Motion in Plane
(a)
1.
Scalar and Vector Quantity
3.
(a) 90º
(c) 45º
(b) 8
(d) –8
Ans. (a) : Given,
| A + B |=| A − B |
Squaring both sides,
EAMCET-1991
| A + B |2 =| A − B |2
A 2 + B2 + 2A ⋅ B = A 2 + B2 − 2A ⋅ B
4A ⋅ B = 0
A⋅B = 0
| A || B | cos θ = 0
For perpendicular vectors
A⋅B = 0
5iˆ + 7ˆj − 3kˆ ⋅ 2iˆ + 2ˆj − ckˆ = 0
)(
)
cos θ = 0
10 + 14 + 3c = 0
θ = 90°
24 = – 3c
4.
When two vectors A and B of magnitude a
c = –8
and b are added, the magnitude of the resultant
The resultant of the vectors A and B depends
vector is always
also on the angle θ between them. The
(a) equal to (a + b)
magnitude of the resultant is always given by
(b) less than (a + b)
(a) A + B + 2AB cos θ
(c) greater than (a + b)
(b) ( A + B + 2AB cos θ )
(d) not greater than (a + b)
EAMCET-1993
2
2
(c) A + B + 2AB cos θ
Ans. (d) : Given,
2
(d) A 2 + B2 + 2ABcos θ
| A |= a, B = b
(
)
EAMCET-1992
Ans. (c) :
| A + B |= a 2 + b 2 + 2ab cos θ
| A + B |max = a 2 + b 2 + 2ab
[For max, θ = 0]
| A + B |max = (a + b)
Hence, magnitude of resultant vector is not greater than
(a + b)
5.
If a unit vector is represented by
0.5iˆ + 0.8jˆ + ckˆ , the value of c is
(a) 1
Resultant vector, R
From ∆DOF,
(OD)2 =(OF)2 + (DF)2
(OD)2 = (OE + EF)2 + (DF)2
R 2 = ( A + B cosθ)2 + ( B sinθ)2
= A2 + B2 cos2θ + 2AB cosθ + B2 sin2θ
= A2 + B2 (cos2θ + sin2θ) + 2AB cosθ
R 2 = A2 + B2 + 2AB cosθ
∴
0.11
0.011
ˆ |= 1
|A
345
If θ is the angle made by the vector with x–axis than,
A
1
cos θ = x ⇒ cos θ =
2
|A|
θ = 45°
9.
7.
The angle between two vectors 6iˆ + 6jˆ - 3kˆ and
(
)(
( a ˆi + a ˆj + a kˆ ).( ˆi − ˆj) = a − a
x
y
z
(1) + ( −1)
2
x
2
y
2
)
(
x
(
y
→
B = 210iɵ + 280kɵ
→
C = 5.1iɵ + 6.8jɵ
)
Objective Physics Volume-I
)
z
→
EAMCET-2008
YCT
=
For vectors A and B making an angle θ which
one of the following relations is correct?
(a) A × B = B × A
(b) A × B = ABsin θ
7iˆ + 4jˆ + 4kˆ is given by
(c) A × B = ABcos θ
(d) A × B = − B × A
 1 
 5 
DCE-2009
(a) cos −1 
(b) cos −1 


Ans. (d) : We know that,
 3
 3
Cross product of vectors A and B
 5
 2 
(c) sin −1 
(d) sin −1 
A × B = ABsin θ



3
 3


Cross product of vectors B and A
EAMCET-1999
B× A = −BA sin θ
Ans. (d) : Given that,
So,
A × B = −B× A
A = 6iˆ + 6ˆj − 3kˆ
10. Given
two
vectors
A = −ˆi + 2jˆ − 3kˆ and
ˆ
ˆ
ˆ
B = 7i + 4 j + 4k
B = 4iˆ − 2jˆ + 6kˆ . The angle made by (A + B)
A ⋅ B =| A || B | cos θ
with x-axis is
(a) 30°
(b) 45°
6iˆ + 6ˆj − 3kˆ ⋅ 7iˆ + 4ˆj + 4kˆ
(c) 60°
(d) 90°
AP
EAMCET(Medical)-2007


=  36 + 36 + 9 ⋅ 49 + 16 + 16  cos θ


Ans. (b) : A = −ˆi + 2ˆj − 3kˆ
42 + 24 −12 = 81 81.cos θ = 9×9cos θ
B = 4iˆ − 2jˆ + 6kˆ
54 = 81cos θ
ˆ + (4iˆ − 2ˆj + 6k)
ˆ
A + B = (−ˆi + 2jˆ − 3k)
54
cos θ =
81
A + B = 3iˆ + 0ˆj + 3kˆ
6 2
α is angle with x–axis
cosθ = =
9 3
x − component of A + B
2
cosα =
sin 2θ = 1− cos 2θ = 1−  2 
|A+B|
 
 3
3
3
cosα =
=
4 5
2
sin θ = 1− =
9+0+9 3 2
9 9
1
cosα =
5
2
sinθ =
3
α = 45o
11. Of the vectors given below, the parallel vectors
5
θ = sin −1
are,
3
→
ˆ
ˆ
ˆ
A = 6iɵ + 8jɵ
8.
The component of vector A = a i + a j + a k
along the direction of ˆi - ˆj is
(a) ax – ay + az
(b) ax – ay
(c) a x − a y / 2
(d) (ax + ay + az)
0.52 + 0.82 + c2 = 1
c2 = 0.11
c = 0.11
R = A 2 + B2 + 2ABcos θ
Objective Physics Volume-I
(b)
(d) 0.39
TS-EAMCET-10.09.2020, Shift-1
EAMCET-1994
Ans. (b) : Â = 0.5iˆ + 0.8ˆj + ckˆ
(c)
( )
Ax = 1, Ay = 1
(b) 60º
(d) 0º
EAMCET-1993
Ans. (d) : A = 5iˆ + 7ˆj − 3kˆ
B = 2iˆ + 2ˆj − ckˆ
(
A and B are vectors such that A + B =
A - B . Then, the angle between them is
If A, B are perpendicular vectors
A = 5iˆ + 7jˆ - 3kˆ
B = 2iˆ + 2jˆ - ckˆ .
The value of c is
(a) –2
(c) –7
2.
The angle made by the vector A = ˆi + ˆj with x- Ans. (c) : Given,
A = a x ˆi + a y ˆj + a z kˆ
axis is
(a) 90º
(b) 45º
Let
B = ˆi − ˆj
(c) 22.5º
(d) 30º
Component of vector A along any vector B
EAMCET-1996
A.B
Ans. (b) : Given that,
=
B
A = ˆi + ˆj
Component of vector A = a x ˆi + a y ˆj + a z kˆ along
2
2
| A |= 1 + 1 = 2
B = ˆi − ˆj
6.
03.
346
D = 3.6iɵ + 6jɵ + 48kɵ
YCT
→
→
→
→
Ans. (a): A = 3 ˆi + ˆj
Angle with x–axis :
(c) A and D
(d) C and D
A
AP EAMCET(Medical)-2006 cosθ = x
|A|
Ans. (b) : If component of vector is same then vectors
will be same.
3
3
cosθ =
=
A = 6iˆ + 8jˆ
2
2
2
3 + (1)
B = 210iˆ + 280kˆ
θ = 30°
C = 5.1iˆ + 6.8jˆ
(a) A and B
→
(b) A and C
→
→
→
( )
π
D = 3.6iˆ + 6ˆj + 48kˆ
θ=
6
1.7
∵ C = 5.1iˆ + 6.8jˆ =
6iˆ + 8jˆ
14. The unit vector parallel to resultant of the
2
→
Hence, it is clear that A and C are parallel and we can
vectors A = 4iˆ + 3jˆ + 6kˆ and B = −ˆi + 3jˆ − 8kˆ is:
1.7
1 ˆ ˆ
1 ˆ ˆ
write as, C =
A
3i + 3j − 2kˆ
(b)
3i + 6j − 2kˆ
(a)
2
7
7
This implies that A is parallel to C.
1 ˆ ˆ
1
(c)
3i + 6 j − 2kˆ
(d)
3iˆ − 6ˆj + 2kˆ
12. A vector Q which has a magnitude of 8 is
49
49
AP EAMCET(Medical)-2000
added to the vector P , which lies along the Xaxis. The resultant of these two vectors is a Ans. (b) : A = 4iˆ + 3jˆ + 6kˆ
third vector R , which lies along the Y-axis and
B = −ˆi + 3jˆ − 8kˆ
has a magnitude twice that of P . The
ˆ + (−ˆi + 3jˆ − 8k)
ˆ
R = A + B = (4iˆ + 3jˆ + 6k)
magnitude of P is:
6
8
ˆ
ˆ
ˆ
R = 3i + 6 j − 2k
(b)
(a)
5
5
R
12
16
unit vector, R̂ =
(c)
(d)
|R|
5
5
AP EAMCET(Medical)-2004
| R |= 32 + 62 + 22
Ans. (b) : Given,
= 9 + 36 + 4
Q = 8 units
= 49
R = 2P
| R |= 7
3iˆ + 6jˆ − 2kˆ
R̂ =
7
1
R̂ = 3iˆ + 6ˆj − 2kˆ
7
15. Pressure is a scalar quantity because
(a) it is the ratio of force to area and both force
and area are vector quantities
Q2 = R 2 + P2
(b) it is the ratio of magnitude of force to area
(c) it is the ratio of component of the force
(8) 2 = (2P) 2 + P 2
normal to the area
= 4P 2 + P 2
(d) it depends on the size of the area chosen
64 = 5P 2
SCRA-2015
64
8
Ans. (c) : Pressure is a scalar quantity because pressure
P2 =
⇒P=
5
is the ratio of normal force to the area and the direction
5
13. Angle (in rad) made by the vector 3 ˆi + ˆj with of force is not required.
16. The component of a vector r along X-axis will
the X-axis:
have a maximum value, if :
π
π
(a)
(b)
(a) r is along positive X-axis
6
4
(b) r is along positive Y-axis
π
π
(c) r is along negative Y-axis
(c)
(d)
3
2
(d) r makes an angle of 450 with the X-axis
AP EAMCET(Medical)-2005
Karnataka CET-2016
(
)
(
)
(
(
Objective Physics Volume-I
347
(
)
)
(
)
)
YCT
Ans. (a) : rx = r cosθ
rx is maximum when θ = 0
r will be along positive X-axis for maximum value.
17. Which of the following is not a vector quantity?
(a) Weight
(b) Nuclear spin
(c) Momentum
(d) Potential energy
Karnataka CET-2014
Ans. (d) : Weight, Nuclear spin and Momentum are
vector quantities because they have both magnitude as
well as direction, whereas potential energy has
magnitude only but no direction, thus it is a scalar
quantity.
18. Two equal forces (P each) act at a point
inclined to each other at an angle of 1200. the
magnitude of their resultant is :
(a) P/2
(b) P/4
(c) P
(d) 2P
Karnataka CET-2004
Ans. (c) : Given, Q = P, θ = 120°
Ans. (c) : A.B = 0 ⇒ A ⊥ B
A.C = 0 ⇒ A ⊥ C
A is perpendicular to both B and C and B × C is also
perpendicular to both B and C .
Therefore, A is parallel to B × C
21. Three forces F1, F2 and F3 together keep a body
in equilibrium. If F1 = 3 N along the positive xaxis, F2 = 4 N along the positive y-axis, then the
third force F3 is
3
(a) 5N making an angle θ = tan −1   with
4
negative y-axis
4
(b) 5N making an angle θ = tan −1   with
3
negative y-axis
3
(c) 7N making an angle θ = tan −1   with
∵ R = P 2 + Q 2 + 2PQ cos θ
4
negative y-axis
= P 2 + P 2 + 2P × P cos120°
4
(d) 7N making an angle θ = tan −1   with
1

