CHAPTER - 5

VECTORS
ur ur
3. Negative of vector : If two vectors A and B are such that they have equal magnitude
but opposite directions, each vector is negative of the other.
ur ur ur ur
Thus A  B or B   A
uur

A


ur uur
B  A

If a vector is given, its negative can be obtained by just reversing the direction.
4. Null vector: “A vector of zero magnitude is called zero vector or null vector”.

It is represented by 0 . The initial point and terminus of the null vector coincide. Its
direction is indeterminate.
5. Co–planar and non-coplanar vectors: Vectors, acting in the same plane are called co–
planar vectors.

A


B

C

r r r
In the diagram A,B and C are coplanar vectors. Vectors, acting in different plane are
called non-coplanar vectors.
6. Angle between two vectors:
To find angle between two vectors, the two vectors from a point are drawn such that
their arrow heads should be away from that point. The angle obtained in this way, is
the angle between the vectors.

A 
 B
B
120°
60°

A
r r
the angle between A and B is not 120°.
*Whenever angle between two vectors is to be taken we must make sure that either
their heads coincide or their tails coincide.
If heads coincide or tails coincide then internal angle is the angle between two
vectors as in figure (1). If heads coincide with tail then external angle is the angle
between the two vectors as in figure (2).

p
q
Figure 1 Figure 2
7. Unit vector : “A vector of unit magnitude is called unit vector”. The unit vector in the
direction of given vector is obtained by dividing the given vector with its magnitude.
It is conventional to denote unit vector with a “cap” instead of “bar” over the symbol.
ur ur
Thus if A is a given vector, the unit vector in the direction of A is written as

A
 

A = A (where A is read as A cap or A hat)

Note: In the right handed cartesian coordinate system, iˆ, ˆj and k̂ are choosen as unit
vectors along, the X-axis, Y-axis and Z-axis respectively.
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Displacement
Displacement is a shortest distance between two points. It is a vector quantity.
Displacement vector
The position of the point Q with reference to the origin is represented by the position

vector r2 . Let the coordinates of the point Q are (x2, y2)

 
Similarly OP represent by a position vector r , let the coordinates of the point P are
1
(x1, y1)
As the displacement vector is the difference of two position vectors
       
r1 = x 1 i + y1 j and r2 = x2 i + y2 j , where i , j are unit vectors along X, Y axes
respectively.
    
Thus, the displacement vector  r  r2  r1 = (x2 – x1) i + (y2 – y1) i

WORK SHEET – 1

Single Answer Type

1. Choose the false statement from the following.
1) If two vectors have same magnitude and direction they are said to be equal vectors.
2) If two vectors have equal magnitude but opposite directions, then each vector is
negative vector of the others.
3) A vector of zero magnitude is called zero vector or null vector.
4) None of these.
2. Find the odd one out from the following
1) distance 2) force 3) speed 4) time
3. Assertion : A vector is represented by a directed line segment
Reason : Length of that line segment is proportional to the magnitude of the
vector and direction of that denotes the direction of vector.
1) Assertion is true, and reason is true and the reason is correct explanation of assertion.
2) Assertion is true and reason is true but the reason is not correct explanation of the
assertion.
3) Assertion is true but reason is false. 4) Assertion is false but reason is true.
4. The addition of one vector to the other result a null vector. Then
1) First vector is the negative vector of the second only
2) Second vector is the negative vector of the first only
3) One is negative vector of the other.
4) Neither is negative vector of the other.

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1  1  
11. If + + c k is a unit vector, then the value of ‘c’ is
2 i 2 j

1) 1/ 2 2) 1 3) 1/2 4) –1/2
  
12. Magnitude of vector 3 i – 12 j – 4 k

1) 13 2) 13 3) 19 4) None
     
 
13. If A = 4 i – 3 j and B = 6 i – 5 j . Then unit vector parallel to B – A is
       
2 i 2 j i j i j i j
1) 2) 3) 4)
2 2 2 2

       
14. If A = 2 i  j and B = j  k then magnitude of 2 A + 3B is

1) 10 2) 5 3) 5 2 4) 20
     
15. By the application of force a particle moves from (2 i  3 j 5 k) m to (i  2 j k) m. Then
its displacement is
        
1)  i  5 j 4 k 2) i  5 j 4 k 3) i  5 j 4 k 4) None of these

  
16. Length of 2 i + 3 j + 4 k in the X Y plane is
1) 13 2) 20 3) 5 4) Zero
  
17. If 0.2 i + 0.3 j + z k is unit vector then z2 =
1) 0.27 2) 0.4 3) 0.87 4) None of these
  

18. If 2 i  y j 3 k = 5, then y =

1) 6 2) 12 3) 15 4) None of these

Subjective Answer Type

19. If the position of a particle changes from (1,2,3) m to (5,4,2)m, then find the displacement
vectors
20. A vector PQ has the initial point P(1,2,–1) and terminal point Q(3,2,2). Write the
displacement vector of PQ and its magnitude.

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
B

A

  
C  AB



A B

5.6 TRIANGLE LAW OF VECTORS : If two vectors are represented in magnitude and
direction by the two sides of a triangle taken in order, the third side of the triangle taken
in reverse order represents their resultant in magnitude and direction.


   
 

R  PQ Q



P

5.7 PARALLELOGRAM LAW OF VECTORS :

Two vector quantities (say, velocity, acceleration, force, etc) can be added using
parallelogram law. This law is useful to find both magnitude and direction of resultant.
Statement: If two vectors are represented in magnitude and direction by the adjacent
sides of a parallelogram drawn from a point, the diagonal passing through that point
represents their resultant both in magnitude and direction.

