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VECTOR 2

CONTENTS 1. DEFINITIONS
2. ADDITION OF TWO VECTORS
3. RESOLUTION OF VECTORS
4. VECTOR PRODUCT

 EXERCISES

 Single Correct Questions
 Previous Years Questions
 Answer Key

VECTOR PHYSICS PART - 1

1. DEFINITIONS

1.1 Scalar Quantity:

A physical quantity which can be described completely by its magnitude only and does not
require a direction is known as a scalar quantity. It obeys the ordinary rules of algebra.
eg: distance, mass, time, speed, density, volume, temperature, current etc.

Note:

Scalar quantities can be added, subtracted and multiplied by simple laws of algebra like numbers.

1.2 Vector Quantity:

A vector physical quantity necessarily requires magnitude and a particular direction, when it is
expressed . eg: displacement, velocity, acceleration, force etc.

Note: The above statement doesn’t define a vector.It merely states that vector quantity has
direction.So, if a physical quantity is a vector it has a direction, but the converse may or may not
be true, i.e. if a physical quantity has a direction, it may or may not a be vector.

Modulus/magnitude of a vector: The modulus of a vector is the number associated with the
vector i.e if we remove dirction from a vector only magnitude will be left.Eg: “4m in east” is
vector whose magnitude is “4m” and has “east” direction . It is always positive and has no

 
direction.M odulus of vector A is represented as |A| or A.

(i) Representation of vector :

Geometrically, the vector is represented by a line with an arrow indicating the direction of vector as

Mathematically, vector is represented by  .Sometimes it is represented by bold letter A .
A

By convention,the length of a vector in diagram is made in proportion to its magnitude/
modulus.

Note:

1. Scalar quantity may be negative e.g. charge, electric current, potential energy, work etc.

2. Scalar quantity are direction independent e.g pressure, electric current, surface tension etc.

3. Vector quantities are direction dependent e.g. force, velocity and displacement.

(ii) Few important vectors:

(a) Zero vector or null vector

A vector whose magnitude is zero is called zero vector. It is represented by  .
0

Note:

One might think what is the “direction of a zero vector”

(i) The direction of a zero vector isn’t physically relevant.

(ii) The direction of a Zero vector can be taken arbitrarily.

(b) Unit vector:

A vector whose magnitude is one unit is called unit vector. A unit vector in the direction of

vector  represented by Aˆ , so Aˆ  A .
A is A

Note:

(i) Any vector can be expressed as   AAˆ .
A

(ii) Unit vectors along the positive x-, y- and z-axes of a rectangular coordinate system are denoted

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VECTOR PHYSICS PART - 1

by ˆi , ˆj and kˆ respectively such that ˆi  ˆj  kˆ = 1.

(iii) Angle between two vectors
Angle between two vectors, bascially angle between their direction, is smaller of the two angles
between the vectors when they are placed tail to tail by displacing either of the vectors parallel to
itself (i.e. 0    ).
Note:
Angle between vectors is independent of their magnitude.

Illustration

 
Three vectors A,B,C are shown in the figure. Find angle between (i) A and B ,
  
(ii) B and C , (iii) A and C .

Sol. To find the angle between two vectors we connect the tails

  
of the two vectors. We can shift B such that tails of A , B and C

are connected as shown in figure.   

Now we can easily observe that angle between A and B is 60º, B and C is 15º and between A and

C is 75º.

(iv) Equality of vectors:

Two vectors are said to be equal if and only if they have equal magnitude and same direction.

Note:

(a) If a vector is displaced parallel to itself it does not change (see Figure) 
A
 
 AB
B

(b) If a vector is rotated through an angle other than multiple
of 2 (or 360º) it changes (see Figure).

Teaser: How many different vectors are drawn here ?

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(v) Multiplication of a vector by a scalar :

When a vector is multiplied by a scalar  , we get a
 C A=B =C
new vector which is  times the vector  i.e.  A .The Transition of a vector
A
 B parallel to itself
direction of resulting vector is that of A if is positive A
 

and opposite to that of A if  is negative .


The unit of resulting vector is the multiplied units of  and A . For example, when mass is multiplied

with velocity, we get momentum. The unit of momentum is obtained by multiplying units of mass

and velocity. 
A

 A

2A   2  

2A   2

Similarly, we can have vector  divided by a scalar . The resulting vector becomes A.
A 

The magnitude of the new vector becomes 1  and direction is same as that of  .
 that of A A

 
A A/ 2   2

Note: Negative of a vector 
The negative of a vector is defined as vector having same A
magnitude as that of the vector but has opposite direction.

A

Illustration

A physical quantity (m = 3kg) is multiplied by a vector a such that   ma . Find the magnitude and
F

direction of F if

 
(i) a = 3m/s2 East wards (ii) a = –4m/s2 North wards

Sol. (i)   ma = 3 × 3 ms–2 East wards = 9 N East wards
F

(ii)   ma = 3 × (–4) N North wards =– 12N North wards = 12 N South wards
F

(vi) Few more terms:

(a) Collinear or parallel vectors

The vectors which act along the same line or along a parallel line are called collinear vectors.

A

  
BA B

a  Like or parallel vectors. 

  A

BA 

 b  Un like or antiparallel B
vectors.

Note: If  and  be two collinear vectors, then there exists a scalar k such that    , the abso-
A B B kA

lute value of k being the ratio of the length of the two collinear vectors.

(b) Coplanar and concurrent vectors :
Vectors originating from same point are known as concurrent vectors.Vector lying in same

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  
plane are called coplanar vectors. In figure, A , B and C are coplanar and concurrent vectors.

(c) Axial vector:
A vector which has rotational effect and acts along axis of rotation is called axial vector. Example:
angular velocity, torque, angular momentum, angular acceleration etc.

Axial vector

Anti clockwise rotation clockwise rotation 
Axial vector B


A

VECTOR OPERATIONS

The possible vector operations are: 
C
(i) Addition or subtraction of vectors (ii) Multiplication of
vectors

Note: (i) Division (of vector/scalar) by a vector is not defined.
(ii) Addition of vector and scalar is meaning less.

2. ADDITION OF TWO VECTORS

2.1 Geometrical method

(a) Triangle law of vector addition : If two non-zero vectors can be represented by the two sides

of a triangle taken in same order , i.e tail of one vector coinciding with head of other vector,

then their resultant is represented by third side of the triangle with tail,head as the free tail,free
 

head respectively. . Consider two vectors A and B and let angle between them be  .

   
 
Finding A  B : First draw vector A OP in the given direction. Then draw vector B starting

  

 from the head of the vector A. Then close the triangle. R OQ will be their resultant

   Q
B R AB

 B

A   
O A P
(a)
(b)

(b) Parallelogram law of vector addition: If two non-zero vectors can be represented by the two

adjacent sides of a parallelogram with common tail( or head ), then their resultant is represented
by the diagonal of the parallelogram passing though the common point of the vectors having

 
tail (or head) as the common point. Suppose two vectors A and B are as shown in figure


B

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A

VECTOR PHYSICS PART - 1

    
   
Finding A  B : Draw vectors A OP and B OQ starting from a common point O in the given


direction. The diagonal OS of the parallelogram OQSP will represent their resultant.