= P 2 + P 2 −  2P × P ×  = P
3
2

negative y-axis
19. The resultant of two forces 3P and 2P is R. If
J&K CET- 2010
the first force is doubled then the resultant is
also doubled. The angle between the two forces Ans. (c) : F1, F2, F3 keep a body in equilibrium then
resultant of force,
is :
(a) 900
(b) 1800
0
0
(c) 60
(d) 120
Karnataka CET-2001
Ans. (d) : Case – I
R=
( 3P )2 + ( 2P )2 + 12P 2cosθ
R = P 13 + 12cos θ
Case -II
R1 =
( 6P )2 + ( 2P )2 + 24P 2 cos θ
= P 40 + 24cos θ
From case (I) and case (II),
R1
=
R
20.
ΣF = 0
40 + 24 cosθ
=2
[∵ R1 = 2R ]
13 + 12cosθ
40 + 24 cosθ = 52 + 48 cosθ
24 cosθ = –12
1
 1
cosθ = − ⇒ θ = cos −1  − 
2
 2
0
θ= 120
Three vectors satisfy the relation A.B = 0 and
A.C = 0 then A is parallel to :
(a) C
(c) B × C
(b) B
(d) B.C
JCECE-2013
COMEDK 2013
Karnataka CET-2003
Objective Physics Volume-I
F1 + F2 +F3 = 0
3 + 4 + F3 = 0
F3 = – 7N
Magnitude of F3 = 7N
θ is angle made with-Y axis
3
tanθ =
4
θ = tan–1 3/4
F3 make angle with negative y-axis.
22. Magnitudes of four pairs of displacement
vectors are given. Which pair of displacement
vectors, under vector addition, fails to give a
resultant vector of magnitude 3 cm ?
(a) 2 cm, 7 cm
(b) 1 cm, 4 cm
(c) 2 cm, 3 cm
(d) 2 cm, 4 cm
J&K CET- 2009
348
YCT
Ans. (a) : The magnitude R of the resultant of two
vectors A and B depends upon the magnitudes of A and
B and the angle θ between them and is given by
R2 = A2 + B2 + 2AB cos θ
When θ = 0, R is maximum and given by
R 2max = A2 + B2 + 2AB = (A + B)2
Rmax = A + B
When θ = 180º, R is minimum and given by
R 2min = A2 + B2 − 2AB = (A − B)2
Rmin = A – B
Thus, the magnitude of resultant will lie between A – B
and A + B.
Now,
Checking option (a)
|A – B| = |2 – 7|= 5
|A + B| = |2 + 7| = 9
So,
5 ≤ R ≤ 9 and R = 4
Hence, the option (a) is the correct answer.
23. A body is under the action of two mutually
perpendicular forces of 3N and 4N. The
resultant force acting on the body is
(a) 7 N
(b) 1 N
(c) 5 N
(d) zero
J&K CET- 2008
Ans. (c) : The two forces be
A = 3N and B = 4N
A is mutually perpendicular to B.
∴ θ = 90°
26.
Two vectors are given by A = 3iˆ + ˆj + 3kˆ and
B = 3iˆ + 5jˆ - 2kˆ . Find the third vector C if
A + 3B - C = 0
(a) 12iˆ +14jˆ +12kˆ
(b) 13iˆ +17jˆ +12kˆ
(d) 15iˆ +13jˆ + 4kˆ
(c) 12iˆ +16jˆ - 3kˆ
J&K CET- 2007
Ans. (c) : Given, A = 3iˆ + ˆj + 3kˆ , B = 3iˆ + 5ˆj − 2kˆ
A + 3B − C = 0
3iˆ + ˆj + 3kˆ + 3 3iˆ + 5ˆj − 2kˆ − C = 0
(
)
30.
3iˆ + ˆj + 3kˆ + 9iˆ + 15jˆ − 6kˆ − C = 0
12iˆ + 16ˆj − 3kˆ − C = 0
C = 12iˆ + 16ˆj − 3kˆ
27.
Vector which is perpendicular to a
(acos θ ˆi + bsin θ ˆj) is
1
1
sinθiˆ − cosθ ˆj
a
b
(c) 5kˆ
(d) all of these
J&K CET- 2006
Ans. (d) : Two vectors are perpendicular if their dot
product is zero i.e., A ⋅ B = 0.
In option (a)
a cos θˆi + bsin θˆj ⋅ bsin θˆi − a cos θˆj
(a) bsinθiˆ − a cosθ ˆj
(
)(
(b)
)
= ab cos θ sin θ − absin θ cos θ = 0
In option (b)
R = 42 + (3) 2 + 2ABcos 90°
1
1
(a cos θˆi + bsin θˆj) ⋅ ( sin θˆi − cos θˆj)
a
b
R = 16 + 9 + 0
= sin θ cos θ − sin θ cos θ = 0
R = 5N
24. If the scalar and vector products of two vectors In option (c)
A,B are equal in magnitude, then the angle (a cos θˆi + bsin θˆj) ⋅ 5kˆ = 0.
between the two vectors is
28. Velocity is
(a) 45°
(b) 90°
(a) scalar
(c) 180°
(d) 360°
(b) vector
J&K CET- 2008
(c) neither scalar nor vector
Ans. (a) : A.B = A × B
(d) both scalar and vector
|A||B|cosθ = |A||B|sinθ
J&K CET- 2002
sin θ A B
Ans. (b) : A vector quantity is defined as the physical
=
quantity that has both magnitude as well as direction.
cos θ A B
Velocity is the directional speed of a object in motion
tanθ = 1
and indication of rate of change in position as observed
θ = 45°
by a particular frame of reference.
25.
A is a vector with magnitude A, then the unit Velocity is a physical vector quantity.
vector  in the direction of A is
29. The sum of two vectors A and B is at right
angles to their difference. Then
(a) AA
(b) A ⋅ A
(a) A = B
A
(b) A = 2B
(c) A × A
(d)
A
(c) B = 2A
J&K CET- 2008
(d) A and B have the same direction
BCECE-2008
A
A
Ans. (d) : Unit vector  =
=
J & K CET - 1998
|A| A
UP CPMT - 2006
R = A 2 + B2 + 2ABcos θ
( )
Objective Physics Volume-I
349
Ans. (a) : Let r1 and r2 be the sum and difference of
vectors A and B respectively i.e.,
r1 = A + B
r2 = A – B
r1 is perpendicular to r2
(given)
Taking the dot product of r1 and r2
r 1. r 2 = ( A + B ) . ( A – B )
0 = A2 – B2
A2 = B2
A=B
YCT
32.
The sum of two vectors A and B is at right
angles to their difference. This is possible if
(a) A = 2B
(b) A = B
(c) A =3B
(d) B =2A
J&K CET- 1998
Ans. (b) : Let, P1 and P2 sum and difference of vectors
A and B ,
P1 = ( A + B )
P2 = ( A – B )
The vectors A and B are such that
A+B = A–B
P1 . P2 = ( A + B ).( A – B )
0 = A2 – B2
The angle between the two vectors is
A2 = B2
(a) 60º
(b) 75º
A=B
(c) 45º
(d) 90º
WBJEE-2016, 33. What is the torque of a force 3iˆ + 7jˆ + 4kˆ about
AIIMS-25.05.2019(E) Shift-2
the origin, if the force acts on a particle whose
J&K CET- 2003, 1999
position vector is 2iˆ + 2jˆ + 1kˆ ?
Ans. (d) : Let angle between A and B be θ
(a) ˆi – 5jˆ + 8kˆ
(b) 2iˆ + 2jˆ + 2kˆ
The resultant of A + B is given by
ˆi + ˆj + kˆ
2
2
(c)
(d) 3iˆ + 2jˆ + 3kˆ
R = A + B + 2ABcos θ
J&K-CET-2014
The resultant of A − B is given by
Ans. (a) : Given that,
R ' = A 2 + B2 − 2ABcos θ
According to the question,
F = 3iˆ + 7ˆj + 4kˆ
R = R'
r = 2iˆ + 2ˆj + 1kˆ
A 2 + B2 + 2ABcos θ = A 2 + B2 − 2ABcos θ
τ
= r×F
A 2 + B2 + 2ABcos θ = A 2 + B2 − 2ABcos θ
ˆi ˆj kˆ
4ABcos θ = 0
31.
θ = 90°
τ= 2 2 1
The vector sum of two forces is perpendicular
3 7 4
to their vector differences. In that case, the
forces
τ = î (8– 7) – ĵ (8 – 3) + k̂ (14 – 6)
(a) are not equal to each other in magnitude
τ = î – 5jˆ + 8kˆ
(b) cannot be predicted
34. The scalar product of two vectors
(c) are equal to each other
(d) are equal to each other in magnitude
A = 2iˆ + 2jˆ – kˆ and B = – ˆj + kˆ is given by
[AIPMT 2003]
(a) A . B = 3
(b) A . B = 4
AIIMS- 2012
(c) A . B = – 4
(d) A . B = – 3
TS-EAMCET - 09.09.2020
J&K-CET-2013
Ans. (d) : Given that,
ˆ B = −ˆj + kˆ = 0iˆ − ˆj + kˆ
A = 2iˆ + 2ˆj − k,
Ans. (d) : Let f1 and f 2 be the two forces
Then sum of forces, a = f1 + f 2
And difference, b = f1 – f 2
( )
The two forces are perpendicular to each other a.b = 0
( f + f ).( f − f ) = 0
1
2
1
2
| f1 |2 − | f 2 |2 = 0
| f1 |2 =| f 2 |2
| f1 |=| f 2 |
In that case both the force are equal and have same
magnitude.
Objective Physics Volume-I
(
)(
A.B = 2iˆ + 2ˆj − kˆ . 0iˆ − ˆj + kˆ
)
= 0 + 2 × (–1) + (−1) × 1
= –2 – 1 = –3
35. The velocity vector of the motion described by
the position vector of a particle r = 2tiˆ + t 2 ˆj is
(a) v = 2iˆ + 2t ˆj
(b) v = 2tiˆ + 2t ˆj
350
(c) v = tiˆ + t 2 ˆj
(d) v = 2iˆ + t 2 ˆj
J&K-CET-2013
YCT
38.
Ans. (a) : Given, r = 2tiˆ + t 2 ˆj
()
dr
v =
dt
Velocity
dr
= 2iˆ + 2tjˆ
dt
36. A certain vector in the xy plane has an xcomponent of 12 m and a y-component of 8 m.
It is then rotated in the xy plane so that its xcomponent is halved. Then its new ycomponent is approximately
(a) 14 m
(b) 13.11 m
(c) 10 m
(d) 2.0 m
J&K-CET-2012
Two forces each of magnitude 'P' act at right
angles. Their effect is neutralized by a third
force acting along their bisector in opposite
direction. The magnitude of the third force is
π


cos 2 = 0


P
(a) P
(b)
2
P
(c) 2P
(d)
2
MHT-CET 2020
Ans. (c) : The third force will have magnitude equal to
their resultant,
Ans. (b) : x – component = 12cm
y – component = 8cm
Length of the resultant vector (R)
= 144 + 64 = 208
Now,
x' =
(x ') + (y ') = 208
2
R + R + 2R 1R 2 cos θ
R=
P 2 + P 2 + 2.P.P cos90°
2
2
R = 2P 2
39.
'
(y ) = 172
(y') = 13.11m
Figure shows three forces F1 , F2 and F3 acting
along the sides of an equilateral triangle. If the
total torque acting at point 'O' (centre of the
triangle) is zero then the magnitude of F3 is
(a) A – B + C
(c) A + B + C
(e) – A – B – C
Q 2 − P 2 = 192
∴ Q2 – (16 – Q)2 = 192
Q2 − 256 − Q 2 + 32Q = 192
32Q = 448
Q = 14 N
Now, from P + Q = 16
P = 16 – 14 = 2N
P = 2N
Kerala CEE - 2016
Ans. (c) : Given, PQ = A, QR = B, RS = C, PS = ?
(a) F1+F2
F −F
(c) 1 2
2
(b) F1-F2
F
(d) 1
F2
(where, ˆi ⋅ ˆi = ˆj ⋅ ˆj = kˆ ⋅ kˆ = 1 )
W = 28 + 4 + 8
W = 40J
44.
( ∵ P = 16 – Q)
Ans. (a) : Force F1 and F2 produce anticlockwise
torque while force F3 produces clockwise torque. The
torque in the two directions balance each other. The
perpendicular distance of the forces from the centre is
the same.
∴ F1r + F2r = F3r or F1+F2 = F3
According to polygon law of vector addition,
PQ + QR + RS – PS = 0
PS = PQ + QR + RS
PS = A + B + C
40. Among the following, the vector quantity is
(a) pressure
(b) gravitation potential
(c) stress
(d) impulse
(e) distance
Kerala CEE 2012
351
YCT
two
vectors
A = 2iɵ + 3jɵ + 4kɵ
and
power exerted is
(a) 4 W
(c) 2 W
(e) 1 W
B = ˆi + 2ˆj − nkˆ
If two vectors are perpendicular then their scalar
product is zero.
∴ A.B = 0
( 2iˆ + 3jˆ + 4kˆ ) ⋅ ( ˆi + 2ˆj − nkˆ ) = 0
(b) 5 W
(d) 8 W
)
)
v = 2iˆ + 2jˆ + 3kˆ
We know that,
P = F⋅v
P = 4iˆ + ˆj – 2kˆ ⋅ 2iˆ + 2jˆ + 3kˆ
(
the
A force
Ans. (a) : Given,
F = 4iˆ + ˆj – 2kˆ
(
(
If
B = ɵi + 2jɵ - nkɵ are perpendicular then the value
of n is :
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5
Kerala CEE 2006
Ans. (b) : Given,
A = 2iˆ + 3jˆ + 4kˆ
Kerala CEE - 2010
MHT-CET 2020
Objective Physics Volume-I
dr = 4iˆ + 2ˆj − 2kˆ
ɵ N acting on a body
(4iɵ + ɵj - 2k)
maintains its velocity at ( 2iˆ + 2jˆ + 3kˆ ) ms–1. The
42.
)
Work done by the forces, W = FR .dr
ˆ
ˆ + 2ˆj − 2k)
ˆ
= (7iˆ + 2ˆj − 4k).(4i
Q = 14N
(b) A + B – C
(d) A – B – C
) (
Displacement, dr = r2 − r1
ˆ − (iˆ + 2ˆj + 3k)
ˆ
dr = (5iˆ + 4ˆj + k)
8 3 = P 2 + Q 2 − 2P 2 ⇒ 8 3 = Q 2 − P 2
R = 2P
In the given diagram, if PQ = A, QR = B and
RS = C, then PS equals
(
FR = 7iˆ + 2ˆj − 4kˆ
 P
8 3 = P 2 + Q 2 + 2PQ  − 
 Q
2
(6)2 + (y ')2 = 208
(y')2 = 208 – 36 = 172
37.
R=
2
1
R
8 3 = P 2 + Q 2 + 2PQcos θ
sin θ
∵ tan 90o =
=∞
P + Qcos θ
P
P + Q cos θ = 0 ⇒ cos θ = −
Q
So,
= x 2 + y 2 = 122 + 82
x 12
= = 6cm
2 2
Resultant will always be constant even after the rotation
So,
Ans. (d) : Impulse is defined as the product of net force 43. A particle acted upon by constant forces
and time interval for which it was applied.
4iɵ + ɵj - 3kɵ and 3iɵ + ɵj - kɵ is displaced from the
i.e. Impulse = Ft
ɵi + 2jɵ + 3kɵ to the point 5iɵ + 4jɵ + k.
ɵ The
point
Force is vector quantity. Therefore, Impulse is a vector
total work done by the forces in SI unit is
quantity. Stress is a tensor quantity.
(a) 20
(b) 40
41. The sum of magnitudes of two forces acting at
(c) 50
(d) 30
a point is 16 N and their resultant 8 3 N is at
(e) 35
90º with the force of smaller magnitude. The
Kerala CEE - 2008
two forces (in N) are
Ans. (b) : Given,
(a) 11, 5
(b) 9, 7
F1 = 4iˆ + ˆj − 3kˆ
(c) 6, 10
(d) 4, 12
(e) 2, 14
F2 = 3iˆ + ˆj − kˆ
Kerala CEE 2012
r1 = ˆi + 2ˆj + 3kˆ
Ans. (e) : Given,
P + Q = 16
r2 = 5iˆ + 4ˆj + kˆ
R=8 3
Now force, FR = F1 + F2
∵ R = P 2 + Q 2 + 2PQ cos θ
F = 4iˆ + ˆj – 3kˆ + 3iˆ + ˆj – kˆ
)(
2 + 6 − 4n = 0
n=2
45. A particle is displaced from a position
ɵ to another position (3iɵ + 2jɵ - 2k)
ɵ
(2iɵ - ɵj + k)
ɵ The
under the action of the force of (2iɵ + ɵj - k).
)
(where, ˆi ⋅ ˆi = ˆj ⋅ ˆj = kˆ ⋅ kˆ = 1 )
P = 8 + 2 − 6 ⇒ P = 4W
Objective Physics Volume-I
352
work done by the force in an arbitrary unit is:
(a) 8
(b) 10
(c) 12
(d) 16
(e) 20
JCECE-2018
Kerala CEE 2005
YCT
48.
Ans. (a) : Given that,
F = 2iˆ + ˆj − kˆ
r1 = 2iˆ − ˆj + kˆ
r2 = 3iˆ + 2ˆj − 2kˆ
The power utilised when a force of
ˆ
(2iˆ + 3jˆ + 4k)N
acts on a body for 4s,
ˆ m, is
producing a displacement of (3iˆ + 4jˆ + 5k)
(a) 9.5 W
(c) 6.5 W
(b) 7.5 W
(d) 4.5 W
W = F.dr
AP EAMCET (21.09.2020) Shift-I
ˆ
ˆ
ˆ
W = (2i + j − k).(r2 − r1 )
∵ dr = r 2 − r1
Ans. (a) : Power = F.v
d
= 2iˆ + ˆj – kˆ ⋅  3iˆ + 2jˆ – 2kˆ − 2iˆ − ˆj + kˆ 
And,
v=