D D C
ur ur ur
Q Q R

 
ur ur
A P B A P B E

figure (d)
ur ur uuur uuur
Explanation: P and Q are two vectors represented by AB and AD . Both vectors act
at the common point A and mutually inclined at angle ‘  ’ as shown in fig (d). If the
parallelogram ABCD is completed taking AB and AD as adjacent sides, then the diagonal
ur
uuur
 
AC represents their resultant R both in magnitude and direction.
ur
Magnitude of the resultant: The line of action of P is extended. The perpendicular
drawn from ‘C’ meets the extension of AB at E.
uuur uuur ur
From the figure, it is obvious that BC  AD  Q and CBE  
ur ur
 Length of AB = magnitude of P = P Length of BC = magnitude of Q = Q
ur
Length of AC = magnitude of R = R

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WORK SHEET – 2

Single Answer Type

1. Arrange the vector addition so that their magnitude in increasing order are
 
a) Two vectors A and B are parallel

 
b) Two vectors A and B are antiparallel

 
c) Two vectors A and B making an angle 600

 
d) Two vectors A and B making an angle 1200

1) b, d, c, a 2) a, b, d, c 3) c, d, b, a 4) c, d, a, b
2. Choose the false one from the following
1) If two vectors are in the same directions the direction of resultant is same as the
individuals vectors.
2) If the two vectors are mutually opposite, the direction of the resultant is same as
that of larger vector
3) If the two vectors are mutually opposite the direction of the resultant is same as
that of smaller vector.
4) None of these
3. The vector sum of two forces 8 N and 6 N can be
1) 18 N 2) 20 N 3) 1 N *4) 8 N
ur ur ur ur
4. If P  4i$ 2j$ 6k$and Q  $
i  3j$ 3k$the P  Q 

1) 5i$ 5j$ 3k$ 2) 3i$ 5j$ 3k$ 3) 5i$ 5j$ 3k$ i  3j$ 3k$
4) $

5. If two forces 12 N and 8 N at on an object in the same direction then the magnitude of
the resultant is
1) 4 N 2) 16 N 3) 20 N 4) 50 N
 
6. If two vectors P and Q making an angle ‘  ’ between them, then the magnitude of the
resultant R of vectors is ___________

1) R= P 2  Q2  2PQ Sin 2) R = P 2  Q2  2PQ Cos

3) R= P 2  Q2  6PQ Cos 4) R = P 2  Q2  6PQ Sin

 
7. If P = Q and if  = 120 0 between them, then select the true answer from the

following
1) P = Q R 2) P = Q = R 3) Both (1) & (2) 4) None of these

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V  V1  V 2  V3  V 4  V5
Note: If a number of vectors simultaneously acting at a point can be represented in
magnitude and direction by the sides of a polygon taken in order, the point is in
equilibrium ie the vectors have zero resultant.

WORK SHEET - 3

Single Answer Type

1. The ratio of maximum and minimum resultant of two forces is 7 : 1. The ratio of the
forces are respectivley
1) 1 : 7 2) 4 : 3 3) 8 : 1 4) 6 : 7
2. Two forces of magnitude 20 N and 25 N respectively act at a point. If the resultant force
39 N, the angle between the forces is

1) 600 2) 450 3) 900 4) 300

3. A and B are two unit vectors and  is the angle between them. Then A + B is a unit
vector if,

1)    /3 2)    /4 3)    /2 4)  = 2  /3

4. At what angle should the two forces 2P and 2P act so that the resultant force is 10 P?
1) 450 2) 600 3) 900 4) 1200
5. The resultant of two forces of equal magnitude is 1414N, when they are mutually
perpendicular.

1) 1000 N 2) 1000 3 N 3) 500 3 N 4) 500 N

6. The greatest and least resultant of two forces at a point are 29N and 5N respectively. If
each force is increased by 3N and applied at right angles on a particle, the new result-
ant force is

1) 35 N 2) 25 N 3) 433 N 4) 423 N
     
7. If three vectors A , B and C are 12, 5 and 13 in magnitude such that C = A + B then
 
the angle between A and B is
1) 600 2) 900 3) 1200 4) None of these
8. The resultant of two vectors (P + Q) and (P – Q) when they at right angles to each other
is

1) 2 P 2  Q2 2) 2  P 2  Q2  3) P 2  Q2 4) (P 2  Q2 )  2

Subjective Answer Type

9. Two forces 4N and F act at 1200 with each other. If their resultant is at right angles to
4N, the value of F is

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5.10 CHANGE OF VELOCITY

(i) Consider a body moving with velocity V1 due east. After a time interval ‘t’ its final

velocity is V 2 due north.
uur uur
 v1 v1

N
r uur
v v2 W E

S

(ii) The change of velocity is obtained by vector subtraction method. The initial velocity

   
vector  V1  is reversed to get  V1 and added to V 2 following the usual procedure for
 
 
vector addition. i.e., the tail of  V1 coincides with head of V 2 .

  
(iii) The vector drawn from tail of V 2 to the head of  V1 represents change of velocity  V
 
both in magnitude and direction. V1 and V 2 are mutually perpendicular as per their
directions mentioned.

