2 Vector

Vector and Scalar  1 ˆ 1 ˆ

7. The Expression  2 i + 2 j  is a
 
1. Which of the following is vector?
(a) Unit Vector (b) Null vector
(a) Electromotive force (b) Surface Tension
(c) Vector of Magnitude 2 (d) scalar
(c) Weight (d) Focal length
2. Which of the following representation is correct?
 Resolution of Vector
(a) Work ( w ) (b) Force (F)
(c) Energy (E) (d) None of these 8. A vector in x-y plane makes an angle of 30 degree with
3. A. Physical quantity which (does not have direction) must y-axis and magnitude of component of vector is 2 3 The
be scalar magnitude of component y vector will be:
B. Physical quantity which have direction must be vector. 1
(a) 3 (b) 2 (c) 6 (d)
(a) Both are true (b) Both are false 3
(c) A true B false (d) A false B true Rotation of Vector
4. Which one of the following statements is true ?
 
(a) A scalar quantity is the one that is conserved in a process. 9. The vector OA where O is origin is given by OA = 2iˆ + 2 ˆj
o
Now it is rotated by 45 anticlockwise about O. What will
(b) A scalar quantity is the one that can never take negative
values. be the new vector?
(c) A scalar quantity is the one that does not very from one (a) 2 2 ˆj (b) 2 ˆj (c) 2iˆ (d) 2 2iˆ
point to another in space. 
10. A vector A = 2iˆ rotated by 90° then new vector becomes
(d) A scalar quantity has the same value for observers with
different orientations of the axes. (a) 2iˆ (b) 2iˆ + 2 ˆj (c) 2 ˆj (d) 2iˆ + 2 ˆj

Types of Vector Magnitude and Direction of Vector

5. Fill in the blank, by magnitude and angle for a given type of 11. Find Magnitude of Vector
vector.
(i) 2iˆ + 3 ˆj (ii) 3iˆ + 4 ˆj
Vector Types Magnitude (Direction) Angle ˆ
(iii) 3iˆ + 4 ˆj + 5k (iv) iˆ + ˆj + kˆ
(i) Equal vector .................. ..................
(ii) Orthogonal vector .................. .................. (v) 6iˆ − 8 ˆj + 10kˆ (vi) 10iˆ + 10 ˆj − 10kˆ
(iii) Parallel Vector .................. .................. 12. Find magnitude of Vector
(iv) Anti parallel .................. .................. (i) 2iˆ + 2 ˆj + 2kˆ (ii) 3iˆ − 4 ˆj + 5kˆ
vector
(iii) 7iˆ − 24 ˆj + 25kˆ (iv) 10i +10j
6. Which of the following is unit vector
(v) 5kˆ + 12 ˆj (vi) 2iˆ + 2 ˆj
1 ˆ 1 ˆ
(a) sin θiˆ − cos θˆj (b) i+ j (vii) i − j + kˆ (viii) 6i − 8 j + 10kˆ
2 2
(c) iˆ − ˆj (d) Both (a) and (b) (ix) i + 3 j + 2kˆ
 
13. If A maxing an angle α, β and γ with x- axis, y- axis & Draw Vector
z- axis respectively then find sin2a + sin2b + sin2γ
   
14. =A 10iˆ + 10 ˆj find Angle between A & x -axis ?? 24. Draw given vector = 2iˆ − 2 Jˆ ,
A B= −3iˆ − 4 ˆj ,

(a) 45° (b) 30°
(c) 60° (d) 0° C= −3iˆ + 3 ˆj

15. The angles which a vector iˆ + ˆj + 2kˆ makes with X,Y and 25. A= 5iˆ + 10 ˆj draw and write with direction
Z axes respectively are 26. Velocity of object 10m/s at 30o North of East. Draw in vector
(a) 60o, 60o, 60o (b) 45o, 45o, 45o form.

o o
(c) 60 , 60 , 45 o (d) 45o, 45o, 60o 27. B = +3iˆ − 4 ˆj draw & write with direction?
 
16. Given A= 2iˆ + 3 ˆj , the angle between A and Y -axis is Vector Addition
−12 −1 2 2 −1 3
(a) sin (b) cos (c) tan −1 (d) tan 2 28. A set of vectors taken in a cyclic order gives a close polygon,
3 3 3
    then resultant of these vector is:
17. Find F1 + F2 + F3 + F4 = ? (a) Unit vector (b) Null vector
F1 = 8 Njˆ (c) Axial vector (d) Parallel vector
29. Which of the following option is correct for given figure?