t
ˆ ˆ + 3jˆ − 3k)
ˆ =2+3+3
= (2iˆ + ˆj − k).(i
3iˆ + 4ˆj + 5kˆ
v=
∴ W = 8 unit
4
46. If a vector A is given as A = 4iɵ + 3jɵ + 12kɵ , then
ˆ ˆ ˆ
ˆ
ˆ
ˆ (3i + 4 j + 5k)
∴ Power = (2i + 3j + 4k).
the angle subtended with the x-axis is
4
4
3
3
4
5
(a) sin −1  
(b) sin −1  
= 2 × + 3× + 4 ×
13 
13 
4
4
4
3
4
6 12 20 6 + 12 + 20
(c) cos −1  
(d) cos −1  
= + +
=
13 
13 
4 4
4
4
COMEDK-2019
38
=
UPSEE - 2015
4
Ans. (d) : Given, A = 4iˆ + 3jˆ + 12kˆ
Power = 9.5 W
Let θ be the angle made by vector A with X-axis.
ˆ
49.
A = 4iˆ + 3jˆ and B = 4iˆ + 2j.
Find a vector
A
∴ cos θ = x
parallel to A but has magnitude five times that
A
of B.
4
4
cos θ =
=
(a) 20 2iˆ + 3jˆ
(b) 20 4iˆ + 3jˆ
2
2
2
13
4 + 3 + 12
ˆ
ˆ
(c) 20 2i + j
(d) 10 2iˆ + ˆj
4
θ = cos −1  
BITSAT-2007
 13 
47. The angle between two vectors A and B is θ. Ans. (b) : Given, A = 4iˆ + 3jˆ , B = 4iˆ + 2ˆj
Vector R is the resultant of the two vectors. If
A
4iˆ + 3jˆ
 =
=
θ
R makes an
with A, then
A
42 + 32
2
(
(
)(
) (
)
)
(
(
)
)
(
(
)
)
B
1
 = 4iˆ + 3jˆ
2
5
(c) A = B
(d) AB = 1
ˆ
UPSEE - 2013 P = 5 B A
1
Ans. (c) : The angle α which the resultant R makes
ˆ
= 5 × 42 + 22 × (4iˆ + 3j)
5
with A is given by–
ˆ
ˆ
Bsin θ
θ
P = 20(4i + 3j)
tan α =
Here, α =
A + Bcos θ
2
50. Given that A + B = R and A2 + B2 = R2 The
θ
Bsin θ
Hence, tan =
angle between Aand B is
2 A + Bcos θ
(a) 0
(b) π/4
θ
(c) π/2
(d) π
sin
2 = 2Bsin(θ / 2) cos(θ / 2)
BITSAT-2009
θ
A
+
Bcos
θ
cos
Ans. (c) : Given, A + B = R and A2 + B2 = R2
2
2
2
2
R = A + B + 2AB cosθ
A + Bcos θ = 2Bcos 2 (θ / 2)
Where, θ = Angle between A & B,
2
2
2
A + B[2cos (θ / 2) − 1] = 2Bcos (θ / 2)
∵ R = A2 + B2
∴ 2
A + B2 = A2 + B2 + 2AB cosθ
2θ
2θ
A + 2Bcos   − B = 2Bcos  
∴cosθ = 0
2
2
A–B=0
π
θ=
∴A=B
2
(a) A = 2B
Objective Physics Volume-I
(
(b) A =
353
)
YCT
51.
The two vectors A and B are drawn from a Ans. (b) : R = xiˆ + yjˆ + zkˆ
common point and C = A + B, then angle
A = 3iˆ + 2ˆj + 5kˆ
between A and B is
(1) 90° if C2 = A2 + B2
(2) greater than 90° if C2 < A2 + B2
(3) greater than 90° if C2 > A2 + B2
(4) less than 90° if C2 > A2 + B2
Correct options are(a) 1,2
(b) 1, 2, 3, 4
(c) 2, 3, 4
(d) 1, 2, 4
BITSAT-2011 54.
Ans. (d) : C = A + B
∵
C2 = A2 + B2 + 2AB cosθ
C 2 − ( A 2 + B2 )
2AB
Case-I, C2 = A2 + B2
A 2 + B2 − ( A 2 + B2 )
cosθ =
=0
2AB
θ = π/2 or 90°
Case-II, C2 < A2 + B2
C 2 − ( A 2 + B2 )
cosθ =
(–ve)
2AB
Then, θ > 90°
Case-III, C2 > A2 + B2
C 2 − ( A 2 + B2 )
cosθ =
(+ve)
2AB
Then, θ < 90°
ˆ The
52. Given P = 2iˆ – 3jˆ + 4kˆ and Q = ˆj – 2k.
magnitude of their resultant is
(a) 3
(b) 2 3
(c) 3 3
(d) 4 3
BITSAT -2018
Ans. (b) : Given, P = 2iˆ − 3jˆ + 4kˆ , Q = ˆj − 2kˆ
cosθ =
R = (2)2 + (−2)2 + (2) 2
R =2 3
53.
A ⋅ R = 3x + 2y + 5z
d(3x) ˆ d(2y) ˆ d(5z) ˆ
i+
j+
k
dx
dy
dz
∇(A ⋅ R) = 3iˆ + 2ˆj + 5kˆ
∇(A ⋅ R) =
∇(A ⋅ R) = A
The direction of A is vertically upward and
direction of B is in north direction. The
direction of A × B will be
(a) Western direction
(b) Eastern direction
(c) At 45º upward in north
(d) Vertically downward
CG PET- 2009
Ans. (a) :
Considering vertically upward direction as z-axis and
north direction as y-axis.
A = akˆ , B = bjˆ
∴ A × B = akˆ × bjˆ = ab(−ˆi)
Thus, it is along negative x-axis
∴ A × B is along west.
55.
R = P+Q
ˆ + (ˆj − 2k)
ˆ
= (2iˆ − 3jˆ + 4k)
R = 2iˆ − 2ˆj + 2kˆ
ˆ ⋅ (3iˆ + 2ˆj + 5k)
ˆ
A ⋅ R = (xiˆ + yjˆ + zk)
If A = B + C and the values of A, B and C are
13, 12 and 5 respectively, then the angle
between A and C will be
(a) cos −1 ( 5 /13)
(b) cos −1 (13 /12 )
(c) π / 2
(d) sin −1 ( 5 /12 )
CG PET- 2009
Ans. (a) : Given that, A = 13,B = 12,C = 5
A2 = B2 + C2 + 2BC cosθ
(13)2 = (12)2 + (5)2 + 2 × 12 × 5 cosθ
cos θ = 0
The position vector of a point is θ = 90°
Hence, it is a right-angle triangle.
R = xiˆ + yjˆ + zkˆ
and another vector is
A = 3iˆ + 2jˆ + 5kˆ . Which of the mathematical
relation is correct?
ˆ Rˆ = 0
(a) ∇ A
( )
(
(
)
)
(b) ∇ A R = A
(c) ∇ A R = R
(d) None of these
Objective Physics Volume-I
So, angle between A and C is,
5
5
cos α =
⇒ α = cos −1  
13
 13 
CG PET- 2009
354
YCT
ˆ where x̂ and ŷ are unit Ans. (b) : r = α cos ωtiˆ + α sin ωtjˆ
If r1 = 2 xˆ , r2 = 2y,
vectors along the X-axis and Y-axis
dr
respectively, then the magnitude of r1 + r2 is
v=
= (−α sin ωt ) ωˆi + (α cos ωt ) ωˆj
dt
(a) 2 2
(b) 2 3
v.r = (−αω sin ωt)iˆ + (αω cos ωt)ˆj .  (α cos ωt)iˆ + (α sin ωt)ˆj
(c) 3 2
(d) 3

 

TS-EAMCET.14.09.2020, Shift-2 = −α 2 ω sin ωt.cos ωt + α 2 ω cos ωt.sin ωt
ˆ
ˆ
Ans. (a) : Given, r1 = 2x, r2 = 2y
v.r = 0
r1 + r2 = 2xˆ + 2yˆ
∵ Dot product of v and r is zero
Magnitude, r1 + r2 = 22 + 22 = 2 2
∴ Both are perpendicular to each other.
60. If a vector A having a magnitude of 8 is added
57. Let A1 + A2 = 5A3, A1 – A2 = 3A3,
to a vector B which lies along x-axis, then the
ˆ then | A1 | is
A 3 = 2iˆ + 4j,
resultant of two vectors lies along y-axis and
| A2 |
has magnitude twice that of B. The magnitude
(a) 4
(b) 8
of B is
(c) 2
(d) 6
6
12
TS-EAMCET.14.09.2020, Shift-2
(b)
(a)
Ans. (a) : Given, A1 + A2 = 5A3
...(1)
5
5
A1 – A2 = 3A3
...(2)
16
8
(c)
(d)
Adding equation (1) & (2),
5
5
2A1 = 8A3 ⇒ A1 = 4A3
JCECE-2012
∴ A1 = 8iˆ + 16ˆj
Ans. (d) : Given,
Subtracting equation (2) from (1),
2A2 = 2A3
∴ A 2 = A 3 = 2iˆ + 4ˆj
56.
A1
82 + 162
=
= 16 = 4
A2
22 + 42
58. The magnitude of x and y components of A are
7 and 6 respectively. Also the magnitudes of x
and y components of A+B are 11 and 9
A = 8 units
respectively. Calculate the magnitude of vector
Since, B is along x - axis and resultant of two vector
B.
(a) 10
(b) 5
C lies on y - axis
(c) 6
(d) 3
JCECE-2018 So, B and C are perpendicular vector.
Hence,
Ans. (b) : Given, A = 7iˆ + 6ˆj
| A |2 =| B |2+ | C |2
Let, B = b1ˆi + b 2 ˆj
∴
∴
7 + b1 = 11 ⇒ b1 = 4
6 + b2 = 9 ⇒ b2 = 3
B = 4iˆ + 3jˆ
∴
61.
B = 4 + 3 = 25 = 5
2
59.
2
| 8 |2 =| B |2+2B |2
64 = 5B2
64
B2 =
5
8
B=
5
It two forces each of 2N are inclined at 60º,
then resultant force is:
(a) 2 N
(b) 2 5N
The position vector of a particle is
(c) 2 3N
(d) 4 2N
JCECE-2006
r = (αcosωt)iˆ + ( αsinωt)jˆ . The velocity vector
Ans. (c) : Let A & B be two forces,
of the particle is
(a) Parallel to position vector
∴ R = A 2 + B2 + 2ABcos θ
(b) Perpendicular to position vector
R = 22 + 22 + 2 × 2 × 2cos 60o
(cos 60o = 1/2)
(c) Directed towards the origin
(d) Directed away from the origin
∴ R = 12 = 2 3N
JCECE-2014
BCECE - 2004 ∴ R = 2 3N
Objective Physics Volume-I
355
YCT
62.
Two vectors have magnitudes 3 and 5. If angle
What is the magnitude of the resultant between
between them is 60º, then the dot product of
these two vectors?
two vectors will be :
(a) 20 3
(b) 35
(a) 7.5
(b) 6.5
(c) 15 3
(d) 10 3
(c) 8.4
(d) 7.9
COMEDK 2015
JCECE-2003
Ans. (d) : The vector representation is as follows,
Ans. (a) : Let A & B be two vector,
∴
A.B = A B cos θ
= 3 × 5 cos60o
1
= 15 ×
2
A ⋅ B = 7.5
63. Calculate the work done when a force ∴ θ = 180 − 30 − 30 = 120
F = 2iˆ + 3jˆ – 5kˆ units acts on a body producing R = A 2 + B2 + 2ABcos θ
a displacement s = 2iˆ + 4jˆ + 3kˆ units :
R = 102 + 202 + 2 × 10 × 20 × cos120o
(a) 1 unit
(b) 20 unit
1