 
Magnitude of change of velocity V  V 2  V1  V1  V2  2V1V2 cos 90
2 2
(iv)

 V  V12  V22

WORK SHEET - 4

Single Answer Type

ur ur
1. Find A  B in the diagram shown in figure. Given A = 4 units and B = 3 units

1) 13 units 2) 7 units 3) 1 unit 4) 13 unit

ur ur ur ur
2. $B  $i  3j$ 2k$then A  B is
If A  3i$ 2j$ 8k,

1) 4i$ 2j$ 6k$ 2) 4i$ 5j$ 6k$ 3) 4i$ 5j$ 6k$ 4) 3i$ 5j$ 6k$
ur ur ur ur ur
3. If R  A  B and angle between A, B is  , then tan  is given by __ (where  is angle
ur ur
made by R with A )
A sin  A sin  Bsin  Bsin 
1) 2) 3) 4)
B  A cos  B  A cos  A  B cos  A  B cos 
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ur ur ur
This equation confirms that A x and A y are the components of A .
In right angled triangle ONP,
Ax Ay
cos   or A x = A cos  ___________ (1) sin   or A y = A sin  ___________
(2)
A A
Squaring and adding (1) and (2), we get
Ax2 + Ay2 = A2 cos2  + A2 sin 2 
or Ax2 + Ay2 = A2 (cos2  + A2 sin 2  )  A x2 + A y2 = A 2 [Q cos2  + sin2  = 1]
2 2
or A  Ax  Ay ___________
(3)

5.12 RESOLUTION OF A VECTOR INTO THREE RECTANGULAR COMPONENTS

 
(I) Consider a right handed three dimensional coordinate system. Let A (= OP ) be a vector

drawn through the origin O. The three rectangular components of A can be determined
as follows:

(ii) Draw PP 1 perpendicular from P upon X–Z plane. From P 1 , draw P 1 P 2 and P 1 P 3
perpendiculars to X–axis and Z–axis respectively.
r  
(iii) OP2 , P1P and OP3 are known as the x-component, y-component and the z-component,
   
respectively, of A . These components are denoted by A x , A y and A z respectively, such that
      
A x  A x i, A y  A y j and A z  A z k
  
(iv) In OP1P3 , A z  A x  OP1
___________
(1)

        

(v) In OP1P, OP1  P1P  OP  A   Az  Ax   Ay  A

 
          
or A  Ax  A y  Az or A  A x i  A y j A z k . ...... (2)

(vi) According to the Pythagoras theorem in three dimensions,

A 2  A 2x  A 2y  A 2z or A  A 2x  A 2y  A 2z ...... (3)

This gives the magnitudes of vector A .

(vii) So, the magnitude of any vector is equal to the square root of the sum of the square of
the magnitudes of its three rectangular components.
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8. If a vector is along the horizontal, its vertical component is
1) maximum 2) minimum 3) zero 4) none
9. If the component of one vector along the direction of the other vector is zero, the angle
between the two vectors is
1) 0° 2) 45° 3) 60° 4) 90°

10. Find the rectangular components (along X and Y axis) of a velocity vector 10 3 m/s, making
an angle of 60° to the X - axis

1) 5 3 , 15 2) 5 3 , 20 3) 10 3 , 20 4) 10 3 , 15

Subjective Answer Type

11. One of the two rectangular components of a force is 25 N and it makes an angle of 60°
with the horizontal (force). Then find the magnitude of the other component
  
12. Three vectors A, B and C are of magnitudes 10 units, 5 unit and 10 unit respectively. If
they make angles 30°, 120° and 300° with X-axis (and all lie along the X - Y plane), then
determine the magnitude of the resultant

SYNOPSIS - 6
5.13 EQUILIBRIUM
i. When a body does not change its state of rest or of uniform motion on the application of
one or more external forces, the body is said to be in equilibrium.
ii. A single force acting on a body or point cannot keep the body in equilibrium.
iii. The minimum number of forces which can keep a point in equilibrium is 2. These
two forces are equal in magnitude and opposite in direction.
5.14 APPLICATION OF TRIANGLE LAW OF VECTORS (LAMI’S THEOREM)
If three vectors, simultaneously acting at a point, have zero resultant, then these
three vectors can be represented in magnitude and direction by the three sides of a
triangle taken in order.
Explanation:
  
(i) P , Q and R are three forces (vectors) which simultaneously act at the point
‘O’ and keep it in equilibrium. Hence they are represented, in magnitude and direction,
by the sides AB, BC and CA of the triangle ABC taken in order.



P Q C

Q

O
B
 
R


P

R A

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
4. Three forces start acting simultaneously on a particle moving with velocity V . The forces
are represented in magnitude and direction by the three sides of a triangle ABC (as
shown). The particle will now move with velocity

A B

 
1) Less than V 2) Greater than V

3) V 4) V remaining unchanged
5. A body of 10 kg is suspended by a rope and it is pulled to a side by means of a horizontal
force so that the rope makes an angle 60° with vertical. Find the horizontal force in the
thread is
[Note: If mass of a body is ‘m’ kg then its weight is ‘m’ kg–wt]

1) 10 3 kg–wt 2) 35 3 kg–wt 3) 40 3 kg–wt 4) 20 3 kg–wt

6. In the above problem the tension in the rope is
[Note: Tension is a force which will always pull the body]

1) 20 kg – wt 2) 10 kg – wt 3) 30 kg – wt 4) 10 3 kg  wt

SYNOPSIS - 7
5.15 TANGENT OF LAW
If the bob of a pendulum is pulled to a side such that the string makes an angle  with
the vertical, F = mg tan  where F is the horizontal force applied. Tension in the string

 mg 
2
T=  F2 (or)
T l 2  x2  T


T mg F x
  F

l l x
2 2 x
mg

Where l is the length of the string and m is the mass of the pendulum.

