RD SHARMA Solutions for Class 12-science Maths Chapter 23 - Algebra of Vectors

Chapter 23 - Algebra of Vectors Exercise Ex. 23.1

Question 1(i)

Represent graphically a dispacement of 40 km, 30° east of north.

Solution 1(i)

Question 1(ii)

Represent graphically a displacement of 50 km, south-east

Solution 1(ii)

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

Question 1(iii)

Represent graphically a displacement of 70 km, 40° north of west.

Solution 1(iii)

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

Question 2

Classify the following measures as scalars and vectors.

(i) 15 kg

(ii) 20 kg weight

(iii) 45°

(iv) 10 metres south-east

(v) 50 m/s2

Solution 2

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

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

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

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

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

Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5

Chapter 23 - Algebra of Vectors Exercise Ex. 23.2

Question 1

Solution 1
Question 2

Solution 2

      

Question 3

Solution 3
Question 4
Solution 4
Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

 

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 23 - Algebra of Vectors Exercise Ex. 23.3

Question 1

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

  

Solution 1

  

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

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

Solution 7

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

 

 


Chapter 23 - Algebra of Vectors Exercise Ex. 23.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

 

Solution 3

Question 4

Solution 4

Question 5

Solution 5


Question 6

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

Solution 6

 

  

 

 

  

Chapter 23 - Algebra of Vectors Exercise Ex. 23.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Find |AB| in each case.

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

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

Question 12

Solution 12

Chapter 23 - Algebra of Vectors Exercise Ex. 23.6

Question 1

Solution 1

Question 2

Solution 2

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

Question 3

Solution 3

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

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

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

Solution 18

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

Question 19

  

Solution 19

  

Chapter 23 - Algebra of Vectors Exercise Ex. 23.7

Question 1

Solution 1

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

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

Solution 12

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

Question 13

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

 , - 10, 3), (1 -1, 3) and (3, 5, 3) are collinear.

Solution 13

  

Chapter 23 - Algebra of Vectors Exercise Ex. 23.8

Question 1

Solution 1

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 2 (iii)

Solution 2 (iii)

Question 2 (iv)

Solution 2 (iv)

Question 3 (i)

Solution 3 (i)

Question 3 (ii)

Solution 3 (ii)

Question 4

Solution 4

 

Question 5 (i)

Solution 5 (i)

Question 5 (ii)

Solution 5 (ii)

Question 6 (i)

Solution 6 (i)

Question 6 (ii)

Solution 6 (ii)

Question 7 (i)

Solution 7 (i)

Question 7 (ii)

Solution 7 (ii)

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 23 - Algebra of Vectors Exercise Ex. 23.9

Question 1

Solution 1

Question 2

 

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7 (i)

Solution 7 (i)

Question 7 (ii)

Solution 7 (ii)

Question 7 (iii)

Solution 7 (iii)

Question 8

Solution 8

Question 9

Solution 9

Question 10

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

Solution 10

Question 11

  

Solution 11

  

Question 12

  

Solution 12

  

Chapter 23 - Algebra of Vectors Exercise MCQ

Question 1

If in a Δ ABC, A = (0, 0), B = (3, 3begin mathsize 12px style square root of 3 end style), = (-3begin mathsize 12px style square root of 3 end style, 3), then the vector of magnitude 2begin mathsize 12px style square root of 2 end style units directed along AO, where O is the circumcentre of Δ ABC is

begin mathsize 12px style open parentheses straight a close parentheses space open parentheses 1 minus square root of 3 close parentheses straight i with hat on top space plus space open parentheses 1 plus square root of 3 close parentheses straight j with hat on top
left parenthesis straight b right parenthesis space open parentheses 1 plus square root of 3 close parentheses straight i with hat on top space plus space open parentheses 1 minus square root of 3 close parentheses straight j with hat on top
left parenthesis straight c right parenthesis space open parentheses 1 plus square root of 3 close parentheses straight i with hat on top space plus space open parentheses square root of 3 space minus space 1 close parentheses straight j with hat on top
left parenthesis straight d right parenthesis space none space of space these end style

Solution 1

Correct option:(a)

  

Question 2

Solution 2

Correct option:(c)