R = 100 + 400 − 200
(c) 5 unit
(d) zero
∵ cos120° = − 
2

JCECE-2003
R = 300
ˆ
ˆ
ˆ
Ans. (a) : Given that, F = 2i + 3j − 5k
∴
R = 10 3 Unit
s = 2iˆ + 4ˆj + 3kˆ
W = F⋅ s
66.
ˆ
ˆ + 4ˆj + 3k)
ˆ
W = (2iˆ + 3jˆ − 5k).(2i
W = 4 + 12 − 15
W = 1 Unit
64. Three forces acting on a body are shown in the
figure. To have the resultant force only along
the y-direction, the magnitude of the minimum
additional force needed along OX is
ˆ and
Two vectors are given by A = (iˆ + 2jˆ + 2k)
ˆ
ˆ
ˆ
B = (3i + 6j + 2k) . Another vector C has the
same magnitude as B but has the same
direction as A . Then which of the following
vectors represents C ?
7 ˆ
3 ˆ
(a)
i + 2 ˆj + 2kˆ
(b)
i − 2 ˆj + 2kˆ
3
7
7 ˆ
9 ˆ
(c)
i − 2 ˆj + 2kˆ
(d)
i − 2 ˆj + 2kˆ
9
7
COMEDK 2018
ˆ
Ans. (a) : Given that, A = (iˆ + 2ˆj + 2k)
(
)
(
)
(
)
(
)
ˆ
B = (3iˆ + 6ˆj + 2k)
Unit vector along direction of A ,
3
N
4
(c) 0.5 N
(a)
(b)
13N
(d) 1.5 N
COMEDK 2019 And, B = 32 + 6 2 + 2 2
2
2
Ans. (c) : ∵ R = (∑ Fx ) + (∑ Fy )
B = 49 = 7
Let the additional force F be directed along the positive
Thus,
vector C is–
x-direction.
ˆ
Taking x-component, the total force should be zero.
C= BA
Let F be the magnitude of minimum force which must
be along x-direction, by resolving the vector we get–
7
ˆ
C = (iˆ + 2ˆj + 2k)
1 × cos60° + 2sin30° + F – 4sin30° = 0
3
1
+1+ F − 2 = 0
67. A particle starts moving from point (2,10,1).
2
Displacement for the particle is 8iˆ − 2jˆ + kˆ . The
F = 1/2 = 0.5 N
final coordinates of the particle is
65. Vector A has a magnitude of 10 units and
(a) (10, 8, 2)
(b) (8, 10, 2)
makes an angle of 30° with the positive x-axis.
(c) (2, 10, 8)
(d) (8, 2, 10)
Vector B has a magnitude of 20 units and
COMEDK 2020
makes an angle of 30° with the negative x-axis.
Objective Physics Volume-I
356
YCT
Ans. (a) : Displacement ∆ r = 8iˆ − 2ˆj + kˆ
Let position vector representing final co-ordinate(x, y,
z).
rf = xiˆ + yjˆ + zkˆ
∴∆ r = rf − ri
ˆ − (2iˆ + 10ˆj + k)
ˆ
8iˆ − 2ˆj + kˆ = (xiˆ + yjˆ + zk)
∴
( 2 + 2 ) = (2 + 2 ) x
2+ 2
2+ 2
x =1
x=1
y − 10 = −2 ⇒ y = 8
)
)
(
R=
)(
(A) + (B) + 2ABcos θ
2
2
)
72.
2
2 + 2 = 2x 2 + 2x 2 ⋅
1
2
Objective Physics Volume-I
(
)
(
)
The angle between P + Q and P – Q will be
(a) 90º only
(b) between 0º and 180º
(c) 180º only
(d) none of these
2 + 2 = ( x ) + ( x ) + 2.x.x cos 45°
2
357
AIIMS-1999
YCT
(
)
Clearly from figure, angle (θ) between P + Q and
(P − Q) between 0 to 180°.
73.
(
)
(
)
7 ˆ ˆ ˆ
i − 2j + 2k
9
(
)
(
)
9 ˆ ˆ
i + 2j + 2kˆ
7
WB JEE 2013
Ans. (a) : Given that, A = ˆi + 2ˆj + 2kˆ
(c)
70. If vectors P = aiˆ + ajˆ + 3kˆ and Q = aiˆ – 2jˆ – kˆ
z −1 = 1 ⇒ z = 2
are perpendicular to each other, then the
rf final co-ordinates are (10, 8, 2).
positive value of a is
(a) zero
(b) 1
68. If two forces of equal magnitudes act
(c) 2
(d) 3
simultaneously on a body in the east and the
AIIMS-2002
north directions then
AP EAMCET (Medical)-1998
(a) the body will displace in the north direction
(b) the body will displace in the east direction
Ans. (d) : If two vector are perpendicular to each other
(c) the body will displace in the north-east then their dot product is zero.
direction
P.Q = 0
(d) the body will remain at the rest
ˆ + ajˆ + 3kˆ . aiˆ − 2ˆj − kˆ = 0
ai
AIIMS-2009
Ans. (c) :
a2 – 2a – 3 = 0
a2 – 3a + a – 3 = 0
a(a – 3) + 1(a – 3) = 0
(a – 3) (a + 1) = 0
a = 3, a = – 1
a = 3, −1
So, positive value of a is 3
71. Two equal vectors have a resultant equal to
either of them, then the angle between
them will be
Let, the force acting on body F1 in north direction and
(a) 110º
(b) 120º
F2 in east direction.
(c) 60º
(d) 150º
∴ So, the resultant force on the body FR = F1 + F2 in
AIIMS-2000, BCECE-2007
North –East direction.
Ans. (b) : Let the two vector be A and B both at angle
69. Two vectors having equal magnitude of x units of θ from each other with a resultant R
acting at an angle of 45º have resultant From the triangle law of vector addition,
R2 = A2 + B2 + 2AB cosθ
2 + 2 units. The value of x is
A = B = R (given)
(a) 0
(b) 1
(A)2 = (A)2 + (A)2 + 2.A.A.cosθ
(c) 2
(d) 2 2
A2 = 2A2 (1 + cosθ)
AIIMS-2009
−1
cos θ =
Ans. (b) : Let two vector is A and B
2
cosθ = cos120°
So, A = x units, B = x units, θ = 45°, R = 2 + 2
θ = 120°
(
The vectors are given by A = ˆi + 2jˆ + 2kˆ and
B = 3iˆ + 6jˆ + 2kˆ . Another vector C has the
same magnitude as B but has the same
direction as A. Then which of the following
vectors represents C?
7 ˆ ˆ
3 ˆ ˆ ˆ
(a)
i + 2j + 2kˆ
(b)
i − 2j + 2k
3
7
2
2
x − 2 = 8 ⇒ x = 10
(
75.
2 + 2 = 2x 2 + 2x 2
x2 =
8iˆ − 2ˆj + kˆ = (x − 2)iˆ + (y − 10)ˆj + (z − 1)kˆ
Ans. (b) :
2 + 2 = 2x 2 + 2x 2
Squaring on both side
(d)
Assertion: If A + B = A – B , then the angle Magnitude of A = 12 + 22 + 22 = 3
between A and B is 90º.
Unit vector along A .
Reason: A + B = B + A
A ˆi + 2ˆj + 2kˆ
(a) If both assertion and reason are true and
 =
=
3
A
reason is the correct explanation of assertion.
(b) If both assertion and reason are true but
ˆ
B = 3i + 6ˆj + 2kˆ
reason is not the correct explanation of
assertion.
Magnitude of
B = 32 + 62 + 22
(c) If assertion is true but reason is false.
(d) If both assertion and reason are false.
= 9 + 36 + 4
AIIMS-26.05.2019(M) Shift-1
B =7
Ans. (b) : Given that, | A + B |=| A − B |
ˆ
So,
C= BA
2
2
2
2
A + B + 2ABcos θ = A + B − 2ABcos θ
ˆi + 2ˆj + 2kˆ
Squaring on both side
C = 7×
3
A 2 + B2 + 2ABcos θ = A 2 + B2 − 2ABcos θ
2
2
2
2
7 ˆ
A + B + 2AB cos θ − A − B + 2AB cos θ = 0
C=
i + 2jˆ + 2kˆ
3
4ABcos θ = 0
θ = 90°
76. The (x,y,z) coordinates of two points A and B
Also, vector addition is cumulative
are given respectively as (0,3,–1) and (–2,6,4).
The displacement vector form A to B may be
Hence, | A + B |=| B + A |
given by :
74. If A = 3iˆ + 4jˆ and B = 7iˆ + 24jˆ then the vector
(a) −2iɵ + 6ɵj + 4kɵ
(b) −2iɵ + 3jɵ + 3kɵ
having the same magnitude as B and parallel to
A is(c) −2iɵ + 3jɵ + 5kɵ
(d) 2iɵ − 3jɵ − 5kɵ
(a) 15iˆ + 20ˆj
(b) 5iˆ − 3jˆ
BCECE-2006
(c) 15iˆ + 13jˆ
(d) 5iˆ + 14jˆ
Ans. (c) : Given,
ˆ
BCECE-2014
rA = 3jˆ − k,
rB = −2iˆ + 6ˆj + 4kˆ
Ans. (a) : Given that,
Displacement vector (rAB) = rB – rA.
A = 3iˆ + 4ˆj and B = 7iˆ + 24jˆ
rAB= −2iˆ + 6ˆj + 4kˆ − 3jˆ − kˆ
to get direction we will find unit vector in the direction
rAB = −2iˆ + 3jˆ + 5kˆ
of a ,
Therefore, the displacement vector from A to B is
A
3iˆ + 4ˆj
3ˆ 4ˆ
 =
=
= i+ j
−2iˆ + 3jˆ + 5kˆ
5
5
|A|
32 + 4 2
So required vector,
77. The resultant of two forces P and Q is of
magnitude P. If P be doubled, the resultant will
ˆ = 72 + 242  3 ˆi + 4 ˆj
|B|A
 5
be inclined to Q at an angle.
5 
(a) 00
(b) 300
3
4 
1
= 625  ˆi + ˆj = 25 × (3iˆ + 4ˆj)
(c) 600
(d) 900
 5
5 
5
UPSEE-2016
ˆ
ˆ
ˆ
| B | A = 15i + 20 j
BCECE-2010
(
(
Objective Physics Volume-I
358
)
) (
)
YCT
80.
Ans. (d) : Let the angle between P and Q be θ.
Since, the resultant of P and Q is P
∴P2 = P2 + Q2 + 2PQ cosθ
− Q2 = 2PQ cosθ
Q = –2P cosθ
According to the question, given P is doubled then
4P2 = 4P2 + Q2 + 4PQ cosθ
Put the value of Q
0 = (–2Pcosθ)2 + 4P(–2Pcosθ).cosθ
0 = 4P2cos2θ – 8P2cos2θ
0 = –4P2cos2θ
cosθ = 0
θ = 90°
78.
∠ABC.
(a) cos–1
5
11

5 
(c)  90° − cos −1


11


(b) cos–1
6
11

5 
(d) 180° − cos −1


11


WB JEE 2018
Ans. (a) :
If a + b = c and a + b = c, then the angle
included between a and b is
(a) 90°
(b) 180°
(c) 120°
(d) zero
WB JEE-2010
Given that, AB = 3iˆ + ˆj + kˆ , AC = ˆi + 2ˆj + kˆ
Ans. (d) : Given that, a + b = c and a + b = c
∴ CB = AB − AC
ˆ
CB = 3iˆ + ˆj + kˆ − (iˆ + 2ˆj + k)
| c | =| a + b |
c2 = a2 + b2 + 2abcosθ
CB = 2iˆ − ˆj
c = a 2 + b 2 + 2ab cos θ
a + b = a + b + 2abcos θ
∴
a2 + b2 + 2ab = a2 + b2 + 2abcosθ
cos θ = 1
θ = 0º
Consider the vectors A = ˆi + ˆj - kˆ , B =
2
79.
In a triangle ABC, the sides AB and AC are
represented by the vectors 3iˆ + ˆj + kˆ and
ˆi + 2jˆ + kˆ respectively. Calculate the angle
2
1 ˆ ˆ
ˆ . What is the
2iˆ − ˆj + kˆ and C =
(i − 2 j + 2k)
5
value of C.(A ×B)?
(a) 1
(b) 0
(c) 3 2
(d) 18 5
WB JEE 2020
ˆ B = 2iˆ − ˆj + kˆ
Ans. (b) : Given, A = ˆi + ˆj − k,
81.
C=
(
1 ˆ ˆ
i – 2 j + 2kˆ
5
)
Now,
ˆi
ˆj kˆ
A × B = 1 1 −1
2 −1 1
A × B = −3jˆ − 3kˆ
According to the question,
1 ˆ ˆ
C. A × B =
i − 2 j + 2kˆ . 0iˆ − 3jˆ − 3kˆ
5
1
=
( 6 − 6)
5
=0
(
)
(
Objective Physics Volume-I
)(
)
∠ABC is angle between AB and CB
(
AB ⋅ CB = AB CB cos θ
(6 – 1) =
83.
5
11
 5 
θ = cos −1 

 11 
Three vectors A = aiˆ + ˆj + kˆ ; B = ˆi + bjˆ + kˆ and
C = ˆi + ˆj + ckˆ are mutually perpendicular ( ˆi, ˆj
and k̂ are unit vectors along X, Y and Z- axis
respectively). The respective values of a, b and
c are
1 1 1
(a) 0, 0, 0
(b) – , – , –
2 2 2
1 1 1
(c) 1, –1, 1
(d) , ,
2 2 2
WB JEE 2017
ˆ
ˆ
ˆ
Ans. (b) : Given that, A = ai + j + k, B = ˆi + bjˆ + kˆ , C =
ˆi + ˆj + ckˆ are mutually perpendicular–
So,
A.B = 0
B. C = 0
A. C = 0
Now, a + b + 1 = 0 …… (i)
359
|C|
θ = cos 

| A|
–1  3 
θ = cos  
5
The magnitudes of vectors A, B and C are 3, 4
and 5 units respectively. If A+B = C, the angle
between A and B is
π
(a)
(b) cos−1(0.6)
2
π
7
(c) tan − 1 
(d)
4
5
[AIPMT 1988]
−1
32 + 12 + 12 × (2)2 + (−1) 2 × cos θ
5 = 11 × 5 × cos θ
5
cos θ =
11 × 5
cos θ =
1 + b + c = 0 …… (ii)
84. Given
A = 2iˆ + 3jˆ
and
B = ˆi + ˆj . The
a+1+c=0
.....(iii)
component
of
vector
A
along
vector
B is
On adding equation (i), (ii) and (iii), we get–
1
3
2(a + b + c) = –3
(b)
(a)
2
2
⇒
a + b + c = –3/2 ….. (iv)
By equation (i) and (iv) we get–
5
7
(c)
(d)
– 1 + c = –3/2
2
2
c = –1/2
HPCET-2018
Similarly, b = –1/2
WB JEE 2011, 2009
a = –1/2
ˆ ˆ
ˆ ˆ
82. If A = B + C have scalar magnitudes of 5, 4, 3 Ans. (c) : Given that, A = 2i + 3j, B = i + j
units respectively, then the angle between A
| B |= 12 + 12 = 2
and C is
(a) cos–1(3/5)
(b) cos–1(4/5)
A ⋅ B = ( 2iˆ + 3jˆ )( ˆi + ˆj) = 2 + 3 = 5
(c) π/2
(d) sin–1(3/4)
Component of vector A along vector B would be
WB JEE 2012
A⋅B 5
Ans. (a) : Here, triangle PQR is a given with vector. A,
=
=
2
B
B, C are its adjacent sides.
85. Given C = A × B and D = B × A . What is the
angle between C and D ?
(a) 30°
(b) 60°
(c) 90°
(d) 180°
WB JEE 2009
5, 4 and 3 makes the triangle right angle triangle.
Ans. (d) : Given, C = A × B , D = B × A
QR
cos θ=
C and D are anti parallel so, A × B = – B × A
PQ
Ans. (a) : Given that, A = 3, B = 4, C = 5
Let θ be the angle between them
A+B=C
Squaring on both side
2
2
2
2
A + B + 2AB = C
YCT
2
On putting the value, we get–
(3)2 +(4)2 + 2× 3× 4 cosθ = (5)2
25+24 cosθ = 25
cosθ = 0
θ=
86.
4
(b) sin–1  
 13 
3
(d) cos–1  
 13 
WB JEE 2008
Ans. (c) : Given that,
A = 4iˆ + 3jˆ + 12kˆ
Angle with x-axis
A
cosθ = x
|A|
cosθ =
4
( 4 ) + ( 3) + (12 )
2
2
2
4
cos θ =
13
…(i)
87.
π
2
Objective Physics Volume-I
The angle subtended by the vector A = 4 î +
3 ĵ + 12 k̂ with x-axis is
3
(a) sin–1  
 13 
4
(c) cos–1  
 13 
2
A + B + 2 A B cos θ = C
)
So, angle between C and D is 180º.
360
4
θ = cos −1  
 13 
Which of the following is a vector quantity?
(a) Temperature
(b) Flux density
(c) Magnetic field intensity
(d) Time
WB JEE-2007
YCT
path of the particle makes with the x-axis an
Ans. (c) : Time and temperature have only magnitude
angle of
so they are scalar. The flux density is the dot product of
(a) 30°
(b) 45°
the field and the area vector so it is also scalar.
Magnetic field intensity (H) has both direction and
(c) 60°
(d) 0°
magnitude so it is a vector quantity.
UP CPMT-2008
88. If a1 and a2 are two non-collinear unit vectors
JIPMER - 2009
and if a1 + a 2 = 3,
Ans. (c) : Let, θ be the angle that particle makes with xaxis.
Then the value of (a1 – a2) . (2a1 + a2) is
(a) 2
(b) 3
1
(c)
(d) a1 , and a 2 1
2
UP CPMT-2009
93.
(a) 6iˆ + 2jˆ − 3kˆ
(c) −18iˆ − 13jˆ + 2kˆ
∴ | a1 | = |a 2 |=1
a12 + a22 + 2a1a2 cosθ =
( 3)
θ = 60°
91.
1 + 1 + 2cosθ = 3
2 cosθ = 3 – 2
2 cos θ = 1
1
cos θ =
2
Now,
= (a1 – a2) . (2a1 +a2)
= 2a12 − 2(a1 ⋅ a 2 ) + (a1 ⋅ a 2 ) − a 22
UP CPMT-2010
τ = r×F
ˆi ˆj kˆ
τ= 8 2 3
−3 2 1
Putting the value of a1 = 1 and a2 = 1 and cosθ =1/2
= 2a12 – a22 – a1 a2 cos θ
1
=2–1–
2
1
(a1 – a2) . (2a1 +a2) = 1 −
2
1
(a1 – a2) . (2a1 +a2) =
2
(d) A × B = AB
UP CPMT-2007
The
A,B and C
vectors
are
( 3, 3) The
361
(b) 30 2 cm
such
that
(d) 15 cm
TS EAMCET 18.07.2022, Shift-I
Ans. (a) : Displacement vector of ants = 10iˆ + 20ˆj
A = B , C = 2 A and A + B + C = 0 .
The
Position vector of point = 2 (cos45o î + sin45o ĵ )
angles between A and B, B and C respectively
Position vector of point = ˆi + ˆj
are
Cross product of Ant displacement and position vector
(a) 45o, 90o
(b) 90o, 135o
(c) 90o, 45o
(d) 45o, 135o
= (10 î + 20 ĵ ).( î + ĵ )
TS EAMCET(Medical)-2015
= (10 + 20)
Ans. (b) : According to figure.
= 30 cm
If force F = 5iˆ + 3jˆ + 4kˆ makes a displacement
of s = 6iˆ – 5kˆ work done by the force is
A vector is given as A = 4iˆ + 7jˆ . What would
be the angle, the vector A makes with y – axis
 7 
 4 
(a) θ = cos −1 
(b) θ = cos −1 