F3 = −6 Niˆ 
F2 = 10 Niˆ C 
B

F4 = 5 Njˆ

A
(a) 2iˆ + 3 ˆj (b) 4iˆ + 3 ˆj (c) 4iˆ + 2 ˆj (d) iˆ + 3 ˆj      
    (a) A = B + C (b) B= A + C
18. Two vectors A = 6iˆ and B = 8 ˆj then find A + B ,       

Vector
   (c) C= A + B (d) A + B + C =
0
direction of A + B from A .
   30. Match the following:
19. If A = 2iˆ + ˆj + 2kˆ and B = iˆ + ˆj + 2kˆ then find vector C List-I List-II
 
magnitude same as A and direction along B . →
   C
   →
20. A= 4iˆ + 3 ˆj and B = 4iˆ + 2 ˆj . Find avector parallel to A a. 0
C − A− B = (i) A
→
but has magnitude five times that of B . B

(
(a) 20 2iˆ + ˆj ) (b) 20 4i + 3 j
ˆ ˆ ( )   
→
C
b. 0
A−C − B = (ii) →
→
10 ( 2iˆ + ˆj )
B
(c) 20 ( 2iˆ + ˆj ) (d) A

21. The component of vector A = a xiˆ + a y ˆj + a z kˆ along the →

C
  
direction of iˆ − ˆj is c. 0
B− A−C = (iii) →
→
A B
(a) ax – ay + az (b) ax – ay
→
(
(c) a x − a y / 2 ) (d) (ax – ay + az ) d.
 
A+ B =

−C (iv) →
A
 C →
22. A makes equal angles with X, Y, and Z -axes. Value of its B

components in terms of magnitude of A will be 31. Two vectors of magnitude 3 and 4 acting at different angle
A A as given below
(a) (b) (i) 0o (ii) 60o
3 2
(iii) 90 o (iv) 120o
(c) 3A (d) A
(v) 180o
23. A person pushes a box kept on a horizontal surface with force
 Then find resultant.
of 100 N. In unit vector notation, force F can be expressed 32. Two forces of magnitude 8 N and 15 N respectively act at
(
(a) 100 iˆ + ˆj ) (b) 100 iˆ − ˆj ( ) a point. If the resultant force is 17 N, the angle between the
forces has to be
(c) 50 2 iˆ − ˆj( ) (d) 50 2 iˆ + ˆj ( ) (a) 60o (b) 45o (c) 90o (d) 30o

19
   
33. Three forces given by vectors 2iˆ + 2 ˆj; +2iˆ − 2 ˆj and −4iˆ 42. Three forces F1, F2 and F3 are represented as shown, Each
are acting together on a point object at rest. The object of them is of equal magnitude.
moves along the direction F1
(a) x-axis (b) y-axis 45°
(c) z-axis (d) Object does not move F2
34. Two vector of magnitude 2 then resultant of these two vector
may be?
F3
(a) 2 (b) 8 (c) 5 (d) 6
     Column-I Column-II
35. If R= A + B and R = A + B then angle between A and B
(Combination) (Approximate Direction)
must be
  
(a) 90o (b) 60o (A) F1 + F2 + F3 (p)
(c) 0o (d) 180o
    
36. Which of the following relation is correct between A, B (B) F1 − F2 + F3 (q)
   
and C if C= A + B   
(a) B + A < C < B – A (b) A ≤ C ≥ B (C) F1 − F2 − F3 (r)
(c) A – B ≤ A + B (d) A – B < C < A + B   
(D) F2 − F1 − F3 (s)
37. Two forces of 10 N and 6 N act upon a body. The direction
of the forces are unknown. The resultant forces on the body
(a) A→R, B→Q, C→P, D→S
may be
(b) A→Q, B→P, C→R, D→S
(a) 15N (b) 3 N (c) A→Q, B→R,C→P, D→S
MR Physics

(c) 17 N (d) 2N (d) A→S, B→P,C→R, D→Q

  
38. A force of 6 N and another of 8 N can be applied to produce 43. Which of the following diagram is correct for A + B =
R
→
the effect of a single force equal to. B
→
(a) 1N (b) 10N R →
W

A
P

(c) 16N (d) 0N (a) → (b)

A
39. Two force 5 N and 2 N acting on the object then net force on →
B
→
object must not be R
→ →
(a) 2N (b) 1N B R
→ →
(c) 6N (d) Both (a) & (b) R
(c) →
(d) B →
 
