AD2 + CD2).Solution 8

Question 9

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

Question 10

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


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