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Vectors
2011
If vectors ɵi − 3 ɵj + 5kɵ and ɵi − 3 ɵj − akɵ are equal vectors, then the value of a is
1.
b) 2
c) -3
d) 4
co
m
a) 5
If a + b = c and a + b = c, then the angle included between a and b is
a) 900
3.
b) 1800
c) 1200
at
io
2.
n.
2010
d) Zero
Three equal masses of 1kg each are placed at the vertices of an equilateral triangle
distance of
2m from each of the vertices of the triangle. The force, in Newton,
acting on the mass of 2kg is
b) 1
c) 1
d) Zero
Find the torque of a force F = 3ɵi + 2 ɵj + kɵ acting at the point r = 8ɵi + 2 ɵj + 3kɵ
b) 4ɵi + 4 ɵj + 6kɵ
ak
a) 14ɵi − 38 ɵj + 16kɵ
c) −14ɵi + 38 ɵj − 16kɵ
d) −4ɵi − 17 ɵj + 22kɵ
A variable force, given by the two dimensional vector F = (3x 2 ɵi + 4 ɵj ) , acts on a
w
.s
5.
sh
i
a) 2
4.
ed
uc
PQR and a mass of 2kg is placed at the centroid O of the triangle which is at a
particle. The force is in Newton and x is in metre. What is the change in the kinetic
energy of the particle as it moves from the point with coordinates (2, 3) to (3, 0)?
w
(The coordinates are in metres)
w
a) – 7 J
6.
b) Zero
c) +7J
d) 19J
The centre of mass of a system of three particles of masses 1 g, 2g and 3g is taken as
the origin of a coordinates system. The position vector of a fourth particle of mass 4g
such that the centre of mass of the four particle system lies at the point (1, 2, and 3) is
α (ɵi + 2 ɵj + 3kɵ ) , where it is constant. The value of α is
a) 10/3
b) 5/2
c) 1/2
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d) 2/5
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2009
If a1 and a2 are two non-collinear unit vectors and if a1 + a2 = 3 , then the value of
7.
(a1 − a ).(2a1 + a2 ) is
3
2
c)
d) 1
There are N coplanar vectors each of magnitude V. Each vector is inclined to the
V
N
b) V
2008
c) Zero
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a)
2π
. What is the magnitude of their resultant?
N
n.
preceding vector at angle
9.
1
2
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8.
b)
co
m
a) 2
d)
N
V
The value of P so that the vectors 2ɵi − ɵj + kɵ , ɵi + 2 ɵj − 3kɵ and 3ɵi + P ɵj + 5kɵ are coplanar
should be
b) -4
maximum value of
sh
i
a) 16
a) 4N
b) Zero
c) 4
d) -8
w
.s
ak
10. Two forces of 12N and 8N act upon a body. The resultant force on the body has a
c) 20N
d) 8N
11. The condition under which vectors (a + b) and (a – b) should be at right angles to
each other is
w
a) a ≠ b
b) a.b = 0
c) a = b
d) a. b = 1
w
12. A car travels 6km towards North at an angle of 450 to the East and then travels
distance of 4kg towards North at an angle 1350 to East. How far is the point from the
starting point? What angle does the straight line joining its initial and final positions
makes with the East?
a) 50km and tan −1
b) 10 km and tan −1 ( 5)
c) 52 km and tan −1 (5)
d) 52 km and tan −1 ( 5)
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13. A train of 150 m length is going towards North direction at a speed of 10ms −1 . A
parrot flies at a speed of 5ms −1 towards South direction parallel to the railway track.
The time taken by the parrot to cross the train is equal to
a) 12s
b) 8s
c) 15s
d) 10s
14. Rain is falling vertically downwards with a velocity of 4kmh −1 . A man walks in the
a) 1kmh −1
b) 3kmh −1
co
m
rain with a velocity of 3kmh −1 . The raindrops will fall on the man with a velocity is
c) 4kmh −1
d) 5kmh −1
n.