Q
S

 R = A+B
B

  P
O A

Note:

(i) Both methods are equivalent i.e they always give same resulltant vector.Intuitively,triangle
law may correspond to addition of displacement and parallelogram law to addition of forces.

(ii) Now, we can define vector quantity as “quantity having both direction, magnitude and

moreover which adds up according to triangle law of addition”.

(iii) Vector addition is commutative i.e a      a (prove it using geometry ).
b b

Teaser: Consider following two statements

(i) Two forces are added using triangle rule because force is a vector quantity.

(ii) Force is a vector quantity because two forces are added using triangle tule.

(a) both are true (b) both are false

(c) (i) is true, (ii) is false (d) (ii) is true, (i) is false

Teaser: Express the vector u in terms of a,b,c.

2.2. Analytical method: Finding   
A B

Q

  Bsin 
R B

  S
O P

A
A B cos 
 
It is clear from the geometry of the figure that resultant of A and B is equal to the resultant of

   
and B sin  . By Pythagoras theorem, we have
A  B cos 

 R2  A  B cos2  B sin 2  A2  B2 cos2   2AB cos  B2 sin2 

or R  A2  B2  2AB cos 

If  is the angle which resultant  makes with  , then: tan   B sin 
R A A  B cos

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Special cases :

(i) For parallel vectors   00 ; Rmax  A2  B2  2AB  A  B
(ii) For anti-parallel vectors   1800 ; Rmin  A2  B2  2AB  A  B

Thus, A  B  R  A  B [*here A,B represent magnitude of vectors not vector itself]

(iii) If A = B, R  A2  A2  2AA cos  = 2A2 1  cos  = 2A2  2 cos2  / 2 = 2A cos  / 2 and

   .
2

Note:

(i) Graphical method is necessary to visualise vector addition but it is useful only in simple
geometry.Analytical method can always be used for calculations.

(ii) Vector quantities that represents the same physical quantity can be added/subtracted together

and the resultant is a vector of the same type.Eg: addition of force vector to displacement

vector is meaningless.


(iii) As previously mentioned that the resultant of two vectors can have any value from (A B) to


(A B) depending on the angle between them and the magnitude of resultant decreases as

angle increases from 0º to 180º.This can easilty be “visualised” from geometrical addition as
compared to analytical method.

(iv) Minimum number of unequal coplanar vectors whose sum can be zero is three.

(v) The resultant of three non-coplanar vectors can never be zero i.e minimum number of non
coplanar vectors whose sum can be zero is four.

Illustration

 
Find A  B in the diagram shown. Given A = 4 units and B = 3 units and angle between them is

60 degrees.

Sol R = A2  B2  2AB cos  = 16  9  2.4.3.cos 600 = 37 and

tan   Bsin  3sin600 
A+Bcos 4+3cos 600 B  

R AB

 0.472

  tan1(0.472)  25.30 
A

 
Thus, the magnitude of resultant of A and B is 37 units at angle 25.3º from A in the direction

shown in figure.

Illustration

Two forces of equal magnitude are acting at a point. The magnitude of their resultant is equal to
magnitude of either. Find the angle between the force vectors.

Sol. Given R =A =B , using R2  A2  B2  2AB cos

A2  A2  A2  2AA cos   cos   1    120o
2

This can be solved orally just by geometrical addition .
Hint: Think of a geometry in which all three sides of a triangle are equal,and remember the
definition of angle between vectors.

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VECTOR PHYSICS PART - 1

2.3 Subtraction of two vectors 
PQ

means addition of -B
    
A OP A
     The subtraction of B PQ
from to

    

A  B = A  B

   
   First draw vector A OP in the given direction. Then draw vector B PQ starting from head of

    will be equal to   .
A-B
the vector A . Then close the triangle. R OQ

The angle between   will be equal to (180o  ) . Q
A and B 
B
Therefore, the resultant will be given by:

 R  A2  B2  2AB cos 1800    A2  B2  2AB cos  O  P
A


   and tan (180o  )
B sin 1800    B sin  B
 A  B cos  R=A 
B
A  B cos 1800  

 Q'

 is the angle made by R with A

Note :
Change in a vector physical quantity means subtraction of initial vector from the final vector.

2.4 Addition of more than two vectors : Polygon law of vector addition

If a number of vectors are represented by the sides of an open polygon taken in the same order, then
their resu ltant is represented by the closing side of the polygon taken in opposite order. Here in the
figure, R (closing side of polygon) represents the resultant of vectors A,B and C .

      
C C R  ABC C

2 1  2    2
A
 2 
B B

1  B 1
A
1  b
A

a

Note:

If n number of vectors makes a closed polygon, their resultant will be zero. Here vectors
    
A,B, C, D andE make closed polygon.

     
 ABCDE0

Illustration

Two non zero vectors  and  are such that  +  = |  –  Find angle between  and  ?
A B |A B| A B |. A B

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Sol.  +  =  –  |
|A B| |A B

 A2 + B2 + 2AB cos  = A2 + B2 – 2AB cos 

 *Add using geometrical method, its very simple.
 4AB cos  = 0  cos  = 0   = 2

Illustration

If the sum of two unit vectors is also a unit vector. Find the magnitude of their difference?

Sol. Let Aˆ and Bˆ are the given unit vectors and Rˆ is their resultant then | Rˆ | = | Aˆ + Bˆ |

1 = (Aˆ )2  (Bˆ )2  2|Aˆ ||Bˆ |cos   1= 1 + 1 + 2 cos   cos  = – 1
2

|  –  | = (Aˆ )2  (Bˆ )2  2|Aˆ ||Bˆ |cos  = 1  1  2  1  1( 1) = 3
A B 2

*********************

Teaser: Find non -zero scalars , such that for all vectors a and 
b,

   α a+2b -βa+ 4b-a =0

Exactly one option must be correct,

(a)   2,  1 (b)   2,  3 (c)   1,  3 (d)   2,  3

3. RESOLUTION OF VECTORS

In general, resolving a vector in n components implies finding n vectors such that their vector sum

or resultant is the given vector and the n vectors (parts) are called components of the given vector.

If  and  be any two nonzero vectors in a plane with different directions and  be another vector
a b as a sum of two vectors - A

in the same plane.  can be expressed  one obtained by multiplying a by
A

a real number and the other obtained by multiplying b by another real number .

 =  a +   (where  and  are real numbers)
A b

We say that  has been resolved into two component vectors namely a and 
A b

 a and   a and  respectively. Hence , one can resolve a given vector into two component
b along b

vectors along a set of two vectors  all the three lie in the same plane.

Therefore, we can say , a given vector can be resolved into n components in infinitely many ways.

3.1 Resolution into rectangular component :
It is convenient to resolve a general vector along axes of a rectangular coordinate system using
vectors of unit magnitude, which we call as unit vectors. iˆ,ˆj,kˆ are unit vector along x,y and z-axis

as shown in figure below:
When a vector is split into components which are at right angle to each other then the components
are called rectangular/orthogonal components or projection of that vector.

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VECTOR PHYSICS PART - 1

 
(i) Let vector a  OA in X - Y plane, make angle  with X-axis. Draw perpendiculars AB and AC

from A on the X-axis and Y-axis respectively.