 11 
 11 
 7 
 4 
(c) θ = cos −1 
(d) θ = cos −1 


 65 
 65 
TS EAMCET 30.07.2022, Shift-I
Ans. (c) : Given, A = 4iˆ + 7ˆj , B = ˆj (y-axis)
(a) 10 units
(b) 122 5 units
(c) 5 122 units
(d) 20 units
Given that, A + B = −C
A + B = −C = C
Squaring both side, we get–
2
A+B = C
(A + B)⋅(A + B) = C
Force (F) = 5iˆ + 3jˆ + 4kˆ
Displacement (s) = 6iˆ − 5kˆ
(∵ cos90° = 0)
2
→
UP CPMT-2003
Hence, two vector are perpendicular if their dot product
is equal to zero.
90. A particle starting from the origin (0, 0) moves
in a straight line in the (x, y) plane. Its
Objective Physics Volume-I
94.
−1
(a) 30 cm
30
(c)
cm
2
ˆ −18 + 20)
v = ˆi(−24 + 6) − ˆj(18 − 5) + k(
v = −18iˆ − 13jˆ + 2kˆ
2
2

∵ C ⋅ C = C 


2
A ⋅ A + A ⋅ B + B⋅ A + B⋅ B = C
We know that,
cos θ =
2
A⋅B
A B
=
(4iˆ + 7ˆj) ⋅ ˆj
4 2 + 7 2 12
 7 
θ = cos 

 65 
−1
2
2
A + A B cos θ + A B cos θ + B = C
The work done is given by the following relation.
A.B =| A || B | cos 90°
coordinates at a later time are
)
kˆ
1
6
96.
Ans. (a) : Given that,
A ⋅ B = A B cos θ
A. B = 0
ˆj
−4
−6
cos φ =
So, θ = 90o, φ = 135o
95. An ant starts from the origin and crawls 10 cm
along the x – axis and then 20 cm along the y –
axis. The dot product of the ant's displacement
vector with the position vector of a point that
makes 45° with the x – axis and has a
magnitude of 2 cm is
= −4iˆ − 17ˆj + 22kˆ
92.
Ans. (c) : We know that,
∴
B2 = −B ⋅ 2 B ⋅ cos φ
= ˆi ( 2 − 6 ) − ˆj ( 8 + 9 ) + kˆ (16 + 6 )
Two vectors are perpendicular, if
(a) A ⋅ B = 1
(b) A × B = 0
(c) A ⋅ B = 0
(b) 4iˆ + 4ˆj + 6kˆ
(d) −4iˆ − 17ˆj + 22kˆ
Ans. (d) : Torque of the force
= 2a12 − (a1 ⋅ a 2 ) − a 22
89.
Find the torque of a force F = −3iˆ + 2jˆ + 1kˆ
acting at the point r = 8iˆ + 2jˆ + 3kˆ .
(a) 14iˆ − 38jˆ + 16kˆ
(c) −14iˆ + 38jˆ − 16kˆ
(
Linear velocity (v) = = ω× r = ω× r
ˆi
v= 3
5
A = B, C = 2 A = 2 B
φ = 135o
We know that
θ = tan –1 3
2
A B cos 90o + B2 = − B C cos φ
UP CPMT-2001
y
3
tanθ = =
= 3
x
3
Let the angle between a1 and a2 is θ
A ⋅ B + B ⋅ B = −B ⋅ C
∵
(b) 18iˆ + 13jˆ − 2kˆ
(d) 6iˆ − 2ˆj + 8kˆ
Ans. (c) : Given that,
ˆ
ω = (3iˆ − 4jˆ + k)
ˆ
r = (5iˆ − 6ˆj + 6k)
Ans. (c ) : Given that, |a1 + a2 | = 3
a1 and a 2 are non-collinear vector
A body is rotating with angular velocity
ˆ . The linear velocity of a point
ω = (3iˆ − 4jˆ + k)
ˆ is
having position vector r = (5iˆ − 6jˆ + 6k)
2
(∵ B = A, C = 2 A)
W = F.d
W = 5iˆ + 3jˆ + 4kˆ . 6iˆ − 5kˆ
(
)(
)
ˆ
ˆ
ˆ
ˆ
ˆ
W = (5i + 3j + 4k ).(6i + 0 j − 5kˆ )
⇒
⇒
⇒
W = 30 + 0 – 20
W = 10 units
97.
A2 + A2cosθ + A2cosθ + A2 = 2A2
(2 + 2cosθ) = 2
cosθ = 0
θ = 90o
Statement (A) : α < β if A < B
Statement (B) : α < β if A > B
Statement (C) : α = β if A = B
A + B = −C
YCT
Objective Physics Volume-I
The resultant of the two vectors A and
B makes angle α with A and β with B .
362
YCT
(a) (A) and (C) are true
100. Find the component of vector P = 2iɵ + 3jɵ along
(b) Only (C) is true
the direction of vector Q = ɵi + ɵj .
(c) (B) and (C) are true
(d) (A),(B),(C) are all true
(a) 2
(b) 2 5
TS EAMCET 08.05.2019, Shift-I
2
5
(c)
(d)
Ans. (c) : A = R cos α
5
2
B = R cos β
TS EAMCET 04.08.2021, Shift-II
Suppose that α < β
HP CET-2018
Ans. (c) :
Then, cosα > cosβ, So A > B
And suppose that α = β
P.Q
Projection of P on Q =
Then, cosα = cosβ, So A = B
Q
98. Vector a = ɵi + 2jɵ + 2kɵ and b = ɵi - ɵj + kɵ . What is
2i + 3j)( i + j)
(
2+3
5
=
=
=
the unit vector along a + b ?
2
2
2
2iɵ + ɵj + 3kɵ
2iɵ − ɵj + 4kɵ
101. The y-component of vector A is +3.0 m if A
(a)
(b)
makes an angle of 30° counter clockwise from
14
20
the positive y - axis, the magnitude of A is
2iɵ + ɵj + 3kɵ
2iɵ + ɵj − 3kɵ
(c)
(d)
(assume A is in x - y plane)
13
10
(a) 2 3m
(b) 11m
TS EAMCET 28.09.2020, Shift-I
(c) 15m
(d) 21m
Ans. (a) : Given that, a = ˆi + 2 j + 2kˆ , b = ˆi – j + kˆ
TS EAMCET 04.08.2021, Shift-II
ˆ + (iˆ – j + k)
ˆ
a + b = (iˆ + 2ˆj + 2k)
Ans. (a) : y - component of vector A = + 3
a + b = 2iˆ + j + 3kˆ
A makes an angle of 30º counter clockwise from the
positive y - axis the magnitude of A is,
a+b
Unit vector of (a + b) =
Ay
a+b
A=
cos30º
ˆ
(2iˆ + j + 3k)
2iˆ + j + 3kˆ
=
=
+3
3
3× 3
4 +1+ 9
14
A=
=
=
=2 3
cos30º
3
3/2
99. Find the angle between the two vectors:
ˆ = 5iˆ + 3jˆ + kˆ
2
a = 3iˆ + 2jˆ + 5k,b
A = 2 3m
 26 
 26 
(a) cos −1 
(b) sin −1 


102.
A
particle
is moving such that its position co1330
1330




ordinates (x, y) are (2m, 3m) at time t = 0, (6m,
26 
26 
−1 
−1 
7m)
at
time
t = 2 s and (13m, 14m) at time t = 5
(c) cos 
(d) tan 


s. Average velocity vector (vav) from t = 0 to t =
 1335 
 1330 
5 s is
TS EAMCET 04.08.2021, Shift-I
1 ˆ
7 ˆ ˆ
ˆ b = 5iˆ + 3jˆ + kˆ
(a)
13i + 14ˆj
(b)
i+ j
Ans. (a) : Given, a = 3iˆ + 2ˆj + 5k,
5
3
2
2
2
11
a = a 2x + a 2y + a z2 = ( 3) + ( 2 ) + ( 5 ) = 38
ˆi + ˆj
(c) 2 ˆi + ˆj
(d)
5
2
2
2
b = b 2x + b2y + b z2 = ( 5) + ( 3) + (1) = 35
[AIPMT 2014]
Ans. (d) : According to figure,
ˆ ⋅ (5iˆ + 3jˆ + k)
ˆ
a.b = (3iˆ + 2ˆj + 5k)
= 15+6+5 = 26
a.b = ab cosθ
It is given that, t = 0 to t = 5 sec
(
)
( )
( )
cosθ =
a.b
a b
=
26
38 35
26
=
26
displacement
time
( rC − rA ) 13iˆ + 14ˆj − 2iˆ + 3jˆ 11iˆ + 11jˆ
=
=
5
5
(5 − 0)
11 ˆ ˆ
vavg =
i+ j
5
vavg =
1330
(
θ = cos–1 1330
 26 
So, Angle between two vector is cos–1 

 1330 
Objective Physics Volume-I
( )
(
363
) (
)
)
YCT
103. If the magnitude of sum of two vectors is equal Ans. (a) : |A × B| = 3 (A.B)
to the magnitude of difference of the two
AB sinθ = 3 AB cosθ
vectors, the angle between these vectors is
(a) 90o
(b) 45o
tanθ = 3
o
o
(c) 180
(d) 0
θ = tan–1( 3 )
[NEET 2016, AIPMT 1991]
θ = 60°
Ans. (a) : There are two vectors A and B
It is given that,
106. If A × B = 3 A.B then the value of A + B is
A +B = A−B
(a) (A2 + B2 + AB)1/2
Let, angle between A and B is φ
1/ 2

AB 
A2 + B2 + 2AB cosφ = A2 + B2 − 2AB cosφ
(b)  A 2 + B2 +

3


cosφ = 0 [∵Α, Β ≠ 0]
(c) A + B
π
o
1/ 2
φ = = 90
(d) A 2 + B2 + 3AB
2
104. If
vectors
A = cosωt ˆi + sinωt ˆj
and
BCECE-2013
(
)
UPSEE - 2006
ωt ˆ
ωt
i + sin ˆj are functions of time,
[AIPMT 2004]
2
2
then the value of t at which they are orthogonal
Ans. (a) : Given that, A × B = 3A.B
to each other, is
π
π
AB sinθ = 3 AB cosθ
(a) t =
(b) t =
4ω
2ω
tanθ = 3
π
θ = 60°
(c) t =
(d) t = 0
ω
We know that,
[AIPMT 2015] Law of parallelogram of addition
Ans. (c) : Given, A = cos ωtiˆ + sin ωtjˆ
| A + B | = A 2 + B2 + 2AB cos θ
ωt ˆ
ωt ˆ
B = cos i + sin
j
1
| A + B | = A 2 + B2 + 2AB ×
2
2
2
If two vector are orthogonal then their dot product will
be zero–
A + B = (A2 + B2 + AB)1/2
A⋅B = 0
107. If a vector 2iˆ + 3jˆ + 8kˆ is perpendicular to the
( cos ωtiˆ + sin ωtjˆ ) ⋅  cos ωt ˆi + sin ωt ˆj  = 0
vector 4jˆ − 4iˆ + αkˆ , then the value of α is

2
2 
1
ωt
ωt
(a) –1
(b)
cos ωt ⋅ cos + sin ωt ⋅ sin
=0
2
2
2
1
ωt 

(d)
1
(c)
–
cos  ωt −  = 0
2

2 
Karnataka CET-2017
[∵ cosAcosB + sinAsinB = cos (A – B)]
JIPMER-2007
ωt 
π

AIPMT-2005
cos  ωt −  = cos

2 
2
Ans. (c) : Given, a = 2iˆ + 3jˆ + 8kˆ , b = −4iˆ + 4ˆj + αkˆ
ωt π
ωt −
=
a and b are perpendicular so,
2 2
ωt π
π
a .b = 0
= or t =
2 2
ω
(
2iˆ + 3jˆ + 8kˆ ) ⋅ ( −4iˆ + 4ˆj + αkˆ ) = 0
105. A and B are two vectors and θ is the angle
–8
+ 12 + 8α = 0
between them. If A × B = 3 A.B , then the
4 + 8α = 0
value of θ is
8α = –4
(a) 60o
(b) 30o
1
(c) 30o
(d) 90o
α= −
2
[AIPMT 2007]
B = cos
(
Objective Physics Volume-I
)
364
YCT
108. If a unit vector is represented
0.5iˆ + 0.8jˆ + ckˆ , then the value of c is
(a) 1
by Ans. (b) : Let A ⋅ (B × A) = A ⋅ C
Ans. (b) : Given that,
Here C = B × A which is perpendicular to both vector
0.11
(b)
A and B .
(d) 0.39
∴
A ⋅C = 0
TS EAMCET (Medical)-2017
∵
C is perpendicular to A and B
[AIPMT 1999]
∴ Angle between A and C is 90°
Ans. (b) : Given, A = 0.5iˆ + 0.8jˆ + ckˆ
A ⋅ C = ACcos θ
It is unit vector so it has magnitude
0.01
(c)
∴
( 0.5 ) + ( 0.8 ) + c2 = 1
2
2
c = 0.11
A ⋅ (B × A) = 0
magnitude of component of vector  along
vector B̂ will be ––––––– m.
109. Which of the following is not a vector quantity?
JEE Main-26.07.2022, Shift-II
(a) Speed
(b) Velocity
Ans. (2) : A = 2iˆ + 3jˆ − kˆ and B = ˆi + 2ˆj + 2kˆ
(c) Torque
(d) Displacement
[AIPMT 1995]
A⋅B
Ans. (a) : The physical quantities for which having both Magnitude of A along B =
B
direction and magnitude is called vector quantity.
Example- force, torque, momentum, acceleration
2iˆ + 3jˆ − kˆ ˆi + 2ˆj + 2kˆ
velocity etc.
2
2
2
(1) + ( 2 ) + ( 2 )
110. The angle between the two vectors
A = 3iˆ + 4jˆ + 5kˆ and B = 3iˆ + 4jˆ − 5kˆ will be
2+6−2 6
= =2
3
(a) 0o
(b) 45o
9
o
o
(c) 90
(d) 180
113. A is a vector quantity such that A = non-zero
[AIPMT 1994]
constant. Which of the following expression is
Ans. (c) : Given,
true for A ?
A = 3iˆ + 4jˆ + 5kˆ and B = 3iˆ + 4ˆj − 5kˆ
(a) A ⋅ A = 0
(b) A × A < 0
A.B
(c) A × A = 0
(d) A × A > 0
cos θ =
| A || B |
JEE Main-25.06.2022, Shift-I
c=
0.11
(
(
cos θ =
cos θ =
(3iˆ + 4ˆj + 5kˆ )( 3iˆ + 4ˆj − 5kˆ )
)(
)
)
Ans. (c) : Given that, A ≠ 0
(3) 2 + (4) 2 + (5)2 (3) 2 + (4)2 + (5) 2
A × A = A A sin 0° nˆ = 0
9 + 16 − 25
So,
A×A = 0
114. Which of the following relations is true for two
0
ˆ and B
ˆ making an angle θ to each
cos θ =
unit vector A
50
other ?
cos θ = 0
ˆ +B
ˆ −B
ˆ = A
ˆ tan θ
(a) A
cos θ = cos90°
2
θ = 90°
ˆ −B
ˆ +B
ˆ = A
ˆ tan θ
(b)
A
111. The angle between A and B is θ. The value of
2
the triple product A. B × A is
θ
ˆ
ˆ
ˆ
ˆ
(c) A + B = A − B cos
2
(a) A2B
(b) zero
(c) A2Bsinθ
(d) A2Bcosθ
ˆ −B
ˆ +B
ˆ = A
ˆ cos θ
(d) A
2
JIPMER-2007
JEE Main-25.06.2022, Shift-I
AIPMT-1989
( 3) + ( 4 ) + ( 5 ) . ( 3) + ( 4 ) + ( 5 )
2
2
(
Objective Physics Volume-I
2
2
2
ˆ =1
B
ˆ +B
ˆ =
A
(1) + (1) + 2cos θ
2
θ 
θ