(
40. Find net force F1 + F2 ) A A

44. Vector →A is 2 cm long and is 60o above the x -axis in the


first quadrant. Vector B is 2 cm long and 60o below the x
 
N
=5

-axis in the fourth quadrant. The sum A + B is a vector of

magnitude
2
F

60° = q
(a) 2 along + y – axis (b) 2 along + x – axis
F1 = 5N (c) 1 along – x – axis (d) 2 along – x – axis
45. If vector sum of two unit vector is unit then angle between
them is:
(a) 90o (b) 60o (c) 120o (d) 0o
(a) 5N (b) 5 3N 46. The ratio of maximum and minimum magnitudes of the
  
(c) 5 2 N (d) 0 resultant of two vectors a and b is 3 : 1 Now, a is equal to:
     
41. If angle between A and B is 60o, then find angle between (a) | b | (b) 2 | b | (c) 3 | b | (d) 4 | b |
   
A and − B & − A and − B . 47. At what angle should the two forces 2P and 2P act so
(a) 30o, 60o (b) 120o, 30o that the resultant force is P 10 ?
(c) 120o, 60o (d) 120o, 120o (a) 45o (b) 60o (c) 90o (d) 120o

20
     
48. The resultant of A and B makes an angle a with A and b 54. Six vectors, a through f have the magnitudes and

with B then correct option is directions indicates in the figure. Which of the following
statements is true?
(a) a > b (b) b < a
(c) a < b if A > B (d) a = b
b
49. Four forces of magnitude P, 2P, 3P and 4P act along the four a c
sides of a square ABCD in cyclic order. Find the resultant force:
f
(a) 2P (b) 3 2P d e
(c) 0 (d) 2 2P      
50. If three force acting on the object as shown in figure. Then (a) b + c = f (b) d + c = f
find net force on object.      
2F = F2 (c) d + e = f (d) b + e = f f
55. Resultant of two vectors must be in the plane of Two
vectors.
(a) True (b) False
120o 120o
56. Which of the following combination of three force can give
zero resultant
120o
F3 = F (a) 2, 4 , 7 (b) 3, 1, 5 (c) 2, 8 , 11 (d) 3, 4, 2
F1 = 2F   A
57. A vector A = 2iˆ + 3 ĵ − 4kˆ , then find −3A and .
51. Three force acting on the body as shown in figure. To 2
have(resultant force only along y-axis, magnitude of 58. The sum of the magnitudes of two forces acting at point is
minimum additional force needed is? 18 and the magnitude of their resultant is 12. If the resultant
y is at 90o with the force of smaller magnitude, what are the
magnitudes of forces?
4N 1N

Vector
(a) 12,5 (b) 14,4 (c) 5,13 (d) 10,8
30o    
60o 59. If A = 2i − 4 ˆj + 3kˆ ; B = 4iˆ − 8 ˆj + 6kˆ find Angle b/w A & B
(a) Zero (b) 9iˆ (c) 6iˆ (d) Can't find
  
2N 60. Find the resultant of three vectors OA , OB and OC shown
in the following figure. Radius of the circle is R.
3
(a) 3N (b) N C
2
B
1
(c) 1.5 N (d) N 45o
2 45o
   
52. A = 2iˆ + 2 ˆj − 3kˆ and B = 3iˆ − 3 ˆj − 2kˆ , find A − B A
O
53. A particle moves along a path ABCD as shown in the figure.
Then the magnitude of net displacement of the particle from
position A to D is:
D (a) 2R (b) R (1 + 2)
5m (c) R 2 (d) R ( 2 − 1)
61. Assertion (A): Minimum number of vectors having unequal
37o
C magnitude in a plane required to give zero resultant is three.
Reason (R): If vector addition of three vectors is zero, then
they must lie in a plane
4m
(a) If both Assertion (A) & Reason (R) are True & the
Reason (R) is a correct explanation of the Assertion (A)
A 3m B (b) If both Assertion (A) & Reason (R) are True but Reason
(R) is not a correct explanation of the Assertion (R)
(a) 10 m (b) 5 2m
(c) If Assertion (A) is True but the Reason (R) is False
(c) 9 m (d) 7 2m (d) If both Assertion (A) & Reason (R) are false.