 
(ii) The length OB is called projection of OA on X-axis or component of OA along X-axis and is

represented by ax . Similarly OC is the projection of  on Y-axis and is represented by ay .
OA

    Y (iˆ)
According to law of vector addition a  OA  OB  OC


Thus a has been resolved into two parts, one along OX and

the other OY, which are mutually perpendicular. C A
a sin
OB 
In OAB,  cos   OB  OA cos   ax  a cos  a
OA

AB  sin   AB  OA sin   OC  ay  a sin   (iˆ)
OA O a cos X
B

If ˆi and ˆj denote unit vectors along OX and OY respectively then

 iˆ and  ˆj
OB  a cos  OC  a sin 

So according to rule of vector addition       ax iˆ  a y ˆj  iˆ  a sin  ˆj
OA  OB  OC a a  a cos

Note :

(i) Component of a vector is a vector itself.

(ii) Magnitude of a component(rectangular) cannot be greater than magnitude of the vector.

(iii) Its clear from above equation that a component of a vector can be positive, negative or zero

depending on the value of .

(iv) A vector  can be specified in a plane by two ways :
A

(a) its magnitude A and the direction  it makes with the x-axis; or

(b) its components Ax and Ay. A = Ax2  A2y ,  = tan1 Ay [should be used carefully
Ax

because tan1  1   tan 1  1  ]
 1   1 

(v) If A = Ax Ay = 0 and if A = Ay  Ax = 0 i.e. component/projection of a vector in a
direction perpendicular to itself is always zero.

*(vi) A vector equation is equivalent to 2 or 3 scalar equations i.e  =  +  is equivalent to
C A B

both Cx = Ax + Bx and Cy = Ay + By

3.2 Rectangular components of a vector in three dimensions

Consider a vector a represented by  as shown in figure. Consider O as origin and draw a
OA ,

rectangular parallelopiped with its three edges along the X, Y and Z axes.

  
Vector a is the diagonal of the parallelopiped whose projections on x, y and z axis are ax , ay and az

respectively.

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VECTOR PHYSICS PART - 1

These are the three rectangular components of  . Y
A C
  
Using triangle law of vector addition, OA  OE  EA ay

    A
Using parallelogram law of vector addition, OE  (OB  OD) a
ax
    O B X
 OA  (OB  OD)  EA

      D az E
 EA  OC  OA  OB  OD  OC
Z
Now,      a x iˆ ,   a y ˆj and   azKˆ
OA a , OB OC OD

   axˆi  ay ˆj  azkˆ Also (OA)2  (OE)2  (EA)2
a

but (OE)2  (OB)2  (OD)2

and EA = OC  (OA)2  (OB)2  (OD)2  (OC)2

or a2  a2x  a2y  a2z  a  a 2  a 2  a2z
x y

3.3 Addition and subtraction of vectors in component form

Let   A x i  Ay j  A z k and   Bx i  By j  Bz k be two vectors whose
A B

addition (or subtraction) has to be done. Then the components in the same  
DC

direction will be added and new vector will be obtained. For example, in
 
this case, A  B  (Ax  Bx )i  (Ay  By )j  (Az  Bz )k  
E B


A

Note:

Resolution of a vector makes addition remarkably easy specially while adding more than 2
vectors,in general.

Illustration


Resolve horizontally and vertically a force F  8N which makes an angle of 45º with the horizontal.

Sol. Horizontal component of  is FV
F

FH = F cos 45º = 8 1
2
 
= 4 2N and vertical component of F is FV = F sin 45º F

1 45º
= (8)  2  = 4 2N FH
Illustration

Resolve a weight of 10 N in two directions which are parallel and perpendicular to a slope inclined
at 30º to the horizontal.

Sol. Component perpendicular to the plane W||
W  W cos 300
3 30º
= (10) 2 = 5 3N W

30º W  10 N

and component parallel to the plane Page # 59
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VECTOR PHYSICS PART - 1

1
W|| = W sin 30º = (10)  2 = 5 N

Illustration

Obtain the magnitude of    if   iˆ  ˆj  2kˆ and   2ˆi  ˆj  kˆ .
2A 3B A B

Sol.     2(ˆi  ˆj  2kˆ )  3(2iˆ  ˆj  kˆ )   4ˆi  5ˆj  7kˆ
2A 3B

 
Magnitude of 2A  3B  (4)2  (5)2  (7)2 = 16  25  49  90

Illustration

The magnitude of a vector  is 10 units and it makes an angle of 300 with the
A

X- axis. Find the components of the vector if it lies in the X-Y plane.

Sol. Components of vector  lying in the X-Y plane are Ax  A cos ,Ay  A sin ,Az 0
A

Thus, Ax  10 cos30o  10 3  8.66 units ; Ay = 10 sin30o = 10  1/2 = 5= 5, Az = 0
2

Illustration


Two forces F1  1N and F2  2N act along the lines x = 0 and y = 0 respectively. Then, find the
resultant force.

Sol. x = 0 means y – axis; y = 0 means x – axis; 1N is acting along y – axis and 2N is acting along x –

axis; So, the force   2ˆi  ˆj
F

Illustration

What vector must be added to the summation of vectors ˆi  3ˆj  2kˆ and 3ˆi  6j  7k so that the

resultant vector is a unit vector along the y-axis.

Sol. ˆi  3ˆj  2kˆ  3iˆ  6ˆj  7kˆ  4iˆ  3ˆj  5kˆ

Now, (4  x)ˆi  (3  y)ˆj  (5  z)kˆ  ˆj So, x = – 4, y = – 2, z = 5

and hence the vector is 4i  2j  5kˆ

 Position vector and displacement vector :

A vector which gives the position of an object with reference
to some specified point in a system is called position vector.
Displacement vector refers to the change of position vectors.

  
Thus displacement vector = final position vector - initial position vector or r  r2  r1

Illustration

A body is displaced from position vector   (2iˆ  3ˆj  kˆ ) m to the position vector   (iˆ  ˆj  kˆ ) m.
r1 r2

Find the displacement vector.


Solution : The body is displaced from r1 to r2 . Therefore, displacement of the body is

      (ˆi  ˆj  kˆ )  (2iˆ  3ˆj  kˆ )  (iˆ  2ˆj)m
S r2 r1

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VECTOR PHYSICS PART - 1

DRILL-1

1 A hall has the dimensions 10 m × 12 m × 14 m. A fly starting at one corner ends up at a diametri-
cally opposite corner. The magnitude of its displacement is nearly

2. A carrom board (4m × 4m) has the queen at the centre. The queen hit by the striker moves to the
front edge, rebounds and goes in the hole behind the striking line.