= 2  1 + 2 cos 2 − 1  = 2 cos
2 
2

θ
ˆ
ˆ
A + B = 2cos
.....(i)
2
ˆ −B
ˆ =
A
2
2
ˆ +B
ˆ B
ˆ −2 A
ˆ cos θ
A
ˆ −B
ˆ = 1 + 1 − 2 cos θ
A
ˆ −B
ˆ = 2 − 2cos θ
A
= 2 (1 − cos θ )
θ

= 2 (1 − 1 − 2 sin 2 
2

θ

= 2  1 − 1 + 2sin 2 
2

θ
θ

= 2  1 − 1 + 2sin 2  = 2sin
2
2

ˆ −B
ˆ = 2sin θ
A
.....(ii)
2
Equation (i) divide by equation (ii), we get –
θ
ˆ −B
ˆ
2sin
A
2
=
θ
ˆ +B
ˆ
A
2cos
2
ˆ −B
ˆ
A
θ
= tan
ˆ +B
ˆ
2
A
ˆ −B
ˆ +B
ˆ =A
ˆ tan θ
A
2
115. Match List I with List II.
List-I
List-II
(A) C−A−B=0
i)
)
365
iii)
(D) A+B=−C
iv)
2
θ 

= 2 + 2  2 cos 2 − 1 
2 

ˆ and B = (iˆ + 2jˆ + 2k)m.
ˆ
112. If A = (2iˆ + 3jˆ - k)m
The
2
ˆ =1
A
2
(C) B−A−C=0
2
ˆ +B
ˆ B
ˆ +2 A
ˆ cos θ
A
= 2 + 2 cos θ
A ⋅C = 0
| A |= 1
2
ˆ +B
ˆ =
A
YCT
(B) A−C−B=0
ii)
Choose the correct answer from the options
given below.
(a) (A) → (iv), (B) → (i), (C) → (iii), (D) → (ii)
(b) (A) → (iv), (B) → (iii), (C) → (i), (D) → (ii)
(c) (A) → (iii), (B) → (ii), (C) → (iv), (D) → (i)
(d) (A) → (i), (B) → (iv), (C) → (ii), (D) → (iii)
JEE Main-25.07.2021, Shift-I
Ans. (b) : Applying triangle law of vectors to the
diagram.
(i) B = A + C
∴ B−A−C = 0
This matches with (C).
(ii) –C = A + B
This matches with (D).
(iii) C + B = A
∴ A−C = B
This matches with (B).
(iv) A + B = C
∴ C−A−B=0
116. Two vectors P and Q have equal magnitudes. If
the magnitude of P+Q is n times the magnitude
of P−Q, then angle between P and Q is
 n −1 
 n −1 
(a) sin −1 
(b) cos −1 


 n +1
 n +1 
2
2




n
−
1
n
−
1
(c) sin −1  2
(d) cos −1  2


 n +1
 n +1 
JEE Main-25.07.2021, Shift-II
JEE Main-20.07.2021, Shift-II
JEE Main-10.01.2019, Shift-II
Ans. (d) : Given that,
P=Q
.....(i)
Let the magnitude of (P + Q) = R
P 2 + Q 2 + 2PQ cos θ
R = |P + Q| =
R = P + P + 2P 2 cos θ
The magnitude of (P – Q) = R'
2
R' = |P – Q| =
2
[from (i)]
P 2 + Q 2 − 2PQ cos θ
R ' =| P − Q |= P 2 + P 2 − 2P 2 cos θ
Given that
R = nR'
[from (i)]
2P 2 + 2P 2 cos θ = n 2P 2 − 2P 2 cos θ
squaring both side
2p2 + 2p2 cosθ = n2(2p2 – 2p2 cosθ)
Objective Physics Volume-I
366
YCT
2p (1 + cos θ) = n 2p (1− cos θ)
5 = ( 3) + ( 5) + 2 × 3 × 5cos θ
1 + cosθ = n2(1 – cosθ)
2
2
5 = 9 + 25 + 2 × 3 × 5cos θ
1 + cosθ = n – n cosθ
Squaring both side
cosθ + n2cosθ = n2 – 1
25 = 9+25+2× 3 × 5 cosθ
cosθ(1 + n2) = n2 – 1
–9
n2 −1
cos θ =
cos θ = 2
2 × 3× 5
n +1
–3
 n2 −1 
cos θ =
θ = cos −1  2 
10
 n +1 
2A1 + 3A 2 . 3A1 − 2A 2
117. If A and B are two vectors satisfying the
relation A ⋅ B = A × B . Then, the value of
= 6 | A1 |2 +9A1 ⋅ A 2 − 4A1 ⋅ A 2 − 6 | A 2 |2
A − B will be
= 6(3) 2 + 9A1.A 2 − 4A1.A 2 − 6 × 25
(a) A 2 + B2
(b) A 2 + B2 + 2AB
= 54 + 5A .A − 6 × 25
2
2
2
2
(
)(
1
A 2 + B2 + 2AB
(d) A 2 + B2 − 2AB
JEE Main-20.07.2021, Shift-I
Ans. (d) : Given that,
(c)
A⋅B = A×B
AB cosθ = AB sinθ
tanθ = 1
θ = tan–1(1)
θ = 45°
The value of A − B is
 a a
Position vector of H is, H  0, , 
 2 2
a
a
OH = ˆj + kˆ
2 2
GH = OH − OG
a  a
a 
a
GH =  ˆj + kˆ  −  ˆi + kˆ 
2 2  2 2 
a
GH = ˆj – ˆi
2
121. If A × B = B × A , then the angle between A and
B is
(a) π
(b) π / 3
(c) π / 2
(d) π / 4
AIEEE-2004
Ans. (a) :
A × B = B× A
2
( )
)
2
= 54 + 5 A1 A 2 cos θ − 150
 −3 
= 54 + 5 × 3 × 5   − 150
 10 
45
= 54 –150 –
2
= – 118.5
120. In the cube of side ‘a’ shown in the figure, the
vector from the central point of the face ABOD
to the central point of the face BEFO will be
| A − B |= A 2 + B2 − 2ABcos 45°
118. If P ×Q =Q ×P, the angle between P and Q is
θ (0°< θ <360°). The value of θ will be
……………°.
JEE Main-25.02.2021, Shift-II
Ans. (180) :
(
(
(
1 ˆ ˆ
a i−k
2
1 ˆ ˆ
(c) a j − k
2
(a)
)
2 P×Q = 0
If P = 0 or Q = 0
The angle between P & Q is 180º (0º<θ < 360º)
So, θ = 180º
)
)
( )
( )
1 ˆ ˆ
a j− i
2
1 ˆ ˆ
(d) a k − i
2
JEE Main-10.01.2019, Shift-I
(b)
(a) −106.5
(c) −99.5
(b) −112.5
(d) −118.5
JEE Main-08.04.2019, Shift-II
Ans. (d) : Given that
A1 = 3, A 2 = 5, A1 + A 2 = 5
A1 + A 2 =
2
2
A1 + A 2 + 2 A1 A 2 cos θ
Objective Physics Volume-I
(
)
(d)
(
)
(
)
46 ˆ ˆ ˆ
6i − j + 3k
29
(
X=
(b) 5 m, 5 m
AB
A
(
)
(
)
(c) 12 N
YCT
)
(a) 5 m, 5 2 m
Now,
)
done by this force in moving the body a
distance 4m along the z-axis is.
(a) −4iˆ + 8ˆj + 12kˆ N (b) −4iˆ + 8ˆj N
a a
Position vector of G is, G  ,0, 
2 2
aˆ a ˆ
OG = i + k
2 2
)(
= 0 + 0 + 12 = 12N
124. The sum of three vectors in the figure below is
zero. The magnitude of OC and OB is
(d) 5 2 m, 5 2 m
29 ˆ ˆ ˆ
Assam CEE-2018
6i − j + 3k
46
Ans. (c) : Given, the sum of three vectors in given
Assam CEE-2020 figure is zero
i.e. 0iˆ + 0ˆj + 0kˆ = 0 ...... (i)
(
367
W = F.d
= –iˆ + 2ˆj + 3kˆ 0iˆ + 0ˆj + 4kˆ
(c) 5 2 m, 5m
46 ˆ ˆ
3i + 4 j + 2kˆ
29
123. A body constrianed to move along the z-axis of
a co-ordinate system is subject to a constant
force F given by F = -iˆ + 2jˆ + 3kˆ N. The work
2
Work done is given by-
B = 6iˆ – ˆj + 3kˆ
Let X be the vector parallel to A whose magnitude is
equal to that of B
X=
( 2A + 3A ) . ( 3A - 2A ) is
1
(b)
3iˆ + 4ˆj + 2kˆ
=
36 + 1 + 9
9 + 16 + 4
A1 + A 2 = 5 . The value of
2
)
29 ˆ ˆ
3i + 4j + 2kˆ
46
Ans. (b) :
119. Let A1 = 3, A 2 = 5 and
1
(
46 ˆ ˆ
3i + 4j + 2kˆ
29
Ans. (a) : Givne that,
A = 3iˆ + 4ˆj + 2kˆ
P × Q = –P × Q
)
)
Find a vector
parallel to A whose magnitude equal to that of
B
(c)
If P × Q = Q × P
(
(
d = 0iˆ + 0ˆj + 4kˆ
A × B = –(A × B)
ABsinθ = –ABsinθ
2ABsinθ = 0
sinθ = 0
θ = 0, π, 2π
ˆ B = 6iˆ - ˆj + 3kˆ
122. A = 3iˆ + 4jˆ + 2k,
(a)
| A − B |= A 2 + B2 − 2AB
Ans. (c) : Given
F = –iˆ + 2ˆj + 3kˆ
Objective Physics Volume-I
(
)
ˆ
(d) 12kN
Assam CEE-2021
368
OA = 5(−ˆj) = −5ˆj
ˆ
OB = OB (i)
OC = OCcos 45°(−ˆi) + OCsin 45° ˆj
=−
OC ˆ OC ˆ
i+
j
2
2
OC  ˆ 
OC  ˆ

R =  OB −
 i +  −5 +
j
2 
2

.....(ii)
YCT
Comparing equation (i) and (ii), we get
OC  ˆ 
OC  ˆ

0iˆ + 0ˆj =  OB −
 i +  −5 +
j
2 
2

OB −
Ans. (b) :
OC
OC
= 0 ⇒ OB =
2
2
= A × (A − B) + B ×(A − B)
OC = 5 2 m
= 0 − A × B + B ×A − 0
The angle β which the resultant = makes with x is given
by
ysinθ
5 2
tanβ =
Then, OB =
= 5m
x + y cosθ
2
y sin θ
θ
θ

∵β = 
125. The unit vector perpendicular to the plane of tan   =
2
 2  x + y cos θ 
A = ˆi – 3jˆ – kˆ and B = 2i + ˆj – kˆ is
sin θ / 2 y × ( 2sin θ / 2.cos θ / 2)
=
4 ˆ
1 ˆ
7 ˆ
cos θ / 2
x + ycos θ
(a)
i−
j+
k
x + y cosθ = 2y .cos2 θ/2
66
66
66
θ 
θ

2 ˆ
1 ˆ
8 ˆ
x + y  2cos 2 – 1 = y.2cos 2
(b)
i−
j+
k
2 
2

66
66
66
x–y=0
4 ˆ
1 ˆ
7 ˆ
x
=
y
(c)
i+
j+
k
66
66
66
127. The
resultant
of
three
vectors
ˆ B (3iˆ − 2jˆ − 2k)and
ˆ
2 ˆ
1 ˆ
8 ˆ
A(2iˆ − ˆj + 3k),
C is a unit
(d)
i+
j+
k
66
66
66
vector along z direction is given by
Assam CEE-2016
(a) C = 3jˆ + 5kˆ
(b) C = 3iˆ + 2kˆ
Ans. (a) : Given,
(c) C = 5iˆ + kˆ
(d) C = −5iˆ + 3jˆ
A = ˆi – 3jˆ – kˆ and B = 2iˆ + ˆj – kˆ
Tripura-2020
(
)
(
Ans. (d) : A = 2iˆ − ˆj + 3kˆ
ˆi
ˆj kˆ
A × B = 1 –3 –1
2
1 –1
Now
16 + 1 + 49 = 66
So, Unit vector perpendicular to the planes A and B is
A×B
ˆj
4iˆ
7kˆ
=
−
+
66
66
66
(
A + B + C = 1 unit along z-direction
2iˆ − ˆj + 3kˆ + 3iˆ − 2jˆ − 2kˆ + xiˆ + yjˆ + zkˆ = 1kˆ
) (
) (
)
(2 + 3 + x )ˆi + (−1− 2 + y)ˆj + (3 − 2 + z) kˆ = 1kˆ
(5 + x )ˆi + (−3 + y)ˆj + (1 + z ) kˆ = 1kˆ
A × B = 42 + (−1) 2 + 7 2
A×B
(
)
C = ( xiˆ + yjˆ + zkˆ )
Let
= 4iˆ − ˆj + 7kˆ
=
)
B = 3iˆ − 2ˆj − 2kˆ
= ˆi ( 3 – ( –1) ) – ˆj ( –1 – ( –2 ) ) + kˆ (1 – ( –6 ) )
Now,
r1 = 4iˆ – 3jˆ – 2kˆ m
= A × A − A × B + B ×A − B × B
)
( )
(
)
r = ( 5iˆ – 4ˆj + 2kˆ ) m
F = 4iˆ + 3jˆ N
(A + B) × (A − B)
OC
−5 = 0
2
(
Ans. (a) : Given data
Ans. (a) : If A and B are two vector
Then,
∵ A × A = 0 