21
   71. Three forces F1, F2 and F3 together keep a body in
62. Two vectors A and B are acting in the same plane and
 equilibrium. If F1 = 3 N along the positive x– axis, F2 = 4 N
vector C is perpendicular to the plane. The resultant of along the positive y-axis, then the third force F3 is
these vectors - 3
(a) 5 N making an angle θ = tan −1   negative y-axis
(a) May be zero 4
(b) can not be zero −1  4 
  (b) 5 N making an angle θ = tan   with negative
(c) lies between A and B y-axis 3
 
(d) lies between A and − B
(c) 7 N making an angle θ = tan −1  3  with negative y -
63. If the Result of n forces of different magnitudes acting at a 4
axis
point is zero, then the minimum value of n is
4
(a) 1 (b) 2 (d) 7 N making an angle θ = tan −1   with negative
(c) 3 (d) 4 y-axis 3
64. If resultant of two vector makes 30o and 60o with these
72. In the given diagram, if PQ = A, QR = B and RS = C, then
vectors and has a magnitude of 40 units, then find magnitude
PS equals
of these vectors. Q

P
60o
→
R R
→
B S
(a) A – B + C (b) A + B – C
30o 90o (c) A + B + C (d) A – B – C
→  
73. Tl two vectors A and B are draw from a common Int and
MR Physics

A
65. Which of the following set of 3 force can keep the object in     
C= A + B , then angle between A and B is
equilibrium?
(i) 90o if C2 = A2 + B2
(a) 3, 4, 8 (b) 2, 5, 1 (ii) Greater than 90º if C2 < A2 + B2
W

(c) 7, 3, 11 (d) 4, 5 ,8 (iii) Greater than 90º if C2 > A2 + B2

P

66. Two vectors acting through a point are in the ratio 3 : 5. If (iv) Less than 90º if C2 > A2 + B2
the angle between them is 60° and the magnitude of their Correct option are –
resultant is 35, find the magnitude of vectors.
(a) 1, 2 (b) 1, 2, 3, 4
67. Two forces acting in opposite direction have resultant 10 N (c) 2,3,4 (d) 1, 2, 4
and when acting perpendicularly have resultant 50 N. Find   
74. Let C= A + B .
the magnitude of forces. 
      (a) | C | is always greater than
68. If a , b and c be three vectors such that a + b + c =0 and    
     (b) It is possible to have | C |<| A | and | C |<| B |
=| a 3,= b 5, c |= 7 | , find the angle between a and b
(c) C is always equal to A + B
69. A particle is being acted upon by four forces of 30 N due
(d) C is never equal to A + B
east, 20 N due north, 50 N due west and 40 N due south.
75. A particle is being acted upon by four forces of 30 N due
The resultant force will be east, 20 N due north, 50 N due west and 40 N due south.
(a) 20 2 N,60 south to west The resultant force will be

(b) 20 2 N, 45 south west (a) 20 2 N,60 south to west

(c) 20 2 N, 45 north to east (b) 20 2 N, 45 south west
(c) 20 2N, 45 north to east
(d) 20 2 N, 45 south to east 
(d) 20 2 N, 45 south to east
70. How many minimum numbers of non-zero vectors in
76. Two vectors of magnitudes 4 and 6 are acting through a
different planes can be added to give zero resultant?
point. If the magnitude of the resultant is R
(a) 2 (b) 3 (a) 4 < R < 6 (b) 4 < R < 10
(c) 4 (d) 5 (c) 2 ≤ R ≤10 (d) 2 ≥ R ≥10

22
 77. The resultant of A and B makes an angle a with A and b
with B, Vector Subtraction
(a) a < b if A > B (b) a < b if A < B
82. Two vector of same magnitude and angle between them is θ
(c) a > b if A > B (d) a < b if A = B then find magnitude of vector subtraction.
     
   83. The vectors A and B are such that A + B = A − B . The
78. Figure shows three vectors a , b , and c , here R is the mid
point of PQ. Which of the following relations is correct? angle between the two vectors is
P (a) 45o (b) 90o (c) 60o (d) 75o
   
84. The Resultant of A + B is R1 On reversing the vector B ,
→   
a R the resultant becomes R 2 what is the value of R12 + R 22
→
c (a) A2 + B 2 (b) A 2 − B2

O →
b
Q ( 2
(c) 2 A + B

2

) 2
(
(d) 2 A − B
2
)
      85. Two vectors A and B have equal magnitudes. If Magnitude
2c    
(a) a + b =c (b) a + b = of A + B is equal to two times the magnitude of A − B ,
       