(i) The magnitude of displacement of the queen from the centre to the hole is

(ii) The magnitude of displacement of the queen from the centre to the front edge is

(iii) the the magnitude of displacement of the queen from the front edge to the hole is


3. The vector sum of two vectors A and B is maximum, then the angle q between two vectors is

4. A set of vectors taken in a given order gives a closed polygon. Then the resultant of these vectors is
a

5. The vector sum of two force P and Q is minimum when the angle  between their positive directions,
is

6. Six forces, 9.81 N each, acting at a point are coplanar. If the angles between neighboring forces are
equal, then the resultant is

7. The minimum number of vectors in different planes which can be added to give zero resultant is

  
8. The vectors A and B are such that | A  B | = | A – B | .The angle between the two vectors is :

9.   = |  | = |  |, then the angle between  and  is
If | A + B| A B A B

10. If |  -  | = |  | = |  |, then the angle between  and  is
A B A B A B

11. Given that     and that  is  to  Further   what is the angle between
AB C C
A. if |A||C|, then

 and 
A B

12. The resultant of two vectors A and B is perpendicular to the vector A and its magnitude is

equal to half the magnitude of vector B. The angle between A and B is

13. Given :  =  +  . Also, the magnitude of  ,  and  are 12, 5 and 13 units respectively. The
C A B A B C
 
angle between A and B is

14. An aeroplane is moving in a circular path with a speed 250 km/hr, what is the change in velocity
in half revolution.

15. A car is moving on a straight road due north with a uniform speed of 50 km h–1 when it turns left
through 90º. If the speed remains unchanged after turning, the change in the velocity of the car in
the turning process is

16. The x and y components of a force are 2 N and – 3 N. The force is

17. Given :  = 2 ˆi + 3 ˆj and  = 5 ˆi – 6 ˆj . The magnitude of  is
A B AB

18. A displacement vector, at an angle of 30º with y-axis has a x-component of 10 units. The magnitude

of the vector is :

19. A displacement vector  has a magnitude of 25m and makes an angle of 210º with the x-axis. Its y-
r

component is : ( iˆ and ˆj are unit vectors along x & y axes)

20. Given :  = 2 iˆ – ˆj + 2 kˆ and  = – iˆ – ˆj + kˆ . The unit vector of  –  is
A B A B

21. Vector  is of length 2 cm and is 60º above the x-axis in the first quadrant. Vector  is of length 2
A   B

cm and 60º below the x-axis in the fourth quadrant. The sum A + B is a vector of magnitude

22. The projection of a vector 3ˆi  4kˆ on y-axis is

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VECTOR PHYSICS PART - 1

23. A truck travelling due north at 20 ms–1 turns west and travels with same speed. What is the
change in velocity ?

24. What is the angle between the forces (x + y) and (x – y) if their resultant is 2 (x2  y2 )

25. If a unit vector lies in yz-plane and makes angles of 30° and 60° with the positive y-axis and z-axis
respectively, then its components along the co-ordinate axes will be

26. Given that 0.4ˆi  0.8ˆj  bkˆ is a unit vector. The value of b is.

27. The rectangular components of a vector are 2, 2. The corresponding rectangular components of

another vector are 1, 3 . The angle between the vector is

28. The following four forces act simultaneously on a particle at rest at the origin of the co-ordinate

system

  2iˆ  3ˆj  2kˆ ,   5iˆ  8ˆj  6kˆ
F1 F2

  4iˆ  5ˆj  5kˆ and   3iˆ  4ˆj  7kˆ
F3 F4

The particle will move in

29. A force vector applied on a mass is represented as   6iˆ  8ˆj  10kˆ N and accelerated the mass at
F

1ms–2. The mass of the body is

30. A 1000 N block is placed on an inclined plane with angle of 300. The components of the weight
parallel & perpendicular to the inclined plane , respectively , are

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4. VECTOR PRODUCT

There are two ways in which vectors can be multiplied. They are termed as dot product and cross
product.

4.1 Scalar product or dot product

Dot product of two vectors is defined as the product of magnitude of one of the two vectors & the

magnitude of the rectangular component of the second vector in the direction of the first vector.The
   
dot product of two vectors A andB is the product of the magnitudes of A andB and cosine of the

 
angle between them. Thus A.B  AB cos  . As A, B and cos  all are scalars, so their product is a

scalar quantity. Dot product is also called scalar product.

(a) Geometrical interpretation of scalar product :

  or  A cosB

We have, A.B  AB cos  = A B cos  B


B A cos 

  O  
O A A
B cos 
 
 
(a) A.B  A B cos 
(b) A.B  A cos   B
= AB cos 
= AB cos 

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VECTOR PHYSICS PART - 1

(b) Properties of scalar product :

(i) The scalar product is commutative i.e     
A.B B.A

      

 (ii) The scalar product is distributive over addition i.e A. B  C  A.B  A.C

   
(iii) If A and B are perpendicular to each other, then A.B  AB cos 900  0

   
(iv) If A and Bare parallel having same direction, then A.B  AB cos 00  AB

(v) If   are antiparallel, then   = A B cos 1800  AB .
A andB A.B

 
(vi) The scalar product of two identical vectors A.A  AA cos 00  A2

(vii) If   are mutually perpendicular unit vectors, then iˆ.ˆi  11 cos 00  1
i , j and k

 ˆi.ˆi  ˆj.ˆj  kˆ.kˆ  1

iˆ.ˆj  11 cos 900  0 ˆi.ˆj  ˆj.kˆ  kˆ .ˆi  0

(viii) If   A1ˆi  A 2 ˆj  A3 kˆ and   B 1iˆ  B2 ˆj  B3kˆ , then  
A B A.B  A1B1  A2B2  A3B3 .

(c) Applications of scalar product :

(i) Work done: If a force  causes displacement  , then work done 
F s w  F.s  Fs cos 

   
(ii) Angle between the vectors : For two vectors A and B , we have A .B  AB cos 

  or, If   A1iˆ  A 2 ˆj  A3kˆ and   B1ˆi  B2ˆj  B3kˆ
 cos   A.B A B

AB

then cos  A1B1  A2B2  A3B3

A12  A 2  A 2 B21  B 2  B 2
2 3 2 3

Component or projection of one vector along other vector:

 
(a) Component of vector A along vector B is given by

 
 A.B 
A cos  Bˆ   AB cos   Bˆ =   Bˆ
 B  
B  
B

 
(b) Component of vector B along vector A is given by

     
 A.B  Bcos  A
B cos Aˆ   AB cos   Aˆ = A Aˆ 
 A 

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VECTOR PHYSICS PART - 1

Illustration

Work done by a force  on a body is W    where  is the displacement of body. Given that
F F .S, S

under a force   (2iˆ  3ˆj  4kˆ )N a body is displaced from position vector   (2iˆ  3ˆj  kˆ )m to the
F r1

position vector   (iˆ  ˆj  kˆ )m , find the work done by this force.
r2

Sol. The body is displaced from  to  . Therefore, displacement of the body is
r1 r2

      (ˆi  ˆj  kˆ )  (2ˆi  3ˆj  kˆ ) = (ˆi  2ˆj) m
S r2 r1



Now, work done by the force is W = F S  (2ˆi  3ˆj  4kˆ)  (ˆi  2ˆj) = (2)(-1) + (3)(- 2) = - 8J

Illustration

Prove that the vectors   2iˆ  3ˆj  kˆ and   iˆ  ˆj  kˆ are mutually perpendicular.
A B

Sol.    (2ˆi  3ˆj  kˆ)  (ˆi  ˆj  kˆ ) = (2)(1) + (-3)(1) + (1)(1) = 0 = AB cos 
A B

cos  = 0 or  = 90º 
or the vectors A andB are mutually perpendicular.

Illustration

  
   Let for two vectors A andB , their sum A  B is perpendicular to the difference A  B . Find the

ratio of their magnitudes (A/B).
 