 B × B = 0 
d = ( 5iˆ − 4ˆj + 2kˆ ) − ( 4iˆ − 3jˆ − 2kˆ )
d = ˆi – ˆj + 4kˆ
(∵ −A × B= B × A )
= B × A + B ×A
(
and r2 = ( 1, −1,1) . The unit vector in the
direction of r1 × r2
ɵi
kɵ
−
2
2
ɵi
kɵ
+
2
2
(c)
ɵi
kɵ
−
1
2
(a)
(b)
2
2
3
3
ɵi
ɵk
(c) 3
(d) 3
(d) −
+
2
2
AP EAMCET-11.07.2022, Shift-II
AP EAMCET-06.07.2022, Shift-I Ans. (a) : A = ˆi + ˆj + kˆ and unit vector parallel to
( ˆi – ˆj + kˆ ) is = AB⋅ B = ( i + j + k ) ⋅ ( i − j + k )
ˆ ˆ
1
ˆ
2
2
1
=
ˆi
ˆj kˆ
r1 × r2 = 1 1 1
1 –1 1
3
132. The component of a vector P = 3iˆ + 8jˆ along the
(
)
= ˆi (1 – ( –1) ) – ˆj (1 – 1) + kˆ ( –1 – 1)
direction ˆi + 2jˆ is
= 2iˆ − 0ˆj – 2kˆ
= 2iˆ – 2kˆ
8
(a)
5
11
(c)
5
r1 × r2 = 22 + (−2)2
(b)
19
(d)
5
10
AP EAMCET-05.07.2022, Shift-I
Ans. (b) : P = 3iˆ + 8jˆ
Q = ˆi + 2ˆj
=2 2
r1 × r2
r × r2
ˆ
P component along Q direction = PQ
2kˆ
=
–
2 2 2 2
ˆi
kˆ
=
–
2
2
P.Q
=
Q
( 3iˆ + 8jˆ )( ˆi + 2ˆj) = 3 + 16 = 19
1 + 22
5
5
133. Which of the following is not true about vectors
A,B and C ?
130. A uniform force of ( 4iˆ + 3jˆ ) newton acts on a
body mass 5 kg. The body is displaced from
( 4iˆ − 3jˆ − 2kˆ ) m to ( 5iˆ − 4jˆ + 2kˆ ) m. Then, the
Objective Physics Volume-I
Objective Physics Volume-I
YCT
ˆ ˆ
1 + ( −1) + (1)
126. The angle between two vectors x and y is θ. If
C = −5iˆ + 3jˆ
the resultant vector z makes an angle θ/2 with
128. If A and B are two vectors, then the value of
x, then which of the following is true?
(A + B) ×(A – B) is
(a) x = 2y
(b) x = y
(a) 2(B × A)
(b) –2(B × A)
y
(c) x = 2y + 1
(d) x =
2
(c) B × A
(d) A×B
Assam CEE-2016
HP CET-2018
369
ˆ
2
r2 = ˆi – ˆj + kˆ
=
(
)
= (−5iˆ + 3jˆ + 0kˆ )
F.d = 1 Joule
131. The dot product of A = ( ˆi + ˆj + kˆ ) and the unit
vector parallel to ( ˆi − ˆj + kˆ ) is
(b) −
Ans. (a) : Given,
r = ˆi + ˆj + kˆ
2iˆ
)
F.d = 4 –3
129. Two position vectors are given by r1 = (1,1,1)
(a)
)(
∴ Work done = F.d = 4iˆ + 3jˆ . i – ˆj + 4kˆ
= 2(B × A)
The unit vector in direction of r1 × r2 is
By comparing both side
5 + x = 0 ⇒ x = –5
y–3=0⇒y=3
1+z=1⇒z=0
Now
C = xiˆ + yjˆ + zkˆ
2
∴ Displacement, d = r2 – r1
work done by the force on the body in joule is
(a) 1
(b) 5
(c) 7
(d) 11
AP EAMCET-11.07.2022, Shift-II
370
( A ⋅ A )( B ⋅ C ) is a scalar value.
(b) ( A × B ) ⋅ ( B × C ) is a scalar value.
(c) ( A × C ) × ( B × C ) is a scalar value.
(d) A × ( B × C ) is a vector value.
(a)
AP EAMCET-04.07.2022, Shift-II
YCT
Ans. (c) : For A,B and C
(
) (
1 −1 −1
3 3
cos θ =
−1
cos θ =
3
cos θ =
=
)
i.e. A × C × B × C is a vector value.
Ans. (b) : Given, A = B
From figure,
a.b
a b
and
∴
1
(a) cos−1  
 3 
−1
−1 
(c) sin  
 3 
 −1
(b) cos−1  
 3 
1
(d) sin  
 3 
JIPMEER-2015
−1
Ans. (b) : Given vectors,
a = ˆi + ˆj – kˆ
b = ˆi − ˆj + kˆ
angle between two linear trans membrane
domains
)(
ˆi + ˆj – kˆ . ˆi − ˆj + kˆ
2
Objective Physics Volume-I
F = 36 + 324 + 100 N
F = 460 N ⇒ F = 2 115 N
Force = Mass × Acceleration
2
115
2 115
⇒ Mass =
kg
4
8
140. For the resultant of two vectors A and B to
maximum. The angle between them should be
perpendicular to the vector A and its
_____.
magnitude is equal to half of the magnitude of
(a) 180°
(b) 0°
(c) 90°
(d) 60°
vector B . Then the angle between A and B is––.
AP EAMCET-23.08.2021, Shift-II
(a) 30º
(b) 45º
Ans. (b) : For maximum Resultant
(c) 150º
(d) 120º
AP EAMCET-06.09.2021, Shift-I FR = F12 + F22 + 2F1F2 cos θ
Ans. (c) : Given that R = B/2
For maximum
Resultant of two forces
cosθ =1
138. The resultant of the two vectors A and B is
... (i)
θ = 0°
Then resultant force
)
2
B = 4iɵ + 2ɵj − 4kɵ
= (2 × 4) + (4 × 2) − (4 × 4)
= 8 + 8 – 16 = 0
2
371
FR = F1 + F2
tan α =
Since A.B = ( 2iˆ + 4ˆj + 4kˆ ) ⋅ ( 4iˆ + 2ˆj − 4kˆ )
(1) + ( −1) + (1)
2
Mass =
FR = F12 + F22 + 2F1F2
So the angle between A and B
180 – θ = 180 – 45
= 135°
π
= 135 ×
180
= 3π radian
4
135. The angle between two linear trans-membranes
domains is defined by following vectors
a = ˆi + ˆj - kˆ and ˆi - ˆj + kˆ
2
8 + 8 − 16
4 + 16 + 16 16 + 4 + 16
(∴A = C )
Bsin θ
(∵ A = B )
A + Bcos θ
sin θ
tan α =
1 + cos θ
θ
θ
2sin .cos
2
2
tan α =
θ
1 + 2 cos 2 − 1
2
θ
θ
2sin .cos
2
2
tan α =
θ
2 cos 2
2
θ
tan α = tan
2
θ
α=
2
137. Find the angle between the vectors
ɵ
A = 2iɵ + 4jɵ + 4kɵ and B = 4iɵ + 2jɵ - 4k.
(a) 0º
(b) 45º
(c) 60º
(d) 90º
AP EAMCET-24.08.2021, Shift-I
Ans. (d) : Given,
A = 2iɵ + 4jɵ + 4kɵ
(1) + (1) + ( −1)
2
R = A 2 + B2 + 2ABcosθ
C
A
A C
cot θ = = = 1
C C
θ = 45°
=
F = 62 + ( −18 ) + 102 N
cosθ = 0
cosθ = cos 90o
θ = 90o
Hence, vectors A and B are perpendicular to each other.
tan θ =
(
Ans. (a) : Magnitude of the force is given as –
A B
The cross product is always vector quantity.
134. If A + B = C and that C is perpendicular to A.
 1
What is the angle between A and B, if |A| = |C|?
θ = cos −1  − 
π
π
 3
(a)
rad
(b)
rad
4
2
136. A and B are two vectors of equal magnitudes
3π
(c)
rad
(d) π rad
and θ is the angle between them. The angle
4
between A or B with their resultant is
JIPMER-2016
θ
θ
Ans. (c) : Given, A = C
(a)
(b)
4
2
(c) 2 θ
(d) Zero
AP EAMCET -2010
cos θ =
A⋅B
YCT
A and R are perpendicular to each other
Bsin θ
∴ tan 90º =
⇒ A + Bcos θ = 0
A + Bcos θ
∴ cos θ = –A/B
put the value of cos θ in equation (i).
For minimum cosθ= –1
θ=180°
Then resultant force
FR = F12 + F22 – 2F1F2 ⇒ FR = F1 − F2
141. The vectors
and C = − ˆi + αˆj + kˆ
B
= A 2 + B2 − 2A 2
4
B2
= B2 − A 2
4
3B
3B2
= A2 ⇒ A =
4
2
3
cosθ = −A / B = −
2
∴ θ = 150º
The angle between A and B is 150º.
(c)
115
kg
2
And
)
)
(b) 1
(d) none of these
AMU-2008
C = −ˆi + αˆj + kˆ
Here, three vector A, B,C are coplanar if
 A BC  = 0


(b) 10 2 kg
115
kg
2
Or
AP EAMCET-24.08.2021, Shift-I
Objective Physics Volume-I
(
are coplanar when the
Ans. (d) : Given,
A = ˆi + ˆj − 2kˆ
B = 2iˆ + 2ˆj − kˆ
139. When a force F given by F = 6iɵ - 18jɵ + 10kɵ acts
on a body, it imparts an acceleration of 8 m.s–2. ∴
Then find the mass of the body ––––
115
kg
4
)
constant α is equal to
(a) 1/3
(c) 3
2
(a)
(
A = ˆi + ˆj − 2kˆ ,B = 2iˆ + 2jˆ − kˆ
(
B
 −A 
= A 2 + B2 + 2AB 

2
 B 
(d)
372
1 1 −2
2 2 −1 = 0
−1 α 1
1(2 + α) −1(2 −1) −2(2α +2) = 0
2 + α −1 − 4 α − 4 = 0
−3 α − 3 = 0
1
α=−
3
YCT
142. If A = ˆi – ˆj and B = 3iˆ + 4jˆ , the vector having 144. The angle between the vectors : a = 3i – 4j
and b = –2i + 3k is
 1
(a) cos –1  − 
 3
 1
(c) cos –1  − 
 2
same magnitude as B but parallel to vector A
can be written as
(a) 5 ˆi - ˆj
(b) 5/ 2 ˆi – ˆj
( )
(c) 2 ( 4iˆ – 3jˆ )
(
(d)
)( )
3 ( ˆi – ˆj)
 1
(b) cos –1  − 
 4
 1
(d) cos –1  − 
 6
AMU-2012
AMU-2019
Ans. (a) : Given
a = 3i − 4 j and b = −2i + 3k
We know
Ans. (b) : Given,
A = ˆi − ˆj
B = 3iˆ + 4jˆ
( ) A
 =
A
=
cos θ =
( ˆi − ˆj) = 1 ˆi − ˆj
( )
1+1
2
( )
Magnitude of B =
2
2
∴
( )
( )
143. Find the component of vector A = 2iˆ + 3jˆ along
( )
the direction ˆi - ˆj .
(a) −
(c)
( )
1 ˆ ˆ
i–j
2
( )
Ans. (a) : A = 2iˆ + 3jˆ
B = ˆi − ˆj
Unit vector B̂
ˆi − ˆj
B
=
B
(1) 2 − (−1) 2
( 3i − 4 j ).( −2i + 3k )
( 3 ) + ( −4 ) ( −2 ) + ( 3 )
2
2
2
(
(a) ˆi + ˆj + kˆ
(c) ˆi + ˆj − kˆ
−1
3
148.
(b) ˆi − ˆj + kˆ
(d) ˆi − ˆj − kˆ
B = ˆi + kˆ
Then
 1  1 ˆ ˆ
( i − j)
= −

 2 2
1 ˆ ˆ
= − ( i − j)
2
Objective Physics Volume-I
between vectors A and B is
(a) 180o
(b) 90o
(c) 45o
(d) 0o
A=
2
2
2
2
2
as we know that vector product is defined as,
A ⋅ B = A ⋅ B cos θ
Let θ be the angle between vector A and B then,
( ˆi + ˆj) ⋅ ( ˆi + ˆj + ckˆ ) = 2 × ( 2 + c ) cos30°
A.B
2
A B
( 4iˆ + 4ˆj − 4kˆ ).( 3iˆ + ˆj + 4kˆ )
cos θ =
42 + 42 + ( −4 )
2
1+1 = 2 ×
( 2 + c ) × 23
2
4
= 2 + c2
6
32 + 12 + 42
12 + 4 − 16
48 26
2 + c2 =
o
16
4
⇒ c2 =
6
6
2
c=±
3
149. Consider the following statements about three
151.
The
angle between A and the resultant of
vectors a,b and c that have non-zero
2A + 3B and 4A − 3B is
magnitudes. It follows b = c
(ii) if a × b = a × c
1 0 1
to c
(a) (i) and (ii) both
(c) (i) only
YCT
1
2
B = 1 + 1 + c = 2 + c2
(i) if a ⋅ b = a ⋅ c
ˆj kˆ
(1) + (1) =
2
B = 3iˆ + ˆj + 4kˆ
cosθ = 0 = cos90
θ = 90o
(d) ±
AMU-2011
AMU-2001
cos θ =
(b) ± 1
Ans. (c) : Given,
A = ˆi + ˆj , B = ˆi + ˆj + ckˆ , θ = 30°
Ans. (b) : Given,
A = 4iˆ + 4ˆj − 4kˆ
A×B = 1 1 0
A × B = ˆi (1 − 0 ) − ˆj (1 − 0 ) + kˆ ( 0 − 1)
A × B = ˆi − ˆj − kˆ
373
2
(c) ±
3
A = 4iˆ + 4jˆ − 4kˆ and B = 3iˆ + ˆj + 4kˆ then angle
cos θ =
Ans. (d) : Given,
A = ˆi + ˆj
ˆi
Find the unknown c.
(a) 0
)
r = −2iˆ + 3jˆ + 5kˆ
AMU-2006
1 ˆ ˆ  1 ˆ ˆ

( i − j)  ( i − j)
= ( 2iˆ + 3jˆ ) ⋅
2

 2
) (
∴ r = −2iˆ + 6ˆj + 4kˆ − 3jˆ − kˆ
 13 = 3.606 


cos θ =
150. The angle between the vectors
A = ˆi + ˆj and B = ˆi + ˆj + ckˆ is 30º .
A = 0iˆ + 3jˆ − kˆ , B = −2iˆ + 6ˆj + 4kˆ
2
( )
A⋅B
ˆ
Component of A along B =
= A⋅B
B
ˆ ]B
ˆ
To get vector form = [ AB
b and c are equal in magnitude so, b and c are not
perpendicular . they are parallel to each other.
So, only (i) is follows.
( b1 − a1 ) ˆi + ( b 2 − a 2 ) ˆj + ( b3 − a 3 ) kˆ
145. Vector A has a magnitude of 5 units, lies in the
xy-plane and points in a direction 120º from the
direction of increasing x. Vector B has a
1
magnitude of 9 units and points along the z(b) − ˆi + ˆj
2
axis. The magnitude of cross product A × B is
(a) 30
(b) 35
1 ˆ ˆ
(d)
i+j
(c) 40
(d) 45
2
AMU-2012
AMU-2013
Ans. (d) : We know that,
A×B = |A| |B| sinθ
Vector B makes angle of 90° with A
A×B = 5×9 sin 90°
A×B = 45
146. If A = ˆi + ˆj , and B = ˆi + kˆ , then A × B is
ˆi − ˆj
B̂ =
2
3
Given,
( )
1 ˆ ˆ
i–j
2
B̂ =
a.b
cos θ =
ˆ = 5. 1 ˆi − ˆj
B .A
2
ˆ = 5 ˆi − ˆj
B .A
2
2
Displacement vector (r) = B − A
−6
5 13
−6
cos θ =
⇒
18
 1
θ = cos −1  − 
 3
∵ Vector having magnitude of B and parallel to A .
1
B = b1ˆi + b 2 ˆj + b3 kˆ
a⋅b
cos θ =
( 3) + ( 4 ) = 9 + 16 = 5
147. The (x, y, z) coordinates of two points A and B Ans. (c) : (i) a.b = a.c
are given respectively as (0, 3, –1) and (–2, 6, 4).
a b cos θ = a c cos θ
The displacement vector from A to B may be
given by
Then
b=c
(a) −2iˆ + 6ˆj + 4kˆ
(b) −2iˆ + 3jˆ + 3kˆ
(ii)
a × b = a ×c
(c) −2iˆ + 3jˆ + 5kˆ
(d) 2iˆ − 3jˆ − 3kˆ
a b sin θ = a c sin θ
AMU-2005
b=c
Ans. (c) : A = a ˆi + a ˆj + a kˆ
Objective Physics Volume-I
(a) 90º
A
(b) tan −1  
B
B
(c) tan −1  
A
 A−B
(d) tan −1 