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7. Forces F1, F2 and F3 act at O as shown. If F2  5 3 N and ‘O’ is in equilibrium, value of

F 1 and F 3 are

F2
F1

60° 30°
O

F3

1) 5N , 20 N 2) 5N , 10 N 3) 5N , 15 N 4) 10N , 15 N

8. A body of weight 3N suspend vertically using a rope is pulled horizontally such that
rope makes an angle 30° to the vertical. Then tension in that rope is
1) 1 N 2) 3 N 3) 2 N 4) 4 N

SYNOPSIS - 8
5.16 MULTIPLICATION OF VECTORS
The multiplication of two vectors can be done in two ways. When two vectors are
multplied the product may be a scalar quantity (or) a vector quantity depending on the
physical quantity to be obtained. So there are two types of products for vectors.
1) Cross product or vector product
2) Dot product or Scalar product

1. CROSS PRODUCT (OR) VECTOR PRODUCT OF TWO VECTORS

If two vectors are multiplied such that their product is again a vector, the product is called
vector product or cross product.
     
The cross product of two vectors A and B is denoted by A B and read as A cross B .
ur ur ur
If C is the resultant of two vectors A and B
  
Then C  A B ,

C  AB sin  , where  is angle between A and B . Here 0     .

ur
The direction of C is determined by right handed screw rule, thus
  
vector C is normal to the plane of A and B and points in the direction in which a right
handed screw would advance when rotated about an axis perpendicular to the plane of

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WORK SHEET - 8

Single Answer Type

1. Two vectors of magnitude 18 units and 13 units are acting at 300 to each other. Then
the magnitude of vector product is
1) 31 units 2) 234 units 3) 117 units 4) 243 units
2. Two forces of magnitude 12N and 14N are acting at a point as shown in figure.

12N
14N 120 0

Then the magnitude of vector product is

1) 42 3N 2) 84 3N 3) 48 3N 4) 64 3N
ur ur ur
3. Given that C  A  B , then the angle between C and B is

1) 00 2) 900 3) 1200 4) 300

ur ur ur
4. If two vectors A and B are enclosing an angle 300 and the magnitude of A is 3 units,
ur
what is the magnitude of B . If the magnitude of vector product is 24 units
1) 16 units 2) 3 units 3) 8 units 4) 21 units
 
5. The numerical value of A A is
1) 0 2) 1 3) A2 4) A
   
6. What is the angle between c d and c  d ?
1) 0° 2) 45° 3) 60° 4) 90°
 
7. If vectors A and B have angle  between them, then the magnitude of their vector
product is
1) ABcos  2) ABsin  3) AB 4) none of these
    
8. Given : c  a b . The angle which a makes with c is
1) 0° 2) 45° 3) 90° 4) 180°
   
9. What is the angle between A B and B A?
1) 0° 2) 180° 3) 45° 4) 90°
     
10. If A B  0 and B C  0 , then the value of A C is


1) 0 2) AC sin  n 3) AC cos  4) AB tan 

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WORK SHEET - 9

Single Answer Type

ur ur
1. The y-component of cross product of P  iˆ  ˆj  kˆ and Q  2iˆ  ˆj  4 kˆ is
1) 4 2) 5 3) 3 4) 2
2. Choose the wrong option
1) iˆ  ˆj  kˆ 2) iˆ  ˆj   kˆ 3) ˆj  kˆ  iˆ 4) kˆ  iˆ  ˆj

  
    

3.  i  j  k  × 2 i  3 j  k  =
   
           
1) 2 i  j  k 2) 2 i  j  k 3) 2 i  j  k 4) 2 i  j  k
ur    ur   
4. Cross product of A  2 i  3 j  k and B   i  3 j  k
           
1) 6 i  3 j  6 k 2) 6 i  3 j  6 k 3) 6 i  3 j  6 k 4) 6 i  3 j  3 k
ur ur
5. If P  iˆ  ˆj  2kˆ , Q  2iˆ  ˆj  kˆ then the angle enclosed by them is
1) 00 2) 450 3) 300 4) 600
6. Choose the correct statement
         
1) j  k  1 2) j k  1 3) j  k  i 4) j k   i
7. Choose the incorrect regarding, null vector.
ur r r ur r r
1) A 0  0 2) A  0  0
3) Null vector direction is indeterminant 4) Magnitude of null vector is ‘0’
   
8. What is the angle between ( A  B ) and ( A  B )?

 3
1) 0° radian 2)3)  radian 4) radian
2 2
ur    ur    ur ur
9. If A  2 i  3 j  6 k and B  3 i  6 j  2 k , then vector perpendicular to both A and B has
  
magnitude K times of 6 i  2 j  3 k . Then K =
1) 1 2) 3 3) 7 4) 9
ur   
10. Magnitude to torque of F  2 i  j  k acting at (–5, 2, 1) about (1,1,1) is
 
[Hint: Torque,  = r  F ]
1) 53 2) 53 3) 24 4) 24
     
11. Area of the parallelogram formed by vectors 3 i  2 j  k and i  2 j  3 k is ____(given
vectors are diagonals)
   
(Note : Area of parallelogram formed by vectors A and B is A  B )
1) 3 8 units 2) 24 sq.units 3) 8 3 sq.units 4) none of these
   
12. If the diagonals of a parallelogram are represented by the vectors P  5 i  4 j  3 k and
   
Q  3 i  2 j  k . The area of the parallelogram is nearly
1) 18 units 2) 4.0 units 3) 8.0 units 4) 13units
IX Class - Physics Page No : 80
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5.19 SCALAR PRODUCT BY COMPONENTS METHOD:

(i) When the vectors are expressed as rectangular components along X, Y and Z-axes, say
ur ur
A  A x iˆ  A y ˆj  A z kˆ and B  B x iˆ  B y ˆj  B z kˆ

Here iˆ, ˆj and k̂ are unit vectors along X, Y and Z axes respectively, then

ur ur
 
A.B  Axiˆ  Ay ˆj  Az kˆ . Bxiˆ  By ˆj  Bz kˆ 
          
 Ax iˆ.Bx iˆ  Ay ˆj.Bx iˆ  Az kˆ.Bx iˆ  Ax iˆ.By ˆj  Ay ˆj.B y ˆj  Az kˆ.By ˆj  
 A iˆ.B kˆ    A ˆj.B kˆ    A kˆ.B kˆ 
x z y z z z

         
 Ax .Bx iˆ.iˆ  Ay .Bx ˆj.iˆ  Az .Bx kˆ.iˆ  Ax .B y iˆ. ˆj  Ay .B y ˆj. ˆj  Az .B y kˆ. ˆj   
   
Ax .Bz iˆ.kˆ  Ay .Bz ˆj.kˆ  Az .Bz kˆ.kˆ  
 Ax Bx 1  Ay Bx  0   Az Bx  0   Ax B y  0   Ay B y 1  Az B y  0   Ax B y  0   Ay Bz  0   Az Bz 1

 Ax Bx  Ay By  Az Bz

 
    A.B A x B x  A y B y  A z Bz
A.B  A B cos   cos     (or)  cos  
A B AB

   
2
A. A  A A  A 2x  A 2y  A 2z  A

 
  A.B
The scalar component of A in the direction of B is expressed as A cos   
B

 
  A.B
The scalar component of B in the direction of A is expressed as B cos   
A

 
A vector cannot have component in a direction perpendicular to it. If P and Q are
 
perpendicular, then P can not have component along Q and vice versa because
 
P .Q= 0

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ur ur ur ur
11. Dot product of A and B is equal to C. Then angle between A and B is
1  AB  1  C  1  B  1  AC 
1) cos   2) cos   3) cos   4) cos  
 C   AB   AC   B 
     
12. If i + 2 j+ n k is perpendicular to 4 i + 2 j+ 2 k , then the value n =
1) 4 2) 2 3) –4 4) –2

  
   
13. When a force  8 i + 4 j  N displaces a particle through  3 i - 3 j  m, the power is 0.6W.
   
Work done Work done
The time of action of force is [Note : Power   time  ]
Time Power
1) 10 s 2) 20 s 3) 15 s 4) none of these
ur ur
14. A point of application of a force F  5i$  3$j  2k µ is moved from r  2i$  7$j  4k
µ to
1
uur ur uur
µ. The work done is___. ( Hint : Work done = W = F.
r2  5i$  2$j  3k r ).
1) 28 units 2) 22 units 3) 11 units 4) 0 units
uur r uuur
15.  
A force of 3i  j  2k N displaced the body from a point (4, –3, –5)m to a point
ur ur
(–1, 4, 3)m. The work done is ____. ( Hint : Work done = W = F . S ).
1) 12 J 2) 14 J 3) 24 J 4) 36 J
ur ur
16. µ and B  4i$  2$j  4k
The vector A= $i  4$j  3k µ are

1) perpendicular 2) parallel 3) inclined at 45° 4) inclined at 60°

WORK SHEET - 11

Single Answer Type

1. A unit vector parallel to the sum of the vectors 2i  4 j  5k , i  2 j  3k are

1)

 3iˆ  6 ˆj  2kˆ  2)

 3iˆ  6 ˆj  2kˆ  3)

 3iˆ  6 ˆj  2kˆ  4)

 3iˆ  6 ˆj  2kˆ 
7 7 7 7
uuur uuur uuur
2. ˆ ˆ ˆ ˆ ˆ ˆ
If OP=2i-3j+k,OQ=i+j+k,then PQ=
1) ˆi + 4jˆ 2) ˆi - 4jˆ 3) -iˆ + 4jˆ 4) -iˆ - 4jˆ
uuur uuur uuur uuur
3. ˆ ˆ ˆ AB=3i-2j+k,
If OA=i+j+k, ˆ ˆ ˆ BC=i+2j-2k,
ˆ ˆ ˆ CD=2i+j+3k
ˆ ˆ ˆ then the position vector of D is

1) 2iˆ  3 ˆj  7 kˆ 2) 7iˆ  2 ˆj  3kˆ 3) 3iˆ  2 ˆj  7 kˆ 4) none

4. The length of the vector 3iˆ  6 ˆj  2kˆ is
1) 3 2) 3 3) 7 4) 14
ur ur uuur
5. If the position vectors of A and B are 3iˆ  2 ˆj  kˆ and 2iˆ  4 ˆj  3kˆ then AB

1) 14 2) 29 3) 43 4) 53

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r r r r rr
20.   0
If a =2, b =3, a,b =120 then a.b =
1) 0 2) -3 3) 16 4) -8
r r r r rr
21. If a +b = a -b then a.b =
1) 0 2) 8 3) 16 4) 10
uuur uuur uuur uuur uuur
22. If OA  iˆ  2 ˆj  3kˆ, OB  3iˆ  ˆj - 2kˆ, OC  2iˆ - 3 ˆj  kˆ then AB. AC 
1) 15 2) 17 3) 0 4) None of these
23. The angle between the vectors 6iˆ  2 ˆj  kˆ, 2iˆ - 9 ˆj  6kˆ is
1) 900 2) 600 3) 450 4) 300