Solution: It is given that (A  B) is perpendicular to (A  B) . Thus,

     
(A  B) . (A  B) = 0 or (A)2  B  A  A  B  (B)2  0

Because of commutative property of dot product :   
A.B  B.A

 A2 – B2 = 0 or A = B

Thus A/B = 1 i.e. the ratio of magnitudes is 1.

Illustration

Find the angle between the vectors 3ˆi  2ˆj  1kˆ and 5iˆ  2ˆj  3kˆ

Sol.  32  22  12  14  52  22  32  38
A B

 = AxBx + AyBy + AzBz =3 x 5 + 2(-2) + (1) (-3) = 8
AB

 8  0.35
cos  A .B  14  38
|A||B|    cos1(0.35)

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DRILL-2

1. If a ,  c are mutually perpendicular vectors of equal magnitudes, then the angle between the
b,

vectors a and a    c is
b

2. (a.ˆi) ˆi  (a.ˆj) ˆj  (a.kˆ ) kˆ 

3. Find the angle between two vectors   2ˆi  ˆj  kˆ and Bˆ  ˆi  kˆ .
A

4. If the vectors 4iˆ  ˆj  3kˆ and 2mˆi  6mˆj  kˆ are mutually perpendicular, find the value of m.

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VECTOR PHYSICS PART - 1

5. What is the angle between ˆi  ˆj  kˆ and ˆi .

*6. Find the magnitude of component of 3iˆ  2ˆj  kˆ along the vector 12iˆ  3ˆj  4kˆ .

7. Find the angle between the vector iˆ  ˆj and ˆi  ˆj .

8. Let   iˆA cos ˆjA sin  be any vector. Another vector  which is normal to A is
A B

9. If a vector  making angles , and  respectively wih the X, Y and Z axes respectively. Then
P

sin2   sin2   sin2  

 
 10. If for two vector A and B , sum (A + B) is perpendicular to the difference A  B . The ratio of their

magnitude is

11. If   2iˆ  3ˆj  kˆ and   ˆi  3ˆj  4kˆ then projection of  on  will be
A B A B

12. The angle between two vector given by 6iˆ  6ˆj  3kˆ and 7ˆi  4ˆj  4kˆ is

13. The angle between two vectors –2iˆ  3ˆj  kˆ and 2iˆ  2ˆj – kˆ is



*********************

4.2 Vector product or cross product

The vector product of two vectors is defined as the vector whose magnitude is equal to the product
of the magnitudes of two vectors and sine of angle between them and whose direction is perpendicular

to the plane of two vectors and is given by right hand thumb rule. Mathematically, if  is the angle
   
between A and B , then A  B  AB sin  nˆ where nˆ is a unit vector perpendicular to the plane of

  direction of unit vector nˆ is given by the right hand thumb rule. The fingers of
vector A and B .The
right hand should be oriented along the direction of  that they curl towards  . The direction
A such B

of outstretched thumb gives direction of nˆ

nˆ   nˆ  
AB AB

  
 B 
A A

  B
BA

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VECTOR PHYSICS PART - 1
(a) Geometrical interpretation of vector product : R

S QT


Bsin  B

  P
O A

Suppose two vectors  and  are represented by the sides OP and OQ of a parallelogram, as
A B
 
shown in figure. The magnitude of vector product AB is

 
|A  B| = AB sin  = A (B sin  ) = area of rectangle OPTS = area of parallelogram OPRQ

Thus the magnitude of the vector product of two vectors is equal to the area of the parallelogram
(OPRQ) formed by the two vectors as its adjacent sides.

= 2  area of triangle OPQ.

 Area of triangle OPQ = 1 (area of parallelogram OPRQ) = 1|A  
2 2 B|.

(b) Properties of vector product:

(i) Vector product is not commutative. It is anticommutative i.e.,    
AB = (B  A) .

      
(ii) Vector product is distributive over addition i.e., A  (B  C)  A  B  A  C .

(iii) Vector product of two parallel or antiparallel vectors is zero i.e
 

if   00 or   A  B  (AB sin )nˆ  0

 

So,Vector product of two identical vectors is also zero i.e A  A  (AA sin 0o )nˆ  0 .

(v) The magnitude of vector product of two mutually perpendicular vectors is equal to the product

of their magnitude.
 

|A  B| AB sin 90o = AB.

(vi) For unit vectors ˆi, ˆj and kˆ  
ˆi  ˆi  (1)(1)sin 0o nˆ  0 B
iˆ  ˆi  ˆj  ˆj  kˆ  kˆ  0
A cos 
and ˆi  ˆj  (1)(1)sin 90o kˆ = kˆ

Similarly, ˆj  kˆ  ˆi and kˆ  iˆ  ˆj   
ˆj  iˆ  kˆ , kˆ  ˆj  iˆ, iˆ  kˆ  ˆj A

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VECTOR PHYSICS PART - 1

(vii) Cross product can be used to find angle between two vectors.
 

According to definition of vector product of two vectors A  B  AB sin nˆ

 
   AB 
So, A  B  AB sin    sin1    
i.e.  A B 
 

However, dot product method is more convenient to find angle between vectors.

To find the cross product of two vectors, the following methods can be used:

Method 1:

    (A x iˆ  Ay ˆj  Azkˆ )  ( Bx ˆi  B y ˆj  Bzkˆ )
A B

 AxBx(ˆi  iˆ)  AyBx(ˆj  iˆ)  AzBx(kˆ  iˆ)
 AxBy(iˆ  ˆj)  AyBy (ˆj  ˆj)  AzBy(kˆ  ˆj)
 AxBz (iˆ  kˆ )  AyBz (ˆj  kˆ )  AzBz (kˆ  kˆ )

[As ˆi  iˆ  0, ˆj  ˆj  0, kˆ  kˆ  0 and iˆ  ˆj  kˆ , ˆj  iˆ  kˆ ,kˆ  iˆ  ˆj,iˆ  kˆ  ˆj,kˆ  ˆj  iˆ]

So, we have

    Ay Bx (kˆ )  AzBx ˆj  A x B y kˆ  AzBy (iˆ)  AxBz (ˆj)  AyBz (iˆ)
A B

 (AyBz  AzBy )iˆ  (AxBz  AzBx )ˆj  (AxBy  AyBx )kˆ

Method 2: Determinant method

 
Cross product of two vectors A and B can be obtained easily by using the determinant method.

   (A x iˆ  A y ˆj  Azkˆ ) (Bxiˆ  B y ˆj  Bzkˆ )  ˆi ˆj kˆ Az
A B Ax Ay

Bx By Bz

iˆ ˆj kˆ iˆ ˆj kˆ iˆ ˆj kˆ
Ax Ay Az Ax Ay
Az Ax Ay Az
= By  By
+
Bx B z Bx
Bz Bx By Bz

Taking down arrow product with positive sign and upward arrow product with negative sign.

 iˆ ˆj kˆ Ay Az Ax Ay
AB By Bz Bx By
i.e  = Ax Ay Az = ˆi  ˆj Ax Az  kˆ
AB = Bx By Bz Bx Bz

iˆ ( A y Bz  Az By )  ˆj ( Ax Bz  Az Bx )  kˆ ( Ax By  Ay Bx )

Note: Don’t miss the -ve sign with ˆj in contrast to iˆ & kˆ .