 

  

Question 3

  1. C is a mid-point of AB
  2. C divides AB in the ratio 2:1
  3. 3 AC = 5CB
  4. 2 AC = 3CB

 

Solution 3

Correct option: (c)

  

  

 

 

Question 4

Solution 4

Correct option: (d)

  

 

Question 5

  1. 40
  2. -40
  3. 20
  4. -20
Solution 5

Correct option: (b)

  

Question 6

Solution 6

Correct option: (b)

  

Question 7

  1. Null vector
  2. Unit vector
  3. Constant vector
  4. None of these
Solution 7

Correct option: (b)

  

Question 8

Solution 8

Correct option: (c)

  

  

Question 9

Solution 9

Correct option: (b)

  

Question 10

Solution 10

Correct option: (b)

  

Question 11

  1. Rhombus
  2. Rectangle
  3. Square
  4. Parallelogram
Solution 11

Correct option: (d)

  

Question 12

Solution 12

Correct option: (a)

NOTE: Answer not matching with back answer.

Question 13

Solution 13

Correct option: (d)

 

 

  

Question 14

  1. form an isosceles triangle
  2. form a right triangle
  3. are collinear
  4. form a scalene triangle
Solution 14

Correct option: (a)

  

Question 15

  1. (2,-3)
  2. (-2,3)
  3. (-2,-3)
  4. (2,3)
Solution 15

Correct option: (a)

  

Question 16

Solution 16

Correct option: (c)

  

 

  

 

Question 17

Solution 17

Correct option: (d)

  

Question 18

If a and b are two collinear vectors, then which of the following are incorrect?

Solution 18

Correct option: (d)

  

Question 19

In Figure 23.67, which of the following is not true?

  

 

Solution 19

Correct option: (c)

  

Chapter 23 - Algebra of Vectors Exercise Ex. 23VSAQ

Question 1

Solution 1

Question 2

Solution 2

Question 3

Define position vector of a point.

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

If begin mathsize 12px style a with rightwards arrow on top comma space b with rightwards arrow on top comma space c with rightwards arrow on top end style are position vectors of the points A, B and C respectively, write the value of begin mathsize 12px style stack A B with rightwards arrow on top plus stack B C with rightwards arrow on top plus stack A C with rightwards arrow on top. end style

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

I f space a with rightwards arrow on top comma b with rightwards arrow on top comma c with rightwards arrow on top space a r e space t h e space p o s i t i o n space v e c t o r s space o f space t h e space v e r t i c e s space o f space a n space
e q u i l a t e r a l space t r i a n g l e space w h o s e space o r t h o c e n t r e space i s space a t space t h e space o r i g i n comma space t h e n space w r i t e space t h e space v a l u e
o f space a with rightwards arrow on top plus b with rightwards arrow on top plus c with rightwards arrow on top

Solution 16

I n space a n space e q u i l a t e r a l space t r i a n g l e comma space t h e space c e n t r o i d comma space o r t h o c e n t r e comma space i n c e n t r e space a n d space c i r c u m c e n t r e
c o i n c i d e.
S i n c e space t h e space o r t h o c e n t r e space i s space a t space t h e space o r i g i n comma space c e n t r o i d space i s space a l s o space a t space t h e space o r i g i n.
P o s i t i o n space v e c t o r space o f space C e n t r o i d space i s space fraction numerator a with rightwards arrow on top plus b with rightwards arrow on top plus c with rightwards arrow on top over denominator 3 end fraction equals 0 with rightwards arrow on top
T h u s comma space w e space h a v e comma space a with rightwards arrow on top plus b with rightwards arrow on top plus c with rightwards arrow on top space equals space 0 with rightwards arrow on top

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

W r i t e space t w o space d i f f e r e n t space v e c t o r s space h a v i n g space s a m e space m a g n i t u d e.