Competitive Galaxy
1. Two cars are 60 km away from each other and are moving towards each other at 116
km/h and 184 km/h. How far are they from each other one end a half minutes before
their collision? (SAT-RAMAIAH-2006]
2. A man moves 100 m 300 East of south then another 100 m 600 North of East. He is at
a distance of ________ from the starting point (SAT-RAMAIAH-1995]
3. A man is 3m to the north of a certain point, another is 4m to the east of the same
point. The first man moves futher north a distance of 9 m and second a distance of 9 m
of weast wards. The distance between them now is (SAT-RAMAIAH-1993]
4. One of the rectangular components of a velocity 50 m/s is 30 m/s. The other component
is (Dr. A.S.Rao Awards Council, Dec-2010)
1) 20 m/s 2) 15 m/s 3) 10 m/s 4) 40 m/s
5. Ramesh walked 10 ft from A to B in the east. Then he turned to the right and
walked 3ft. Again he turned to the right and walked 14ft. How far is he from A ?
(UNIFIED CYBER OLYMPIAD - 2011)
a) 4 ft b) 5 ft c) 24 ft d) 27 ft
6. The distance travelled by a body is given by  3S  15t  9t  m.
2
Find the uniform
ecceleration of the body in m S 2 ? ( SLSTE-AP - 2011)
a) 9 m / s 2 b) 6 m / s 2 c) 1 m / s 2 d) 3 m / s 2

KEY & HINTS

WORKSHEET – 1 (KEY)
1) 4 2) 2 3) 1 4) 3 5) 4

6) 3 7) 4 8) 3 9) 1 10) 1

11) 1 12) 1 13) 2 14) 3 15) 1

16) 1 17) 3 18) 2

1) First three options are correct.

2) Except force other quantities are scalars.

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(iˆ  2jˆ  k)
ˆ (2iˆ  3jˆ  5k)
ˆ

15) Displacement vector of the particle, S = –

= iˆ  5 ˆj  4kˆ

16) Length of 2iˆ  3 ˆj  4kˆ in XY - plane = (2)2  (3)2 = 13

17) Given vector may written as (0.2)2 + (0.3)2 + Z2 = 1

or 0.04 + 0.09 + Z2 = 1
or Z2 = 1 – 0.13 = 0.87

18) (2)2  (y)2  (3)2 = 5 or 4 + y2 + 9 = 25 or y2 = 12 or y = 12

19) Displacement vector of the particle,

   
S = (x2 – x1) i +(y2 – y1) j + (z2 – z1) k

  
= (5 – 1) i + (4 – 2) j + (2 – 3) k

  
=4i + 2 j – k

   
20) The initial and final position of vectors may be written as r1 = i + 2 j – k

   
and r2 = 3 i + 2 j + 2 k

uuur          
Displacement vector of PQ = r2 – r1 = (3 i + 2 j + 2 k ) – ( i + 2 j – k ) = 2 i + k
uuur
Magnitude of displacement vector, PQ = (2)2  (3)2 = 13

WORKSHEET – 2 (KEY)
1) 1 2) 3 3) 4 4) 1 5) 3

6) 2 7) 2 8) 4 9) 2

 
1) If two like vector A and B are parallel

  
a) C  A  B for like vector = A2 + B2 + 2AB

  
b) C  A  B for unlike vector = A2 + B2 – 2AB

  2  2  
2AB
c) C  A  B  2 A B Cos60 0 = A2 + B2 + = A2 + B2 + AB
2
IX Class - Physics Page No : 88
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Q Sin  Q Sin 
11) tan  = =
Q + Q Cos  Q [1 + Cos ]

Sin 

= 2 Cos2   = Tan
 2  2

 
[Q Sin  =2 sin cos ]
2 2

 
 Tan  = Tan   =
2 2

WORKSHEET – 3 (KEY)
1) 2 2) 1 3) 4 4) 1 5) 1

6) 2 7) 2 8) 2

PQ 7
1) 
PQ 1

P  Q  7P  7Q

P 4
6P  8Q  
Q 3

2) R 2  P 2  Q 2  2PQ cos 

 39    20    25   2  20  25  cos 
2 2 2

62 1
cos   ;    600
125 2

 
3) Given, A  B = A2 + B2 + 2AB cos  and A = 1, B = 1

 
1
The required condition is A  B = 1  1 = 2 + 2 cos  = cos  =
2

2
  = 1200 or
3

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 F1  F2 = 25
2 2

F12  F22  2F1F2 cos1200 = 13

 F1  F2  2F1F2 × 1/2 = 13
2 2

 F1F2 = 12
(F1 – F2)2 = F12 + F22 – 2F1F2
= 25 – 2 × 12 = 1
 F1 – F2 = 1  (1) and F1 + F2 = 7
 (2) and from (1) and (2) we get
F1 = 4N and F2 = 3N

WORKSHEET – 4 (KEY)
1) 4 2) 2 3) 3 4) 2 5) 4

6) 4 7) 4 8) 2 9) 1

1. R  A 2  B2  2ABcos 

 42  32  2  4  3 cos 60

 16  9  12  13 units
ur ur
2.    
A  B 3i$ 2$j  8k$ $i  3j$ 2k$

 3i$ 2j$ 8k$ $i  3j$ 2k$

 4i$ 5j$ 6k$

3. Conceputal
4.

v

B A
O

v
Change in velocity

  
 v    v   2 v = 2 × 10 = 20ms–1
 

IX Class - Physics Page No : 92

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= 2V 2  V 2  3V 2

= 3V
9. Figure shows the conditions of the problem.
uur
Initial velocity; v1 = 5 ms–1 along east = 5 ˆi
uur
Final velocity, v 2  5 ms 1 along north = 5 ˆj
N
uuur
 v1

uur
45° v 2

W E
O

S
uur uur uur
Change in velocity, v  v 2  v1  5 ˆj  5 ˆi
uur
52   5   5 2
2
 v 
uur
The direction of v is north - west.
Average acceleration,
uur
ur v 5 2 1
a    ms 2 north - west
t 10 2
8.