(c) a − b =
c (d) a − b =2c then angle between A and B will be:
−1 −1
79. If two equal magnitude force F0 are acting on particle and (a) sin ( 3 / 5 ) (b) sin (1 / 3)
angle between them is given in column-I and their resultant −1 −1
in column-II then match the column and select correct (c) cos ( 3 / 5 ) (d) cos (1 / 3)
option. 86. If the sum of two unit vectors is a unit vector, then the
magnitude of their difference is:
Column-I Column-II 1
(a) 3 (b) 2 (c) 5 (d)
(A) 0o (P) 2
F0 3 87. Which of the following relations is True for two unit vectors
 
(B) 90o (Q) F0  A B

Vector
  and   making an angle θ to each other?
(C) 60o (R) 2F0  A B
(D) 120o (S) ˆ +Bˆ = Â − B
ˆ tan(θ / 2)
F0 2 (a) A

A B C D ˆ −B
ˆ = Â + B
ˆ tan(θ / 2)
(b) A
(a) R Q P S
(b) R P S Q
ˆ +B
(c) A ˆ −B
ˆ= A ˆ cos θˆ / 2( )
(c) R S Q P (d) A ˆ −Bˆ =A ˆ +B ˆ cos ( θ / 2 )
(d) R S P Q     
   88. For two vector A and B vector addition R= A + B and
80. Assertion (A): If vector A, B and C are coplanar, then     
   vector subtraction D= A − B then angle between A and B .
A + B + C must be zero.
   Condition
Reason (R): C always cancel resultant of A and B .     
Choose correct alternative : (i) A + B = R = A = B .......
   
(a) If both are correct and Reason (R) is correct explanation (ii) A= B= A − B .......
for Assertion (A).  
(iii) A2 + B 2 =| A − B |2 =D 2 .......
(b) If both are correct and Reason (R) is not correct
explanation for Assertion (A). (iv) A2 + B 2 = R 2 .......
   
(c) If Assertion (A) is correct, Reason (R) is incorrect.
(v) | A + B | = | A − B |= θ .......
(d) If both Assertion (A) and Reason (R) are incorrect.    
81. What displacement must be added to the displacement (vi) A + B = A + B .......
( )
−9iˆ + 3 ˆj to produce a net displacement of 6 m, pointing
   
(vii) A − B = A + B .......
in the negative x-direction?    
(a) 3iˆ + 3 ˆj (b) 3iˆ − 3 ˆj (viii) A + B = A − B .......
   
(c) −3iˆ + 3 ˆj (d) −3iˆ − 3 ˆj (ix) A − B = A − B .......

23
       99. A particle moving with a speed of v changes direction by an
89. The vectors A and B are such that A + B = A = B , then
angle q, without change in speed.
 
A − B may be equated to (i) The change in the magnitude of its velocity is zero.
(ii) The change in the magnitude of its velocity is 2 v sin q/2.
3  
(iii) The magnitude of the change in its velocity is 2 v sin q/2.
(a) A (b) A
2
(iv) The magnitude of the change in its velocity is v
(c) 3 2 A (d) 3 A (1 – cos q).
  (a) (i), (ii) (b) (i), (iii)
90. When vector A = 2iˆ + 3 ĵ + 2kˆ is subtracted from vector B ,
 (c) (i), (ii), (iii) (d) all
it gives a vector equal to 2 ĵ . Then magnitude of vector B
100. A truck travelling due north at 20 m/s turns west and travels
will be:
at the same speed. What is the change in the velocity?
(a) 3 (b) 33 (a) 40 m/s NW (b) 40 m/s SW
(c) 6 (d) 5 (c) 20 2 m / s NW (d) 20 2 m / s SW
91. Majnu Majedar is moving with 6 m/s in east and Ramlal is 101. A particle moves towards east with a speed 5 m/s. After 10
moving with 6 m/s at 30o east of North, then find relative s its direction changes towards north with the same speed.
velocity of Ramlal w.r.t. MJ. The average acceleration of the particle is
92. Initial velocity of Ramlal is 5 m/s in north after some time it 1
(a) Zero (b) m / s2 N − W
is moving 5 m/s in east then find 2
(i) Change in velocity 1 1
(c) m / s2 N − E (d) m / s2 S − W
(ii) Magnitude of change in velocity 2 2
(iii) Change in magnitude of velocity
   
 102. For two vectors A and B, A + B = A - B is always true
93. If initial velocity of object u= 3iˆ + 4 ˆj and after some
MR Physics

 when :
time it’s velocity v= 4iˆ + 3 ˆj then final
(i) Magnitude of change in velocity (i) |A| = |B| ≠ 0
(ii) Change in Magnitude of velocity (ii) A ⊥ B
 