15. A proton in a cyclotron changes its velocity from 30kms −1 North to 40kms −1 East in
20s. What is the magnitude of average acceleration during this time?
b) 12.5kms −2
d) 32.5kms −2
ed
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2006
c) 22.5kms −2
at
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a) 2.5kms −2
16. A police jeep is chasing with velocity of 45kmh −1 , a thief in another jeep moving with
velocity 153kmh −1 . Police fires a bullet with muzzle velocity of 180ms −1 . The velocity
b) 27ms −1
c) 450ms −1
d) 250ms −1
ak
a) 150ms −1
sh
i
with which will strike the car of the thief, is
2005
w
.s
17. Minimum of unequal vectors which can gives zero resultant are
a) Two
b) Three
c) Four
d) More than four
w
18. If a vector 2ɵi + 3 ɵj + 8kɵ is perpendicular to the vector 4 ɵj − 4ɵi + α kɵ , then the value of α is
w
a) -1
b)
1
2
c) −
1
2
d) 1
19. The vectors from origin to the points A and B are A = 3ɵi − 6 ɵj + 2kɵ and B = 2ɵi + ɵj − 2kɵ
respectively. The area of the triangle OAB is
a)
5
17
2
b)
2
17
5
c)
3
17
5
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d)
5
17
3
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Key
2) d
3) d
4) d
5) c
6) b
7) c
8) c
9) b
11) c
12) c
13) d
14) d
15) a
16) a
17) b
18) c
19) a
Hints
1.
Given vectors
B = ɵi − 3 ɵj − akɵ
at
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n.
A = ɵi − 3 ɵj + 5kɵ
According to problem both vectors are equal then A = B, so that the value of a =-5
We have a + b = c and c = a + b
⇒ c = a 2 + b2 + 2ab cos θ
⇒ a + b = a 2 + b 2 + 2ab cos θ
ed
uc
2.
Given, OP = OQ = OR =
2m
ak
3.
sh
i
⇒ a 2 + b 2 + 2ab = a 2 + b 2 + 2 ab cos θ
⇒ cos θ = 1 ⇒ θ = 0 0
w
.s
The gravitational force on the mass 2kg due to the 1kg mass at P is
2 ×1
= G along OP
( 2) 2
w
w
FOP = G
Similarly, FOQ = G1
10) c
co
m
1) c
2 ×1
= G1 along OQ and
( 2) 2
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FOR = G1
2 ×1
= G1 along OR
( 2) 2
FOQ cos 300 and FOR cos 300 are equal and acting in opposite directions, thus they cancel out.
Hence, the resultant force on the 2kg mass at O, is
co
m
F = FOP = ( FOQ sin 300 + FOR sin 300 )
G G 
= G1 −  1 + 1 
2 
 2
Torque of the force, τ = r × F
ɵi
ɵj kɵ
So, τ = 8
2 3
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4.
n.
=0
= ɵi(2 − 6) − ɵj (8 + 9) + kɵ (16 + 6)
= −4ɵi − 17 ɵj + 22kɵ
Given two dimensional force
sh
i
5.
w
.s
dr = dxiɵ + dy ɵj
ak
F = 3x 2 ɵi + 4 ɵj
r = xiɵ + y ɵj
ed
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−3 2 1
Kinetic energy = work done
w
W = ∫ F .dr
(3,0)
(3 x 2 ɵi + 4 ɵj ).(dxiɵ + dy ɵj )
w
=∫
(2,3)
3
= ∫ (3x 2 dx + 4dy )
2
= [ x 3 ]32 + 4[ y ]30 = (27 − 8) + 4( −3)
= 19 -12 = 7J
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6.
The coordinates (x, y, z) of masses 1g, 2g, 3g and 4g are ( x1 = 0, y1 = 0, z1 = 0) ,
( x2 = 0, y2 = 0, z2 = 0)
=
4α
1+ 2 + 3 + 4
=
4α
10
⇒α =
5
2
⇒ zCM =
m1 z1 + m2 z2 + m3 z3 + m4 z4
m1 + m2 + m3 + m4
12α
=3⇒α =5/2
10
ak
=
m1 y1 + m2 y2 + m3 y3 + m4 y4 8α
=
=2
m1 + m2 + m3 + m4
10
ed
uc
⇒ yCM =
at
io
5
2
sh
i
⇒α =
n.