N
V1 = 50
V2 kmph
W

V

-V1 = 50
kmph
S
uur
v1 = 50 km h –1 due north
uur
v 2 = 50 km h–1 due west.
uur uur uur
Also, v  v 2   v1  
2 2
v  v1  v2  50 2 km h1
This confirms S-W direction.

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5. Maximum value of resultant, P + Q = 7,
Minimum value of resultant, P – Q = 3,

PQ 7 P 10 5
  4P  10Q or  
PQ 3 Q 4 2

3P 3
6. Let P cos  = or cos  =
5 5
4
 sin  
5
4P
Other component of force P sin  =
5
7. X – Component of force, Fx = F cos 
3
= 100 × cos30° = 100 × = 50 3N
2
Y – Component of force, Fy = F sin 

1
= 100 × sin30° = 100 × = 50 N
2
8. Let v the vector, vertical component V sin  = V sin 0°  zero
 
9. Let one vector is A and other is B .
The given condition is Ab = Acos  = 0

 cos  = 0 or  = 90°
10. The rectangular components of the vector 10 3 , are : Component along X axis
Y

vy 10 3

60°

vx X

Vx = Vcos  = 10 3 cos 60° = 5 3 ms –1 and component along Y-axis :

Vy = Vsin  = 10 3 sin 60° = +15 ms–1

25
11. Given that, P cos  = 25 or P =
cos 60
25
  50N
1/2

Other component of force = Psin 

3
= 50 sin 60° = 50 × = 25 3N
2

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1. The minimum number of forces which can keep a point in equilibrium is 2. These two
forces are equal in magnitude and opposite in direction.
  
2. The magnitude of P, Q and R are proportional to the lengths of AB, BC and CA
respectively.

P Q R
i.e.,    K (cons tan t)
AB BC CA
3. Whenever a triangle is formed to represent three forces keeping a body in equilibrium,
then the length of each side of the triangle to proportional to the sin of the angle opposite
to it.
4. From triangle law of vectors, the resultant force acting on a particle is zero and hence
there is no change in its velocity.
5. P = 10 Kg - wt,  = 60°
The body is in equilibrium at B under the action of three forces
(i) weight of the body
(ii) horizontal force and
(iii) tension ‘T’ in the string. These three forces are parallel to three sides of the triangle OAB.
0

60°
T

150°
B
A

120°

According to Lamis’ theorem.

P F T
 
sin150 sin120 sin 90
P F T
  
sin 30 sin 60 sin 90

2F
 2P  T
3

F= 3 P= 3 × 10 Kg – wt

 mg 
2
6. Tension in the rope, T =  F 2 , Here F = mg tan is the horizontal force F = 10

× 9.8 × 3 = 10 3 Kg – wt.

 
2
 T= 102  10 3 = 20 Kg – wt.

(or)

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By Lami’s theorem

T2
T1
60° 30° 90°

T1 T2 150° 120°

T3
80 Kg

T1 T2 T3
 
sin120 sin150 sin 90

sin120 80  3 /2
 T1  T3  
sin 90 1

 40 3 Kg  wt

sin150 1
 T2  T3   80  =40 kg - wt
sin 90 2
5. Horizontal force or applying force, (F) = mg tan = 2 × 10 × tan60 = 20 3 N

F    mg 
2 2
Tension in the string, T 

 20 3 
2
  2  10   40N
2


Horizontal force F 20 3
  
Tension in t he string T 40

3

2
Alternative method

T
T sin30°
30°
F
T cos30°

2 kgwt

T 3
F = T Cos30° = --(1)
2
2 = Tsin30°  T = 4 kg – wt --(2)

Substitute (2) in (1) we get, F = 2 3 kg – wt

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WORKSHEET – 8 (KEY)
1) 3 2) 2 3) 2 4) 1 5) 1

6) 4 7) 2 8) 3 9) 2 10) 1

11) 4

1. A  B  AB sin 
1
 18 13  sin 300  9 13   117 units
2
2. A  B  AB sin 

3
 AB sin 600  14  12   84 3 N
2
3. C is perpendicular to A to B
1
4. 24  A  B  sin 300  24  3  B 
2
 B  16 Units
5. Cross product of equal vectors is zero.
       
6. c  d is in the plane of c and d . Also c d is perpendicular to the plane of c and d . So,
   
angle between c  d and c d is 90°
7. According to definition of cross product.
8. The result follows from the definition of cross product.
   