W

 
( ) (iii) |A| = |B| ≠ 0 and A and B are parallel or antiparallel
P

94. If two vector A & B such that A + B is perpendicular to

 (iv) when either |A| or |B| is zero.
  A
the difference A − B then ratio of magnitude  (a) Only (ii) is correct.
B
(b) Only (iv) is correct.
95. Find position vector of A(1,2,1) w.r.t. point B(-1, 2, 0) and
(c) Both (ii) and (iv) are correct.
(1,4,5) w.r.t. (2,3,5).
(d) (i), (ii) and (iv) are correct.
96. If a student moves as given below:
(i) 50 m in East
Lami’s Theorem
(ii) 20 m in North
(iii) 40 2 m in S-E direction. Find net Displacement? 103. As shown in figure tension in the horizontal cord is 30N.
    The weight W and tension in the string OA in Newton are
97. If A + B = A − B then which of the following option will be
correct? A

(a) Magnitude of A may be zero. T1
 30o
(b) Magnitude of B must be zero.
 
(c) Angle between A and B must be 90o
  30N
(d) Angle between A and B may be 90o O
98. The length of the second’s hand in a watch is 1 cm. The
magnitude of the change in the velocity of its tip is 15 s is W
(a) Zero (b) π / 30 cm / s (a) 30 3,30 (b) 30 3,60
(c) ( π / 30 ) 2 cm / s (d) π / 30 2 cm / s (c) 60 3,30 (d) None of these

24
   
0 find value of F3?
104. If F1 + F2 + F3 = Application of Dot Product
F3
   
113. If A= iˆ, B= iˆ + 3 ˆj , then find angle between A and B
  
45o 114. If A = 2iˆ − 2 ˆj then find angle between A
= 2iˆ + 2 ˆj and B
F1 = 50 N 
and B
   
115. If A = 2iˆ + 6 ĵ + 3kˆ and B = 4iˆ . Find angle between A and B .

F2 = 50 N 116. The angle between the two vectors A = 3iˆ + 4 ˆj + 5kˆ and

B = 3iˆ + 4 ˆj − 5kˆ will be
Multiplication of Vector with Scalar (a) 90o

(b) 180o (c) Zero (d) 45o

  A 0.5iˆ + 0.4 ˆj − αkˆ then find a if A is unit vector.
117. If =
105. If A and αA will be parallel to each other then a must be 
(a) Positive (b) Negative 118. If=A 0.6iˆ + βˆj is a unit vector then, find value of ‘b’
(c) Zero (d) None of these (a) 0.4 (b) 0.8 (c) 0.7 (d) 0.6
  
106. A is a vector of magnitude 2.7 units due east. What is the ωt ωt ˆ
 119. If vector A = cosωt iˆ + sinωt ˆj =
and B cos iˆ + sin j
magnitude and direction of vector 4A ? 2 2
are function of time, then the value of t is at which they are
1 units due east
(a) 4 units due east (b) 4 units due west orthogonal to eachother is
(c) 2.7 units due east (d) 10.8 units due east (a) t = p/w (b) t = 0
(c) t = p/4w (d) t = p/2w
Dot Product ˆ
120. If a vector 2iˆ + 3 ˆj + 8k is perpendicular to the vector
3 ˆj − 4iˆ + αkˆ , then the value of a is

Vector
   

107. If A = 2 and B = 4, θ= 60 then find A ⋅ B .
(a) 1/2 (b) -1/2 (c) 1 (d) -1
   
108. If A = 2iˆ + 3 ˆj − 4kˆ , B =−2iˆ + 4 ˆj + 2kˆ , then find A ⋅ B (e) -1/8
  
109. If A & B is perpendicular to each other (orthogonal) then 121. What will be the projection of Vector A = iˆ + ˆj + kˆ on

find their dot product Vector B= iˆ + ˆj ?