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VECTOR PHYSICS PART - 1

Illustration:

If   3ˆi  ˆj  2kˆ and  2ˆi  2ˆj  4kˆ
A B

 
(i) Find the magnitude of A  B


(ii) Find a unit vector perpendicular to both A and B

(iii) Find the cosine and sine of the angle between the vectors  and 
A B

Sol.   ˆi ˆj kˆ  8iˆ  8ˆj  8kˆ    (8)2  (8)2  (8)2  8 3
(i) A  B  3 1 2 |A  B| 

2 2 4

(ii) nˆ     8iˆ  8ˆj  8kˆ  1 (iˆ  ˆj  kˆ )
A  B
|A  B| 8 3 3

  1 (iˆ  ˆj  kˆ )
There are two unit vectors perpendicular to both A and B , they are nˆ  
3
 
AB
(iii) sin   AB  8 3  2  cos  3 [calculate the same using dot product]
14 24 7 7

Illustration

Let a force  be acting on a body free to rotate about a point O and let r be the position vector of
F

any point P on the line of action of the force. Then torque of this force about point O is defined as
 
  Given, F  (2iˆ  3ˆj  kˆ )N and r  (iˆ  ˆj  6kˆ )m ,find the torque of this force.
: rF

   ˆi ˆj kˆ
Solution:   r  F  1 1 6  iˆ(1  18)  ˆj( 1  12)  kˆ(3  2)  17iˆ  11ˆj  5kˆ

2 3 1

Illustration

What is the value of linear velocity, if w  3ˆi  4ˆj  kˆ and r  5iˆ  6ˆj  6kˆ

(a) 6iˆ  2ˆj  3kˆ (b) –18iˆ – 13ˆj  2kˆ (c) 4iˆ – 13ˆj  6kˆ (d) 6ˆi – 2ˆj  8kˆ

Sol. Angular velocity    3ˆi  4ˆj  kˆ and distance from centre  r  5iˆ  6ˆj  6kˆ .

  r iˆ ˆj kˆ
 
We know that linear velocity ( ) = × = 3 –4 1 = – 18 ˆi – 13 ˆj + 2 kˆ

5 –6 6

Illustration

Find the area of the parallelogram whose diagonals are represented by 3iˆ  ˆj  kˆ and ˆi  ˆj  kˆ .

Sol. Let the sides of the parallelogram be  and  a    3ˆi  ˆj  kˆ and a    iˆ  ˆj  kˆ
a b . Then b b

We have (a    (a    2 (a  
b) b) b)

Now, a     1 ˆi ˆj kˆ =  1 (0iˆ  4ˆj  4kˆ )  0iˆ  2ˆj  2kˆ  |a   22  22  2 2
b 3 1 1 2 b|
2
1 1 1

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VECTOR PHYSICS PART - 1

Illustration

Two vectors,   2iˆ  2ˆj  pkˆ and   iˆ  ˆj  kˆ are given. Find the value of p if (i) the two vectors are
A B

perpendicular (ii) the two vectors are parallel.

Sol. (i)  and  will be perpendicular if  = 0
A B AB

Now  = AxBx + AyBy + AzBz =2x1+2x1+px1 =4+p
AB

For AB = 0, we must have 0 = 4 + p or p = –4

(ii)    = 0 ;
A and B will be parallel if AB

  iˆ ˆj kˆ
AB=
Now 2 2 p = iˆ(2  p)  ˆj(p  2)  kˆ(2  2) = ˆi(2  p)  ˆj(p  2)

111

For    = 0, we must have each component to be zero. That is 0 = 2 – p, and 0 = p – 2
A B

(both conditions are similar). Thus p = 2.

*********************

DRILL-3

1. The torque of a force    3iˆ  ˆj  5kˆ acting at the point   7ˆi  3ˆj  kˆ is:
F r

2.  is directed vertically downwards and  is directed along the north.What is the direction of
A B

BA

3. Given :  is the angle between  and  . Then |Aˆ  Bˆ | is equal to
A B

4. The magnitude of the vector product of two vectors is 3 times their scalar product. The angle

between vectors is

5.   then angle between vector  and  is.
If |A B| = |A .B| A B

6. Find a unit vector perpendicular to both   2iˆ  ˆj  iˆ  2ˆj .
A and B 

    
P Q R R P. Q
 7. iˆ 4ˆj, 3ˆj iˆ
Three vector are given as    and  2 . Calculate the value of .

8. Find the area of parallelogram formed by two vectors  iˆ  2ˆj  kˆ and   ˆi  5ˆj  6kˆ respectively.
A B

9. Find a unit vector perpendicular to a  ˆi  2ˆj  kˆ and   –3iˆ  4ˆj  kˆ both.
b

  
10. The linear velocity of a rotating body is given by v=×r .

If =iˆ-2ˆj+2kˆ and r=4ˆj-3kˆ , then what is the magnitude of  ?
v

11. Vector  makes equal angles with x, y and z axis. Value of its compoents (in terms of magnitude
A

of its components (in terms of magnitude of A ) will be

 

   12. The angle between vector A  B and B  A us

  
13. If A = 5 units, B = 6 units and A  B  15 units, then what is the angle between A and B ?

14. The area of a parallelogram whose diagonals are p  2iˆ  3ˆj and   iˆ  4ˆj is
Q

****************************************************************************

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VECTOR PHYSICS PART - 1

EXERCISE

SINGLE CORRECT TYPE QUESTIONS

1. Two forces of 12 N and 8 N act upon a body. The resultant force on the body has maximum value
of

(a) 4 N (b) 0 N (c) 20 N (d) 8 N

2. A force of 5 N acts on a particle along a direction making an angle of 60° with vertical. Its vertical
component be

(a) 10 N (b) 3 N (c) 4 N (d) 2.5 N

3. Five equal forces of 10 N each are applied at one point and all are lying in one plane. If the angles
between them are equal, the resultant force will be

(a) Zero (b) 10 N (c) 20 N (d) 10 2N

4. The angle made by the vector A  iˆ  ˆj with x- axis is

(a) 90° (b) 45° (c) 22.5° (d) 30°

5. Two forces, each of magnitude F have a resultant of the same magnitude F. The angle between the
two forces is

(a) 45° (b) 120° (c) 150° (d) 60°

6. Three concurrent forces of the same magnitude are in equilibrium. What is the angle between the
forces ? Also name the triangle formed by the forces as sides

(a) 60° equilateral triangle (b) 120° equilateral triangle

(c) 120°, 30°, 30° an isosceles triangle (d) 120° an obtuse angled triangle

7. Two force, each equal to F, act as shown in the following Fig. Their resultant is :

(a) F/2 (b) F (c) 3 F (d) 5 F

8. Two vectors acting in the opposite directions have a resultant of 10 units. If they act at right angles
to each other, then the resultant is 50 units. Calculate the magnitude of two vectors