4α
=1
10
Hence,
Since, a1 and a2 are non=collinear
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.s
7.
m1 x1 + m2 x2 + m3 x3 + m4 x4
m1 + m2 + m3 + m4
co
m
⇒ xCM =
∴ a1 = a2 = 1
w
And a1 + a2 = 3
w
⇒ a12 + a22 + 2a1a2 cos θ = ( 3)2
⇒ 1 + 1 + 2 cos θ = 3 ⇒ cos θ =
1
2
Now (a1 + a2 ).(2a1 + a2 )
= 2a12 − a22 − a1a2 cos θ = 2 − 1 −
1 1
=
2 2
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8.
Since each of N-coplanar vectors is inclined at
2π
to the preceding hence, they will form
N
a closed polygon. Therefore, their resultant must be zero
For coplanarity
2 −1
1
1
2
−3 = 0
3
P
5
co
m
9.
Or 2(10 +3P) +1(5 + 9) + 1 (P – 6) =
n.
Or 20 + 6P + 5 + 9 + P – 6 = 0
at
io
Or 7P + 34 – 6 = 0
Or 7P + 28 = 0
Or 7P = -28
28
= −4
7
ed
uc
⇒P=−
10. When the two forces of 12N 8N act upon a body, the resultant force on the body has
= 12N + 8N = 20N
sh
i
maximum value when resultant force
⇒ a2 − b2 = 0
w
.s
⇒a =b
ak
11. The dot product of two vectors should be equal to zero is (a + b). (a – b) =0
12. Net movement along X-direction
w
S x = (6 − 4) cos 450 ɵi
1
= 2km
2
w
= 2×
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Net movement along Y-direction
S y = (6 + 4) sin 450 ɵj
= 10 ×
1
= 5 2km
2
co
m
Net movement from starting point
S = S x2 + S y2 = ( 2) 2 + (5 2) 2
Angle which resultant makes with the east direction
at
io
=
y − component
x − component
5 2
2
θ = tan −1 (5)
ed
uc
tan θ =
n.
= 52km
13. Relative velocity of the parrot w.r.t the train
sh
i
= [10 − (−5)]ms −1 = 15ms −1
150
= 10 s
15
ak
Time taken by the parrot to cross the train
w
.s
14. Relative velocity of man w.r.t rain
vrm = vr − vm
w
= 4 ɵj − 3ɵi
w
= −3ɵi + 4 ɵj
Or = 5kmh −1
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vrm = (−3) 2 + (4) 2
= 9 + 16
= 25 = 5
16. Change in velocity = (40) 2 + (30) 2
50
20
= 2.5kms −2
ed
uc
Average acceleration =
at
io
n.
co
m
= 50kms −1
= 108kmh −1
5
= 30ms −1
18
ak
= 108 ×
sh
i
17. The relative velocity of thief’s jeep with respect to police jeep = 153 – 45
Therefore, striking speed = relative speed of bullet with respect to thief’s car
w
.s
= 180 – 30 = 150ms −1
18. By triangle law of vectors minimum three vectors are required to give zero resultant
w
19. Let a = 2ɵi + 3 ɵj + 8kɵ
w
b = 4ɵi − 4 ɵj + α kɵ
= −4ɵi + 4 ɵj + α kɵ
Given a ⊥ b
⇒ a.b = 0
⇒ (2ɵi + 3 ɵj + 8kɵ ).(−4ɵi + 4 ɵj + α kɵ ) = 0
⇒ −8 + 12 + 8α = 0
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⇒ 8α = −4
4
1
=−
8
2
w
w
w
.s
ak
sh
i
ed
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at
io
n.
co
m
∴α = −
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