B
(
(a) 2 iˆ + ˆj + kˆ ) (b) iˆ + ˆj ( )
(c) (
2 iˆ + ˆj ) ( ˆ
(d) 2 iˆ + ˆj + k )
90
( ) (
122. A particle moves from point −2iˆ + 5 ˆj to 4iˆ + 3 ˆj when )

A ( )
a force of 4iˆ + 3 ˆj N is applied. How much work has been
done by force?
110. Which of the following vectors is/are perpendicular to the
(a) 5j (b) 2j (c) 18j (d) 11j
vector 4iˆ − 3 ˆj ?
123. A body, constrained to move in y - direction, is subjected to
(a) 4iˆ + 3 ˆj (b) 6iˆ 
( )
a force given by F = −2iˆ + 15 ˆj + 6kˆ N. The work done by
(c) 7 k̂ (d) 3iˆ − 4 ˆj this force in moving the body through a distance of along
    
111. Three vectors A, B and C satisfy the relation A ⋅ B = 0 10 ĵ m y -axis, is
 
and A ⋅ C =0 . The vector A is parallel to (a) 105 J (b) 20J (c) 190 J (d) 150 J
 
      124. Two forces F1 =i + 2 j − 2k and F2 = 2iˆ + 2 ˆj + 3kˆ are
ˆ ˆ ˆ
(a) B (b) C (c) B ⋅ C (d) B × C
acting on a particle and it’s displacement is −iˆ + 2 ˆj + kˆ .
(
112. Vector which is perpendicular to acosθiˆ + bsinθˆj is ) Find work done on the particle
1 1 (a) 2J (b) 6J (c) -3J (d) Zero
(a) bcosθiˆ − asinθˆj (b) sinθiˆ − cosθˆj   
a b ˆ ˆ ˆ ˆ ˆ ˆ
125. If a =2i + 2 j + k , b =−i + j + 2k and c
= 3iˆ + ˆj such that
(c) 5k̂ (d) All of these  
a + λb̂ is perpendicular to c , find λ.

25
      
Cross Product 133. a = 3 ˆj − 4kˆ and c = iˆ + ˆj + kˆ . Find (a × b ) ⋅ c
3iˆ + 2kˆ, b =

 134. Two adjacent sides of a parallelogram are represented by

126. Position of particle is given by r =iˆ + 2 ˆj − kˆ and
 the two vectors iˆ + 2 ˆj + 3kˆ and 3iˆ − 2 ˆj + kˆ . What is the
momentum p = 3iˆ + 4 ĵ − 2kˆ . The angular momentum is
area of parallelogram?
perpendicular to
(a) x - axis (a) 8 (b) 8 3
(b) z - axis
(c) 3 8 (d) 192
(c) y - axis   
(d) line at equal to all three axis 135. If A × B =
C , then which of the following statements is wrong
    
127. If F is the force acting on particle having position vector (a) C ⊥ A (b) C ⊥ B
 
r and τ to the torque of this force about origin, then      
       
(a) r ⋅ τ > 0 and f ⋅ τ < 0 (b) r ⋅ τ = 0 and f ⋅ τ ≠ 0
(
(c) C ⊥ A + B ) (d) C ⊥ A × B ( )
     136. The magnitude of the vector product of two vectors A and B
(c) r ⋅ τ = 0 and f ⋅ τ = 0 (d) τ ⋅ τ ≠ 0 and f ⋅ τ = 0
  may not be
128. If A and B are two vectors satisfying the relation
      (a) greater than AB (b) equal to AB
A ⋅ B = | A × B | Then the value of | A − B | will be:
(c) less than AB (d) equal to zero
(a) 2 2
A + B + 2 AB (b) A2 + B 2  
137. A vector A points vertically upward and B points towards
 
(c) A2 + B 2 − 2 AB (d) A 2 + B2 + 2AB north. The vector product A × B is
 
129. A is a vector Quantity such that | A |= non zero constant (a) Zero (b) Along west

which of the following expression is true for A ? (c) Along east (d) Vertically downward
       
(a) A ⋅ A = 0 (b) A × A = 0 ( ) (
138. The value of A + B × A − B is )
MR Physics

   
(c) A × A < 0 (d) A × A > 0    
    (a) 0 (b) A2 − B 2 (c) B × A (d) 2( B × A)
130. Assertion (A): A × B is perpendicular to both | A + B | as
  139. The scalar product of two vectors is 2 3 and the magnitude
well as | A − B |
    of their cross product is 2. The angle between them is
Reason (R): A + B as well as A − B lie in plane
W
P

    (a) 30o (b) 45o (c) 60o (d) 90o

containing A and B , but A × B lies perpendicular to the
   
plane containing A and B . 140. A vector A points vertically upward and B points
(a) Assertion (A) is false but Reason (R) is true  
towards north. The vector product A × B is
(b) Both Assertion (A) and Reason (R) are true but Reason
(R) is NOT the correct explanation of Assertion (A) (a) Zero (b) Along west
(c) Assertion (A) is true but Reason (R) is false (c) Along east (d) Vertically downward
(d) Both Assertion (A) and Reason (R) are true and Reason    
141. If P × Q = P × R then which of the following is correct?
(R) is the correct explanation of Assertion (A)  
 