Solution 35

L e t space u s space c o n s i d e r space t h e space t w o space v e c t o r s space a with rightwards arrow on top space a n d space b with rightwards arrow on top space s u c h space t h a t space
a with rightwards arrow on top equals 3 i with hat on top minus j with hat on top plus 3 k with hat on top space a n d
b with rightwards arrow on top equals i with hat on top plus 3 j with hat on top minus 3 k with hat on top
T h e space m a g n i t u d e space o f space a space i s space open vertical bar a with rightwards arrow on top close vertical bar equals square root of 3 squared plus open parentheses minus 1 close parentheses squared plus 3 squared end root equals square root of 19
T h e space m a g n i t u d e space o f space b space i s space open vertical bar b with rightwards arrow on top close vertical bar equals square root of 1 squared plus 3 squared plus open parentheses minus 3 close parentheses squared end root equals square root of 19
I t space i s space c l e a r space t h a t space space open vertical bar a with rightwards arrow on top close vertical bar equals open vertical bar b with rightwards arrow on top close vertical bar

Question 36

W r i t e space t w o space d i f f e r e n t space v e c t o r s space h a v i n g space s a m e space d i r e c t i o n.

Solution 36

L e t space u s space c o n s i d e r space t h e space t w o space v e c t o r s space a with rightwards arrow on top space a n d space b with rightwards arrow on top space s u c h space t h a t space
a with rightwards arrow on top equals i with hat on top minus j with hat on top plus 3 k with hat on top space a n d
b with rightwards arrow on top equals 2 i with hat on top minus 2 j with hat on top plus 6 k with hat on top
I t space i s space c l e a r space t h a t space space
b with rightwards arrow on top equals 2 i with hat on top minus 2 j with hat on top plus 6 k with hat on top equals 2 open parentheses i with hat on top minus j with hat on top plus 3 k with hat on top space close parentheses equals k a with rightwards arrow on top
T h u s comma space a with rightwards arrow on top space i s space p a r a l l e l space t o space b with rightwards arrow on top space a n d space h e n c e space i n space t h e space s a m e space d i r e c t i o n.

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

begin mathsize 12px style table attributes columnalign left end attributes row cell text If   end text straight a with rightwards arrow on top text  =  end text straight x straight i with hat on top plus 2 straight j with hat on top minus straight z straight k with hat on top text   and   end text straight b with rightwards arrow on top equals text 3 end text straight i with hat on top minus straight y straight j with hat on top plus straight k with hat on top text   are   end text end cell row cell text two   equal   vectors ,  then   write   the   value   of   end text straight x plus straight y plus straight z text. end text end cell end table end style

Solution 44

begin mathsize 12px style table attributes columnalign left end attributes row cell straight a with rightwards arrow on top text  =  end text straight x straight i with hat on top plus 2 straight j with hat on top minus straight z straight k with hat on top text   and   end text straight b with rightwards arrow on top equals text 3 end text straight i with hat on top minus straight y straight j with hat on top plus straight k with hat on top text   are   two   equal   vectors. end text end cell row cell text For   equal   vectors ,  the   components   are   equal. end text end cell row cell text Hence ,  end text straight x equals 3 comma text   end text 2 equals negative straight y comma text   end text minus straight z equals 1 end cell row cell straight x plus straight y plus straight z equals 3 minus 2 minus 1 equals 0 end cell end table end style

Question 45

begin mathsize 12px style table attributes columnalign left end attributes row cell text Write   a   unit   vector   in   the   direction   of   the   sum   of   the   vectors   end text end cell row cell straight a with rightwards arrow on top text  =  2 end text straight i with hat on top plus 2 straight j with hat on top minus 5 straight k with hat on top text   and   end text straight b with rightwards arrow on top equals text 2 end text straight i with hat on top plus straight j with hat on top minus 7 stack straight k. with hat on top end cell end table end style