A+B
b must be perpendicular
(b) neither (i) nor (ii)
(d) (ii) only
AMU-2001
374
(e) 0º
Kerala CEE 2020
YCT
154. When a particle moved from point A(2, 2, 3) to
point B(6, 6, 9), its displacement vector is ____.
(a) 4iɵ + 4ɵj + 6kɵ
(b) 8iɵ + 8jɵ + 12kɵ
Ans. (e) : Consider A and B,
(c) 4iɵ + 8jɵ + 6kɵ
(d) 8iɵ + 4ɵj + 6kɵ
AP EAMCET-23.08.2021, Shift-I
Ans. (a) : Given,
The particle is moving from point
A (2, 2, 3) to the point B (6, 6, 9)
Displacement vector ( r ) = ∆xiˆ + ∆yjˆ + ∆zkˆ
Resultant of 2A + 3 B and 4 A − 3 B,
r = ( 6 – 2 ) ˆi + ( 6 – 2 ) ˆj + ( 9 – 3) kˆ
( 2A + 3 B) + ( 4 A − 3 B) = 2A + 3 B + 4 A − 3 B = 6 A
r = 4iˆ + 4ˆj + 6kˆ
Angle
between
A
and
resultant
vector 155. The position of a particle x (in meters) at a time
t seconds is given by the relation
2A + 3 B and 4 A − 3 B .
r = 3tiˆ - t 2 ˆj + 4kˆ . Calculate the magnitude of
The angle between A and 6A is 0°, because they are
velocity of the particle after 5 seconds.
parallel vectors.
(a) 3.55
(b) 5.03
ˆ 6kˆ to
152. A particle moves from position 3iˆ + 2j+
(c) 8.75
(d) 10.44
AMU-2010
14iˆ +13jˆ + 9kˆ due to a uniform force of
Ans. (d) : Given,
4iˆ + ˆj + 3kˆ newton. Find the work done if the
2
r = 3tiˆ − t ˆj + 4kˆ
displacement is in meters.
dr
(a) 16 J
(b) 64 J
v = = 3iˆ − 2tjˆ + 0
(c) 32 J
(d) 48 J
dt
J&K CET- 2007
2
| v |= 9 + (−2t )
ˆ
ˆ
ˆ
Ans. (b) : Given, F = 4i + j + 3k
2
= 9 + 4t
ˆ − (3iˆ + 2ˆj + 6k)
ˆ
dS = (14iˆ + 13jˆ + 9k)
Magnitude of velocity of particle t = 5 secˆ
ˆ
ˆ
= 11i + 11j + 3k
2
v = 9 + 4 ( 5)
Work done W = F ⋅ dS
| v |= 9 + 100
∴ W = 4iˆ + ˆj + 3kˆ ⋅ 11iˆ + 11jˆ + 3kˆ
= 10.44
156. The velocity of a moving particle at any instant
W = 44 + 11 + 9
is ˆi + ˆj. The magnitude and direction of the
∴ W = 64J
velocity of the particle are
153. What is the linear velocity, if angular velocity
(a) 2 units and 45° with the x-axis
vector ω = 3iˆ − 4jˆ + kˆ and position vector
(b) 2 units and 30° with the z-axis
ˆ
ˆ
ˆ
r = 5i - 6j + 6k ?
(c) 2 units and 45° with the x-axis
(a) 6iˆ - 2jˆ + 3kˆ
(b) -18iˆ -13jˆ + 2kˆ
(d) 2 units and 60° with the y-axis
(e) 2 units and 60° with the x-axis
(c) 18iˆ +13jˆ + 2kˆ
(d) 6iˆ - 2jˆ + 8kˆ
Kerala CEE 2021
[AIPMT 1999]
Ans. (c) : Given,
Ans. (b) : Given,
Velocity of moving particle isω= 3iɵ − 4ɵj + kɵ
v = ˆi + ˆj
ɵ
ɵ
ɵ
r = 5i − 6 j + 6k
Then, magnitude | v |= 12 + 12
ɵj
i
kɵ
v= 2
v = ( ω× r ) = 3 − 4
1
A
1
1
Direction,
cos
θ= x =
=
5 −6
6
|A|
2
12 + 12
θ = 45o
= ɵi ( −24 + 6 ) − ɵj (18 − 5) + kɵ (−18 + 20)
So, the magnitude of velocity is 2 and direction 45°
v = − 18iɵ − 13jɵ + 2kɵ
with the x-axis.
(
) (
(
)
(
)
)
(
)(
Objective Physics Volume-I
)
375
YCT
157. A certain vector in the xy-plane has an x1
SFinal – Sinitial = ut + at 2
component of 4 m and a y-component of 10m.
2
It is then rotated in the xy-plane so that its xcomponent is doubled. Then its new yˆ + 4.0ˆj = 2 5.0iˆ + 4.0ˆj + 1 4.0iˆ + 4.0ˆj (2)2
S
−
2.0i
Final
component is (approximately)
2
(a) 20m
(b) 7.2m
SFinal = 12iˆ + 12ˆj + 8iˆ + 8jˆ
(c) 5.0m
(d) 4.5m
AP EAMCET -2011
SFinal = 20iˆ + 20ˆj
Ans. (b) : Given that,
| SFinal |= 202 + 202
Initially : X – component = 4 m
Y – component = 10 m
|SFinal| = 20 2 m
Finally :
X – component = 2 × 4 = 8 m
160. A particle is moving with a velocity
Y – component = y
The magnitude of vector do not change by its rotation
v = k ( yiˆ + xjˆ ) , where k is a constant. The
So,
general equation for its path is
2
2
2
2
4 + 10 = 8 + y
(a) y = x2 + constant
(b) y2 =x + constant
y = 52
(c) xy = constant
(d) y2 = x2 + constant
y = 7.2 m
JEE Main-09.01.2019, Shift-I
158. A particle has an initial velocity 3i +4j and an
BITSAT- 2016
acceleration of 0.4i +0.3 j. Its speed after 10 s is
AIEE-2010
(a) 10 units
(b) 7 2 units
Ans. (d) : Given,
(c) 7 units
(d) 8.5 units
AIPMT-2010 v = kyiˆ + kxjˆ
…(i)
AIEEE-2009
ˆi + v ˆj
General
equation
v
=
v
…(ii)
x
y
Ans. (b) : Given data,
Comparing eqn (i) & (ii) we get
u = 3iˆ + 4ˆj
vx = ky, vy = kx
a = 0.4iˆ + 0.3jˆ
dx
t = 10 sec
= ky
…(iii)
dt
we know that,
v = u + at
dy
= kx
…(iv)
dt
v = 3iˆ + 4ˆj + 0.4iˆ + 0.3jˆ × 10
n
Now, eq (iv) / (iii)
v = 7iˆ + 7ˆj
dy
2
2
dt = kx
So, magnitude of v = 7 + 7 = 7 2
dx
ky
dt
159. A particle moves from the point ( 2.0iˆ + 4.0jˆ ) m
(
(
)
(
)
(
)
)
at t=0 with an initial velocity ( 5.0iˆ + 4.0jˆ ) ms −1 .
It is acted upon by a constant force which
produces
a
constant
acceleration
( 4.0iˆ + 4.0jˆ ) ms −2 . What is the distance of the
particle from the origin at time 2 s ?
(a) 5m
(b) 20 2m
dy x
=
dx y
∫ y dy =∫ x dx
y2 x 2
= + constant
2
2
(d) 15m
Or y 2 = x 2 + constant
JEE Main-11.01.2019, Shift-II
161. In three dimensional system, the position
Ans. (b) : Given,
coordinates of a particle (in motion) are given
Sinitial = 2.0iˆ + 4.0ˆj
below
u = 5.0iˆ + 4.0ˆj
x = a cosωt, y = a sinωt, z = aωt
The velocity of particle will be
a = 4.0iˆ + 4.0ˆj
Now,
(a) 2aω
(b) 2aω
1
∆ s = ut + at 2
(c) aω
(d) 3aω
2
JEE Main-09.01.2019, Shift-II
(c) 10 2m
Objective Physics Volume-I
376
YCT
Ans. (a) : r ( t ) = 15t 2 ˆi + ( 4 − 20t 2 ) ˆj
Ans. (a) : Given,
x = a cos ωt
dx
vx =
= −a sin ωt.ω
dt
Similarly
dy
vy =
= a cos ωt.ω
dt
dz
vz =
= aω
dt
dy
= bx
dt
dy
dy
bx
= dt =
dx dx
a
dt
dy bx
=
dx
a
b
dy
=
xdx
a
20iˆ + y 0ˆj = 5t 2ɵi + (5t + 2t 2 )ˆj
Integrating both side,
compare x component of the position we get,
y
x b
20 = 5t2
∫0 dy = ∫0 a x dx
t = 2 sec
b x2
compare y component of the position,
y=
y0 = 5t + 2t2
a 2
y0 = 5 × 2 + 2 × 22
b
y = x2
y0 = 18 m
2a
163. The position vector of particle changes with 165. A particle moves over a xy plane with a
time
according
to
the
relation
constant
r(t) = 15t 2 ˆi + ( 4 − 20t 2 ) ˆj.
What
is
the
acceleration a = 4.0m/s 2 ˆi + 4.0m/s 2ˆj . At time
magnitude of the acceleration (in ms−2) at t =1?
t = 0, the velocity is 4.0m/s 2 ˆi. The speed of
(a) 50
(b) 100
(c) 25
(d) 40
the particle when it is displaced by 6.0 m
JEE Main-09.04.2019, Shift-II
parallel to the x-axis is
80 × 80
60 m/s
a=
2 × ( 0.40 )
(d) 20 m/s
a = 8000 m/s2
TS EAMCET 04.08.2021, Shift-I
According to Newton’s second law
Ans. (a) : Use equation of motion
F = ma
Apply second equation of Motion–
= 0.05 × 8000
Given that,
= 400 N
t = 0, u = 4 î
167. A particle is moving in the x-y plane and its
ux=4
coordinates at any time t are given by
a = 4iˆ + 4ˆj m / s 2
x = 5 cos ωt
y = 5 sin ωt
a x = 4 m / s 2 ,a y = 4m / s 2
π
Where (ω) =
rad/s. The direction of force it
1
4
sx = ux t + ax t2
experiences at t = 3s is.
2
∧ ∧
∧ ∧
1
(a) i + j
(b) i − j
6 = 4 t + ×4×t2
2
∧
∧
2t2 + 4t –6 = 0
(c) i
(d) j
2
t + 2t – 3 = 0
TS EAMCET (Medical) 09.08.2021, Shift-I
t2 + 3t – t – 3 = 0
Ans. (b) : Given,
t(t + 3) –1(t + 3) = 0
π
ω=
t+3=0
4
t–1=0
x = 5 cos ωt
t = 1 sec, t = –3 (not possible)
dx
So,
= v x = −5 ω sin ωt
dt
t=1
dv x
Velocity of particle–
= a x = −5ω2 cos ωt
dt
v = u + at
ax at t = 3
v = (4iˆ + 0ˆj) + (4iˆ + 4ˆj) ×1
2
π
ax = – 5   cos (135º)
= 4iˆ + 4iˆ + 4ˆj
4
2
v = 8iˆ + 4ˆj
π  1 
ax = – 5    −

Speed of particle2
4 
2
2
2
v = (8) + ( 4 )
5 π
ˆ
ax = +
  along x (i)
24
= 64 + 16
y = 5 sin ωt
v = 80
dy
= v y = 5cos ωt
= 4× 4×5
dt
dv y
v = 4 5 m/s
= a y = −5ω2 sin ωt
dt
166. A bullet of mass 0.05 kg moving with a speed of
ay at t = 3
80 ms–1 enters a wooden block and is stopped
2
π
after a distance of 0.40 m. The average resistive
ay = –5   sin(135º )
force exerted by the block on the bullet is
4
2
(a) 300 N
(b) 20 N
π  1 
ay = – 5   
(c) 400 N
(d) 40 N

4


 2
(e) 200 N
2
Kerala CEE - 2008
5 π
ˆ
ay = –
  along y (− j)
Ans. (c) : Given that, m = 0.05 kg, u = 80 m/s
24
v = 0, s = 0.4m
Direction of net accl..
We know that,
= + î – ĵ
2
2
v = u – 2as
= î – ĵ
0 = (80)2 – 2a (0.40)
Objective Physics Volume-I
Objective Physics Volume-I
(
2
2
2
= a ω +a ω
2
2
= 2 aω
162. Starting from the origin at time t =0, with
initial velocity 5jˆ ms −1 , a particle moves in the
xy-plane with a constant acceleration of
(10iˆ + 4jˆ ) ms−2 . At time t, its coordinates are
(20m, y0 m). The values of t and y0 respectively,
are
(a) 2 s and 18 m
(b) 5 s and 25 m
(c) 2 s and 24 m
(d) 4 s and 52 m
JEE Main-04.09.2020, Shift-I
Ans. (a) : Given data.
u = 5jˆ m / s
(
)
a = 10iˆ + 4ˆj m / s2
Position coordinate (s) at time 't' = (20, y0) ⇒ 20iˆ + y0ˆj
we know that,
second equation of motion is given by –
1
s = ut + at 2
2
1
20iˆ + y 0ˆj = 5tjˆ + 10iˆ + 4ˆj t 2
2
20iˆ + y 0ˆj = 5tjˆ + 5t 2 ˆi + 2t 2 ˆj
(
(
a = 302 + ( −40 ) = 50 m / s 2
2
2
)
dv(t)
= a(t) = 30iɵ − 40ɵj
dt
= (−aω sin ωt ) + (aω cos ωt ) + (aω)
164. Consider a particle moving in the xy plane with
ˆ where î and ĵ are the unit
velocity v = aiˆ + bxj,
vectors along x and y axes and 'a' and 'b' are
constants. If the initial position of the particle
is x = y = 0. The equation to describe the
particle's trajectory in the plane is
b 2
(a) y =
x
(b) y = x2
2a
b
2b 2
(c) y = x
(d) y =
x
a
a
TS EAMCET 08.05.2019, Shift-II
Ans. (a) : Given,
velocity v = aiˆ + bxjˆ
Initial position , x = 0 , y = 0
Here, vx = a
dx
=a
dt
v y = bx
)
(
377
(b)
(c) 3 10 m/s
v ( t ) = 30tiˆ − 40tjˆ
∴ v = v 2x + v 2y + v 2z
2
(a) 4 5 m/s
dr(t)
= v(t)
dt
v ( t ) = 30tiˆ + 0 − 40tjˆ
(
) (
)
)
YCT
)
378
YCT