131. For three vector A = (− xiˆ − 6 ˆj − 2kˆ), B = (−iˆ + 4 ˆj + 3kˆ) (a) Q must be equal to R
     
and C =(−8iˆ − ˆj + 3kˆ) if A.( B × C ) = 0 then value of x (b) | Q | must be equal to | R |
is__________.
      (c) There may be infinite solution for the equation
132.=If | a | = 26, | b | 7 , and | a × b | =35 . Find a ⋅ b
(d) Only one solution is possible

26
 Answer Key
1. (c) 2. (c) 3. (c) 4. (d)

5. Magnitude: (i) Must be equal, (ii) May be equal, (iii) May be equal, (iv) May be equal, Direction Angle: (i) 0o, (ii) 90o, (iii) 0o, (iv) 180o

6. (d) 7. (a) 8. (b) 9. (a) 10. (c) 11. (i) 13 , (ii) 5, (iii) 5 2 , (iv) 3 , (v) = 10 2 , (vi) 10 3

12. (i) 2 3 , (ii) 5 2 , (iii) 25 5 , (iv) 10 2 , (v) 13, (vi) 2, (vii) 3 , (viii) 10 2 , (ix) 2 2 13. 2 14. (a) 15. (c)

 3 ˆ ˆ
16. (c) 17. (b) 18. 6iˆ + 8 ˆj ,53 C
19. =
6
( )
i + 3 + 2kˆ 20. (b) 21. (c)

y N
x –x C =3 2 y A=11.1 u =10 m/s
22. (a) 23. (c) 24. 45o 53o 25. 26.

45o tan–1(2) 30o

–y B=5 –y x x E
A=2 2
3
27. x 28. (b) 29. (d) 30. a-iv, b-iii, c-i, d-ii 31. (i) 7, (ii) 37 , (iii) 5, (iv) 13 , (v) 1
53o
4
5

–y

Vector
32. (c) 33. (d) 34. (a) 35. (c) 36. (c) 37. (a) 38. (b) 39. (d) 40. (a) 41. (c)

42. (c) 43. (d) 44. (b) 45. (c) 46. (b) 47. (a) 48. (c) 49. (d) 50. Fnet = F 51. (d)

 A ˆ
52. −iˆ + 5 ˆj − kˆ 53. (d) 54. (c) 55. (a) 56. (d) 57. −3 A = ˆ
−6i − 9 j + 1=ˆ
2k , i + 1.5 ˆj − 2kˆ
2
58. (c) 59. (a) 60. (b) 61. (b) 62. (b) 63. (c) 64. A = 20 3 , B = 20 65. (d) 66. (25, 15)

67. (30N, 40N) 68. 60° 69. (b) 70. (c) 71. (c) 72. (c) 73. (d) 74. (b) 75. (b)

θ
76. (c) 77. (a) 78. (b) 79. (d) 80. (d) 81. (b) 82. D = 2 A sin   83. (b) 84. (c)
2
85. (c) 86. (a) 87. (b) 88. (i) 120°, (ii) 60°, (iii) 90°, (iv) 90°, (v) 90°, (vi) 0, (vii) 180°, (viii) 180°, (ix) 0° 89. (d)

90. (b) 91. 6 m/s 92. (i) 5 2 (South - East), (ii) 5 2 (iii) 0 93.   (i) 5 (ii) 0 94. Division of vectors is not allowed

95. 5, 2 96. 90iˆ − 20 ˆj 97. (b) 98. (c) 99. (b) 100. (d) 101. (b) 102. (c) 103. (b)

104. 50 2 105. (a) 106. (d) 107. 4 108. 0 109. 0 110. (d) 111. (d) 112. (d) 113. 60°

−1 2
114. 90° 115. cos   116. (a) 117. α = 0.59 118. (b) 119. (a) 120. (e) 121. (b)
7
122. (c) 123. (d) 124. (b) 125. l = 4 126. (a) 127. (c) 128. (c) 129. (b) 130. (d) 131. x = 4

132. 7 133. (15) 134. (b) 135. (d) 136. (a) 137. (b) 138. (d) 139. (a) 140. (b) 141. (c)

27