Solution 45

begin mathsize 12px style table attributes columnalign left end attributes row cell text   end text straight a with rightwards arrow on top text  =  2 end text straight i with hat on top plus 2 straight j with hat on top minus 5 straight k with hat on top text   and   end text straight b with rightwards arrow on top equals text 2 end text straight i with hat on top plus straight j with hat on top minus 7 straight k with hat on top end cell row cell straight a with rightwards arrow on top plus straight b with rightwards arrow on top equals left parenthesis 2 plus 2 right parenthesis straight i with hat on top plus left parenthesis 2 plus 1 right parenthesis straight j with hat on top minus left parenthesis 5 plus 7 right parenthesis straight k with hat on top end cell row cell equals 4 straight i with hat on top plus 3 straight j with hat on top minus 12 straight k with hat on top end cell row cell text Unit   vector   in   the   direction   of   end text straight a with rightwards arrow on top plus straight b with rightwards arrow on top end cell row cell equals fraction numerator straight a with rightwards arrow on top plus straight b with rightwards arrow on top over denominator vertical line straight a with rightwards arrow on top plus straight b with rightwards arrow on top vertical line end fraction equals fraction numerator 4 straight i with hat on top plus 3 straight j with hat on top minus 12 straight k with hat on top over denominator square root of 4 squared plus 3 squared plus 12 squared end root end fraction equals fraction numerator 4 straight i with hat on top plus 3 straight j with hat on top minus 12 straight k with hat on top over denominator square root of 169 end fraction equals 4 over 13 straight i with hat on top plus 3 over 13 straight j with hat on top minus 12 over 13 straight k with hat on top end cell end table end style

Question 46

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   the   value   of  ' end text straight p text '  for   which   the   vectors   end text end cell row cell text 3 end text straight i with hat on top plus 2 straight j with hat on top plus 9 straight k with hat on top text   and   end text straight i with hat on top minus 2 straight p straight j with hat on top plus 3 straight k with hat on top text   are   parallel. end text end cell end table end style

Solution 46

begin mathsize 12px style table attributes columnalign left end attributes row cell text If   two   vectors   end text straight a with rightwards arrow on top text   and   end text straight b with rightwards arrow on top text   are   parallel ,  then   end text straight a with rightwards arrow on top text  =  end text straight lambda straight b with rightwards arrow on top. end cell row cell text Here   end text straight lambda text   is   a   constant. end text end cell row cell text 3 end text straight i with hat on top plus 2 straight j with hat on top plus 9 straight k with hat on top text   and   end text straight i with hat on top minus 2 straight p straight j with hat on top plus 3 straight k with hat on top text   are   parallel. end text end cell row cell text Hence ,  3 end text straight i with hat on top plus 2 straight j with hat on top plus 9 straight k with hat on top equals straight lambda open parentheses straight i with hat on top minus 2 straight p straight j with hat on top plus 3 straight k with hat on top close parentheses end cell row cell rightwards double arrow text 3 end text straight i with hat on top plus 2 straight j with hat on top plus 9 straight k with hat on top equals straight lambda straight i with hat on top minus 2 λp straight j with hat on top plus 3 straight lambda straight k with hat on top end cell row cell text Equating   the   components   we   get end text end cell row cell straight lambda equals 3 text   and   end text minus 2 λp equals 2 rightwards double arrow 6 straight p equals negative 2 rightwards double arrow straight p equals negative 1 third end cell end table end style

Question 47

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   a   vector   end text straight a with rightwards arrow on top text   of   magnitude   5 end text square root of text 2 , end text end root text   making   an   agle   of   end text fraction numerator straight pi over denominator text 4 end text end fraction text   with   x-axis ,  end text end cell row cell fraction numerator straight pi over denominator text 2 end text end fraction text   with   y-axis   and   an   acute   angle   end text straight theta text   with   z-axis. end text end cell end table end style