(a) P  20,Q  30 (b) P  30,Q  40 (c) P  50,Q  70 (d) P  40, Q  20

9. The sum of the magnitudes of two force acting at a point is 16 N. The resultant of these force is
perpendicular to the smaller force and has magnitude of 8N. If the smaller force is of magnitude x,

then the value of x is

(a) 2N (b) 4N (c) 6 N (d) 8N

10.  
A vector A which has magnitude of 6 is added to a vector B which is long x-axis. The resultant of
tmwaognvietcutdoersofA Bainsd 
 lies along y-axis and has magnitude twice that
B of the magnitude of B . The

(a) 2 6 (c) 3 (d) 8
6 (b) 5 6

   
11. If |V1  V2||V1  V2| and V2 is finite, then

(a) V1 is parallel to V2  
(b) V1  V2
(c) V1 and V2 are mutually perpendicular
 
(d) |V1||V2|

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VECTOR PHYSICS PART - 1

12. Two equal forces (P each) act at a point inclined to each other at an angle of 120°. The magnitude
of their resultant is

(a) P/2 (b) P/4 (c) P (d) 2P

13. The vectors 5i  8j and 2i  7 j are added. The magnitude of the sum of these vector is

(a) 274 (b) 38 (c) 238 (d) 560

14.  are such that    . Then
Two vectors A and B A BA B (d) B  0

 (b)  (c) 
(a) A.B  0 AB0 A0

15. The unit vector in the direction of vector A  5iˆ  12ˆj , is

(a) ˆi (b) ˆj  (c) iˆ  ˆj /13  (d) 5iˆ  12ˆj /13

16. How many minimum number of coplanar vectors having different magnitudes can be added to
give zero resultant

(a) 2 (b) 3 (c) 4 (d) 5

17. 100 coplanar forces each equal to 10 N act on a body. Each force makes angle  / 50 with the

preceding force. What is the resultant of the forces

(a) 1000 N (b) 500 N (c) 250 N (d) Zero

18. The expression  1 iˆ  1 ˆj  is a
 2 2 
 

(a) Unit vector (b) Null vector

(c) Vector of magnitude 2 (d) None

19. Given vector   2iˆ  3ˆj, the angle between  and y-axis is
A A

(a) tan1 3 / 2 (b) tan1 2 / 3 (c) sin1 2 / 3 (d) cos1 2 / 3

20. If   4ˆi  3ˆj and   6iˆ  8ˆj then magnitude and direction of   be
A B A  B will

(a) 5, tan1(3 / 4) (b) 5 5 , tan1(1 / 2) (c) 10, tan1(5) (d) 25, tan1(3 / 4)

21. A body is at rest under the action of three forces, two of which are   4iˆ ,   6 ˆj , the third force is
F1 F2

(a) 4iˆ  6ˆj (b) 4iˆ  6ˆj (c) 4iˆ  6ˆj (d) 4iˆ  6ˆj

22. The maximum and minimum magnitude of the resultant of two given vectors are 17 units and 7
unit respectively. If these two vectors are at right angles to each other, the magnitude of their
resultant is

(a) 14 (b) 16 (c) 18 (d) 13

23. The vector sum of two forces is perpendicular to their vector differences. In that case, the forces

(a) Are equal to each other in magnitude (b) Are not equal to each other in magnitude

(c) Cannot be predicted (d) Are equal to each other

24. If the sum of two unit vector is a unit vector, then the magnitude of their difference is

(a) 3 (b) 2 (c) 5 1
(d) 2

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VECTOR PHYSICS PART - 1

25. When two vectors of magnitude P and Q are inclined at an angle  , the magnitude of their result-
ant is 2P. When the inclination is changed to (180  ) the magnitude of the resultant is halved. Find
the ratio of P to Q.

(a) 1, 2 (b) 5 , 7 (c) 2 , 3 (d) 7 , 2

26. The angle  between the vector p  iˆ  ˆj  kˆ and unit vector along X-axis is

(a) cos 1  1 (b) cos1  1 (c) cos1  3 (d) cos1  1 
     2   2 
 3   2  

27. Two forces 3N and 2 N are at an angle  such that the resultant is R. The first force is now increased

to 6N and the resultant become 2R. The value of  is

(a) 30° (b) 60° (c) 90° (d) 120°

28. Consider a vector  4iˆ  3ˆj. Another vector that is perpendicular to  is
F F

(a) 4iˆ  3ˆj (b) 6iˆ (c) 7kˆ (d) 3ˆi  4ˆj

29. Which of the sets given below may represent the magnitudes of vectors adding to zero?

(a) 3, 2, 4, 8 (b) 1, 4, 8, 16 (c) 1, 2, 4 (d) 0.5, 1, 2

30. Which of the following is incorrect.

(a) A vector necessarily change when rotated by some angle.

(b) sum of two unit vectors is always unit vector

(c) 3m magnitude in north is greater than 2 m magnitude in East.

(d) Magnitude of component of a vector can never be greater than magnitude of the vector.

31. The resultant of   is  On reversing the vector  the resultant becomes  What is the
AB R1. B, R2.

value of R 12  R 2
2

(a) A2 +B2 (b) A2 - B2 (c) 2(A2  B2 ) (d) 2(A2  B2 )

32. The unit vector parallel to the resultant of the vectors   4ˆi  3ˆj  6kˆ and   ˆi  3ˆj  8kˆ is
A B

(a) 1 (3ˆi  6ˆj  2kˆ) (b) 1 (3ˆi  6ˆj  2kˆ) (c) 1 (3ˆi  6ˆj  2kˆ) (d) 1 (3ˆi  6ˆj  2kˆ)
7 7 49 49

33. Position of a particle in a rectangular-co-ordinate system is (3, 2, 5). Then its position vector will be

(a) 3iˆ  5ˆj  2kˆ (b) 3iˆ  2ˆj  5kˆ (c) 5iˆ  3ˆj  2kˆ (d) None of these

34. If a particle moves from point P (2,3,5) to point Q (3,4,5). Its displacement vector be

(a) iˆ  ˆj  10kˆ (b) ˆi  ˆj  5kˆ (c) iˆ  ˆj (d) 2iˆ  4ˆj  6kˆ

35. If A  3iˆ  4ˆj and B  7iˆ  24ˆj, the vector having the same magnitude as B and parallel to A is

(a) 5ˆi  20ˆj (b) 15iˆ  10ˆj (c) 20ˆi  15ˆj (d) 15iˆ  20ˆj

36. VofecAto)rwAillmbaekes equal angles with x, y and z axis. Value of its components (in terms of magnitude

A A (c) 3 A (d) 3
(a) 3 (b) 2 A

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VECTOR PHYSICS PART - 1

37. A vector is represented by 3ˆi  ˆj  2 kˆ .Length of its projection in XY plane is

(a) 2 (b) 14 (c) 10 (d) 5

38. With respect to a rectangular cartesian coordinate system, three vectors are expressed as a  4iˆ  ˆj ,

  3iˆ  2ˆj and c  kˆ where ˆi,ˆj,kˆ are unit vectors, along the X, Y and Z-axis respectively. The
b

unit vectors ˆr along the direction of sum of these vector is

(a) rˆ  1 (ˆi  ˆj  kˆ ) (b) rˆ  1 (i  ˆj  kˆ ) (c) rˆ  1 (ˆi  ˆj  kˆ ) (d) rˆ  1 (iˆ  ˆj  kˆ )
3 3 3 2