9. A  B and B  A are oppositely directed. So, angle is 180°.
   
10. A B = 0  A P B
   
B C = 0  B P C
 
 A PC
 
 A C  0
 
11. A B = (15)(8) sin 60° = 60 3 units.

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 3iˆ  3 ˆj  3kˆ
ur ur
 3    3 
2 2
 PQ   32

 9  9  9  27  3 3
ur
P  12   1  22  6 and
2

ur
Q  22  12  12  6
ur ur
PQ 3 3 3
 sin   ur ur      600
P Q 6 6 2

       
8. A  B is perpendicular to the plane of A and B . Also, A  B is in the plane of A and B .
   
So, A  B is perpendicular to A  B .


  radian.
2

ˆi ˆj ˆ
k
 
9. A  B 2 3 6
3 6 2

  
 i  6  36   j  4  18   k  12  9 

  
   

 42 i  14 j  21k = 7  6 i  2 j 3 k 
 
 
   

Given A  B  k  6 i  2 j 3 k 
 
K=7
r ur r
10. Torque,   r  F Here r  5iˆ  2 ˆj  kˆ

 
 iˆ  ˆj  kˆ =  6iˆ  ˆj  0kˆ

iˆ ˆj kˆ
  6 1 0
2 1 1

 iˆ  1  0   ˆj  6  0   kˆ  6  2 

 iˆ  6 ˆj  4kˆ
Magnitude of torque

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ˆi ˆj ˆ
k
 
A  B 3 1 2
2 2 4

  
 i  4  4   j 12  4   k  6  2
  
 8( i  j k)

1   1   1  8

 
2 2 2
AB  2
8 3

   
 
8  i  j k 
 Unit vector perpendicular to plane containing A and B   
8 3
  
i  j k

3

WORKSHEET – 10 (KEY)
1) 4 2) 1 3) 2 4) 1 5) 1

6) 1 7) 3 8) 2 9) 4 10) 3

11) 2 12) 3 13) 2 14) 1 15) 3

16) 1

SOLUTIONS
ur ur
2. 
A.B  5iˆ  2 ˆj  3kˆ . 2iˆ  8 ˆj  5kˆ  
 10  16  15  11 units
   
3. P .Q  PQ cos  and P Q  PQsin 

As per problem, PQ cos  = PQ sin  or tan  = 1   =450

 
4. A.B  AB cos 270
= AB cos (360° – 90°)
= AB cos 90° = 0
   
5. P .Q  0  P  Q or   90
   
P Q  PQ sin 90° = PQ or P Q

6. The dot product of two equal vectors is equal to the square of the magnitude of either of

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12 120
  = 20sec
0.6 6
  
   
    

14.  r  r2  r1   5 i  2 j  3 k    2 i  7 j  4 k 
   
  
 3 i  5 j k
 
   
    

 W  F .  r   5 i  3 j  2 k  . 3 i  5 j  k 
   
= 15 + 15 – 2 = 28 units
   
15. S = (x2 – x1) i + (y2 – y1) j + (z2 – z1) k
     
= (–1 –4) i +[4 –(–3)] j + [3 –(–5)] k = –5 i + 7 j + 8 k
 
Work done, W = F .S

   
    

=  3 i  j  2k  .  5 i  7 j  8 k 
   
= 15 – 7 + 16 = 24 joule.
ur ur r ur ur r r ur
16.  
A . B = i + 4k + 3k . 4 i + 2 j  4k 
 
From given data we get A.B = 4 + 8 – 12 = 0. As dot product is zero these vectors are
perpendicular.

WORK SHEET -11 (KEY)

1) 1 2) 3 3) 2 4) 3 5) 4

6) 1 7) 4 8) 2 9) 3 10) 2

11) 1 12) 1 13) 2 14) 2 15) 2

16) 2 17) 4 18) 3 19) 4 20) 2

21) 1 22) 2 23) 1

r r
1.    
a  b  2iˆ  4 ˆj  5kˆ  iˆ  2 ˆj  3kˆ  3iˆ  6 ˆj  2kˆ


r r
ab
r r  
3iˆ  6 ˆj  2kˆ
 
3iˆ  6 ˆj  2kˆ
.
 
Required unit vector =
ab 9  36  4 7
uuur uuur uuur
2. PQ  OQ  OP  iˆ  4 ˆj

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16. If a,b,c are the sides of the triangle respectively then
r
a  2iˆ  3 ˆj  6kˆ  4  9  36  49  7
r
b  iˆ  2 ˆj  3kˆ  36  4  9  7,
r
c  3iˆ  6 ˆj  2kˆ  9  36  4 49  7.
 Perimeter = 7+7+7=21
r r r
17.    
c  5iˆ  9 ˆj  3 iˆ  4 ˆj  2iˆ  3 ˆj  3a  b
uuur
18. AB=iˆ  2 ˆj  2kˆ  D.r.'s are 1,-2,2   D.c.'s are 1/3,-2/3,2/3
rr
19. a.b  2  12  2  8.
rr r r r r
20.  
a.b  a b cos a, b   2  3 cos1200  3
r r r r r r 2 r r 2 rr rr
21.    
a  b  a  b  a  b  a  b  2a.b  0  a.b  0.
uuur uuur uuur uuur uuur uuur uuur uuur
22. AB  OB  OA  2iˆ  ˆj  5kˆ. AC  OC  OA  i  5 j  2k , AB. AC  2  5  10  17.
rr r r r r
23.  
a.b  12  18  6  0  a  b  a, b  900

Competitive Galaxy
60
1. Time taken t meet t   0.2 hr  12 min
300
distance 3/2 min before collision 300  3 /120  7.5 km
2. Two given displacement form two sides of equilateral triangle, then closing side equal
to one of the side = 100 m.

100 100

3. 13m, 12.6
4. Let 50sin   30
sin   3 / 5

52  32  4
The other component is given by 50cos 

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