Solution 47

begin mathsize 12px style table attributes columnalign left end attributes row cell text Vector   end text straight a with rightwards arrow on top text   of   magnitude   5 end text square root of text 2 end text end root text   makes   an   angle   of   end text fraction numerator straight pi over denominator text 4 end text end fraction text   with   the   x-axis. end text end cell row cell text Hence ,  the   component   along   the   x-axis   is   end text 5 square root of 2 text cos end text fraction numerator straight pi over denominator text 4 end text end fraction text  =  end text 5 square root of 2 cross times fraction numerator text 1 end text over denominator square root of text 2 end text end root end fraction text = 5   end text end cell row cell text Vector   end text straight a with rightwards arrow on top text   of   magnitude   5 end text square root of text 2 end text end root text   makes   an   angle   of   end text fraction numerator straight pi over denominator text 2 end text end fraction text   with   the   y-axis. end text end cell row cell text Hence ,  the   component   along   the   y-axis   is   end text 5 square root of 2 text cos end text fraction numerator straight pi over denominator text 2 end text end fraction text  =  end text 5 square root of 2 cross times 0 text   =  0 end text end cell row cell text Vector   end text straight a with rightwards arrow on top text   of   magnitude   5 end text square root of text 2 end text end root text   makes   an   acute   angle   end text straight theta text   with   the   z-axis. end text end cell row cell text Hence ,  the   component   along   the   z-axis   is   end text 5 square root of 2 text cos end text straight theta. end cell row cell left parenthesis text 5 end text square root of text 2 end text end root right parenthesis squared equals 5 squared plus 0 squared plus left parenthesis 5 square root of 2 text cos end text straight theta right parenthesis squared equals 50 equals 25 plus 50 cos squared straight theta end cell row cell rightwards double arrow cos squared straight theta equals 25 over 50 equals 1 half end cell row cell rightwards double arrow cosθ equals fraction numerator 1 over denominator square root of 2 end fraction... left parenthesis because straight theta text   is   acute end text right parenthesis end cell row cell text Hence ,  end text straight a with rightwards arrow on top equals 5 straight i with hat on top plus 5 straight k with hat on top end cell end table end style

Question 48

begin mathsize 12px style table attributes columnalign left end attributes row cell text Write   a   unit   vector   in   the   direction   of   end text PQ with rightwards arrow on top comma text   where   end text straight P text   and   end text straight Q text   end text end cell row cell text are   the   points  ( 1 , 3 , 0 )  and  ( 4 , 5 , 6 )  respectively. end text end cell end table end style

Solution 48

begin mathsize 12px style table attributes columnalign left end attributes row cell straight P text   and   end text straight Q text   are   the   points   with   co-ordinates  ( 1 , 3 , 0 )  and  ( 4 , 5 , 6 )  respectively. end text end cell row cell stack text PQ end text with rightwards arrow on top equals left parenthesis 4 minus 1 right parenthesis straight i with hat on top plus left parenthesis 5 minus 3 right parenthesis straight j with hat on top plus left parenthesis 6 minus 0 right parenthesis straight k with hat on top end cell row cell 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 text Unit   vector   in   the   direction   of   end text stack text PQ end text with rightwards arrow on top equals fraction numerator 3 straight i with hat on top plus 2 straight j with hat on top plus 6 straight k with hat on top over denominator square root of 3 squared plus 2 squared plus 6 squared end root end fraction equals fraction numerator 3 straight i with hat on top plus 2 straight j with hat on top plus 6 straight k with hat on top over denominator square root of 49 end fraction end cell row cell equals 3 over 7 straight i with hat on top plus 2 over 7 straight j with hat on top plus 6 over 7 straight k with hat on top end cell end table end style

Question 49

begin mathsize 12px style table attributes columnalign left end attributes row cell text Find   a   vector   in   the   direction   of   vector   2 end text straight i with hat on top minus 3 straight j with hat on top plus 6 straight k with hat on top text   end text end cell row cell text which   has   magnitude   21   units. end text end cell end table end style

Solution 49

begin mathsize 12px style table attributes columnalign left end attributes row cell text Unit   vector   in   the   direction   of   vector   2 end text straight i with hat on top minus 3 straight j with hat on top plus 6 straight k with hat on top end cell row cell equals fraction numerator text 2 end text straight i with hat on top minus 3 straight j with hat on top plus 6 straight k with hat on top over denominator square root of 2 squared plus 3 squared plus 6 squared end root end fraction equals fraction numerator text 2 end text straight i with hat on top minus 3 straight j with hat on top plus 6 straight k with hat on top over denominator 7 end fraction end cell row cell text Hence ,  a   vector   along   the   above   vector end text end cell row cell text which   has   magnitude   21   units end text equals 21 open parentheses fraction numerator text 2 end text straight i with hat on top minus 3 straight j with hat on top plus 6 straight k with hat on top over denominator 7 end fraction close parentheses end cell row cell equals 6 straight i with hat on top minus 9 straight j with hat on top plus 18 straight k with hat on top end cell end table end style