39.   2iˆ  ˆj, B  3ˆj  kˆ and   6iˆ  2kˆ . Value of    would be
A C A  2B  3C

(a) 20iˆ  5ˆj  4kˆ (b) 20ˆi  5ˆj  4kˆ (c) 4iˆ  5ˆj  20kˆ (d) 5ˆi  4ˆj  10kˆ

40. The sum of magnitudes of two forces acting at a point is 16 N. If their resultant is normal to the
smaller force and has a magnitude of 8 N, the forces are :

(a) 6N, 10N (b) 8N, 8N (c) 4N, 12N (d) 2N, 14N

41. The component of a vector is :

(a) always less than its magnitude (b) always greater than its magnitude

(c) always equal to its magnitude (d) none of these

42. A displacement vector, at an angle of 30º with y-axis has a x-component of 10 units. The magnitude
of the vector is

(a) 5.0 (b) 10 (c) 11.5 (d) 20


43. A displacement vector r has a magnitude of 25m and makes an angle of 210º with the x-axis. Its y-

component is : ( ˆi and ˆj are unit vectors along x & y axes)

(a) 21.7 (b) –21.7 ˆj (c) 12.5 ˆj (d) –12.5 ˆj

44. Resultant of which of the following may be equal to zero?

(a) 10N; 10N; 30N (b) 10N; 20N; 30N; 40N

(c) 5N; 10N; 20N; 40N (d) none of these

45. The resultant of three vectors of magnitudes 1, 2 and 3 units and whose directions are along the
sides of an equilateral triangle taken in the same order is

(a) 1unit (b) 6 units (c) 3 units (d) 14 units

46. In the above equation the resultant vector makes an angle of :

(a) 210º with the first vector (b) 30º with the first vector

(c) 60º with the first vector (d) 120º with the second vector

47. A body is constrained to move only in Y-direction under the action of a force   2iˆ  15ˆj  6kˆ N.
F

If the body moves a distance of 10m then the work done is :

(a) 150J (b) 190J (c) 163J (d) 185J

48. If   i  j &   i  j , then a vector  perpendicular to both   and having a magnitude equal
A B C A& B

to 3 is

(a) 3kˆ  (b) 3 iˆ  ˆj  (c) 3 iˆ  2kˆ  (d) 3 ˆi  ˆj

49. A vector  points vertically downward &  points towards east, then the vector product  is
A B AB

(a) along west (b) along east (c) along north (d) along south

IIT ASHRAM UG–1 & 2, Concorde Complex, Above OBC Bank. R.C. Dutt Road., Alkapuri Vadodara. 390007 Page # 73

VECTOR PHYSICS PART - 1

50. Five equal forces of 10N each are applied at one point and all are lying in one plane. If the angles
between any two adjacent forces are equal, the resultant of these forces will be

(a) zero (b) 10N (c) 20N (d) 10 2 N

51. Forces proportional to AB, BC and 2CA act along the sides of triangle ABC in order. Their resultant
represented in magnitude and direction as

(a) CA (b) AC (c) BC (d) CB

52. The magnitude of resultant of two forces of magnitudes 3P and 2P is R. If the first force is doubled,

the magnitude of the resultant is also doubled. The angle between the two forces is

(a) 30o (b) 60o (c) 120o (d) 150o

53. The resultant of two vectors  and  is perpendicular to the vector  and its magnitude is equal to
u v u
  
half of the magnitude of vector v , The angle between u and v in degrees is

(a) 120 (b) 60 (c) 90 (d) 150

54. A force of 6 N and another of 8 N can be applied to produce the effect of a single force equal to

(a) 1 N (b) 10 N (c) 16 N (d) 0 N

QUESTIONS ASKED IN PREVIOUS AIEEE / JEE MAIN

1. Two forces P and Q of magnitude 2F and 3F, respectively, are at an angle  with each other. If the

force Q is doubled, then their resultant also gets doubled. Then, the angle is :[JEE MAINS 2019]

(a) 30º (b) 60º (c) 90º (d) 120º

 
 2. Two vectors A andB have equal magnitudes. The magnitude of A  B is 'n' times the magnitude

   
of A  B The angle between A andB . [JEE MAINS 2019]

(a) sin1  n2  1  (b) cos1  n  1  (c) cos1  n2  1 (d) sin1  n  1 
 n2  1   n  1   n2    n  1 
   1 

3. In the cube of side 'a' shown in the figure, the vector from the central point of the face ABODto the

central point of the face BEFO will be: [JEE MAINS 2019]

 (a)1a 1 a ˆj  iˆ 1 a kˆ  iˆ  (d)1a
2 2 2 2
iˆ  kˆ  (b)  (c) ˆj  kˆ

4.  is equal to magnitude of  . Find the angle between
Magnitude of resultant of two vectors P and Q P

 
Q and resultant of 2P andQ . [JEE MAINS 2020]

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IIT ASHRAM UG–1 & 2, Concorde Complex, Above OBC Bank. R.C. Dutt Road., Alkapuri Vadodara. 390007 Page # 74

VECTOR PHYSICS PART - 1

ANSWER KEY

DRILL-1

1. 21 m 2. 2 10m (iii) 4 10m 3. 0º
(i) 2 2m (ii) 3 3

4. Null vector 5.  6. 0 N 7. 4

8. 90º 9. 120º 10. 60º 11. 3 radian
4

12. 150º  14. 500 km/hr
13. 2 16. 2iˆ  3ˆj

15. 50 2km h1S  W direction 3iˆ  kˆ 17. 58units
20. 10
18. 20 19. –12.5ˆj 21. 2 along + x-axis
25. 0, 3 , 1
22. zero 23. 20 2ms01south west 24. 90º
22
26. 0.2 27. 15º 28. YZ plane 29. 10 2kg

30. 500N, 500 3N DRILL-2
3.  = 30º
1. cos1(1 / 3 ) 2.  4. m  3/14
a

5. θ  cos1  1  6. 2 7. 90º 8. ˆiB sin   jB cos 
 3 
 3
11. 26
sin 1  5
9. 2 10. 1 DRILL-3 12.  
13. obtuse 3. sin   3 
7. Not possible
1. 14ˆi  38ˆj  16kˆ 2. west 4. /3
A 8. 347 m
5. 45° 6. kˆ or - kˆ 11. 3
12. 
9.   3ˆi  2ˆj  kˆ 10. 29 units

14

13. 30º 14. 2.5 square units

SINGLE CORRECT TYPE QUESTIONS

1. c 2. d 3. b 4. b 5. b 6. b

7. b 8. b 9. c 10. b 11. c 12. c

13. a 14. d 15. d 16. b 17. d 18. a

19. b 20. b 21. d 22. d 23. a 24. a

25. c 26. a 27. d 28. c 29. a 30. b

31. c 32. a 33. b 34. c 35. d 36. a
37. c 38. a 39. b 40. a 41. d 42. d

43. d 44. b 45. c 46. a 47. a 48. a

49. d 50. a 51. a 52. c 53. d 54. b

QUESTIONS ASKED IN PREVIOUS AIEEE / JEE MAIN

1. d 2. c 3. b 4. 90

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IIT ASHRAM UG–1 & 2, Concorde Complex, Above OBC Bank. R.C. Dutt Road., Alkapuri Vadodara. 390007 Page # 75