Vectors Rg-Phy.

XI-2020-22
PREFACE
Physics is a science based on experimental observations and mathematical analysis. It is
possible to explain the behavior of various physical systems using relatively few fundamental
laws and physical quantities. These chapters in this module are common for both JEE Main &
Advanced. Here we deal with the categorizations of these physical quantities into two. i.e. one
with direction and the other without direction and calculus. Calculus is a mathematical tool
which helps us understand and calculate instantaneous values of a physical quantity. Once a
teacher was teaching divisions to elementary school students without the help of a board &
chalk, he asked the students to distribute ten fruits equally among a group of ten students.
Each student was left with one fruit. After making the students practice this with several
numbers, he generalised this concept & said if ‘the number of students equals to the number of
fruits, every student will get one fruit’. A boy got up and asked the teacher, is this true for all
numbers? For which the teacher said ‘yes’. He then asked the teacher if there are no students
and no fruits even then will each student get one fruit?. This boys name is ‘Srinivas
Ramanujam’ who later showed to the world that (0/0) is a singularity and cannot be
determined.
This booklet consists of summarized text coupled with sufficient number of solved examples of
varying difficulties, which enables the students to develop problem solving ability along with
emphasis on physical concept.
The end–of–chapter problems are categorized into four section, namely Exercise – I (objectives
where only one of the option is correct) Which emphasizes on JEE (main) pattern, Exercise – II
(objectives where more than one option may be correct), Exercise – III (matrix matches and
paragraph type questions), Exercise – IV (subjective questions), to help the student assess his
understanding of the concept and further improvise on his problem solving skills. Solutions to
all the questions in the booklet are available and will be provided to the students (at the
discretion of the professor). Every possible attempt has been made to make the booklet
flawless. Any suggestions for the improvement of the booklet would be gratefully accepted and
acknowledged.
(Dept. of Physics)
IIT –ian’s PACE

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE
 Vectors Rg-Phy. XI-2020-22
IIT–JEE SYLLABUS

Scalars and vectors, vector addition and Subtraction, Zero vector, Scalar and vector products, Unit vector,
Resolution of a Vector.

CONTENTS


No. Section Page No.



1. THEORY 1 – 16

2. JEE MAIN EXERCISE 17 – 19

3. EXERCISE # I 19 – 21

4. EXERCISE # II 22 – 23

5. EXERCISE # III 23 – 23

6. EXERCISE # IV 24 – 25

7. ANSWER KEY 26 – 26 




Vector Analysis : Tentative Lecture Flow
(Board Syllabus & Booklet Discussion Included)
Lecture 1 Definition of Scalar & Vector. Representation of Vector, Geometrical representation,
Cartisian Co - ordinate system. Position vector, unit vector, coplanar & collinear vector
Addition & Subtraction of vectors, triangle law, parallelogram law, resolution of vector,
Lecture 2 polygon law, solved examples involving addition & subtraction of vector.

Lecture 3 Dot product and cross product. Solved examples.



COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE
 Vectors Rg-Phy. XI-2020-22

VECTORS
INTRODUCTION
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.

Vector Quantity:
A physical quantity which requires magnitude and a particular direction, when it is expressed is
known as vector quantity
eg: displacement, velocity, acceleration, force etc.
A vector is represented by a line headed with an arrow. Its length is
Q
proportional to its magnitude, with respect to a suitably chosen scale. 
   A
A is a vector and A=PQ
 
Magnitude of A =|A|or A 
P
Modulus of a vector:
The modulus of a vector means the length of the vector. It is always positive and has no direction.
 
Modulus of vector A is represented as |A| or A.

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.

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

Zero vector or null vector

A vector whose magnitude is zero and has any arbitrary direction is called zero vector. It is represented

by 0 . The need of a zero arises in the situations:
      
(i) If A  B, then A  B  0 (ii) If   , then      A  0

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 1
 Vectors Rg-Phy. XI-2020-22
Unit vector: j

A vector whose magnitude is one unit is called unit vector. A unit


vector in the direction of vector A is represented by  , and
 i
A
is given by
A
 k
Any vector can be expressed as A  AA ˆ.

ˆi, ˆj and kˆ are unit vectors along x, y and z-axes.

Equal vectors:
Two vectors are said to be equal if they have same magnitude and same direction.

A
  
B AB

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

Multiplication of a vector by a scalar :


When a vector is multiplied by a scalar  , we get a new vector which is  times the vector A i.e.
 
A. The direction of resulting vector is that of A .

If  has negative value, then we get a vector whose direction is opposite of A . 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 isobtained by multiplying units of mass and velocity.
 
A A
 
2A   2 2A    2 

 A
Similarly, we can have vector A divided by a scalar  . The resulting vector becomes .


1  
The magnitude of the new vector becomes that of A and direction is same as that of A .

 
A A / 2    2

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 2
 Vectors Rg-Phy. XI-2020-22
Collinear or parallel vectors :
The vectors which act along the same line or along a parallel line are called collinear vectors.

A
  
B A B

 a  Like or parallel vectors. 

  A
B A

B
 b  Unlike or antiparallel vectors.
   
Note: If A and B be two collinear vectors, then there exists a scalar k such that B  kA , the absolute
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 plane are called coplanar vectors. In figure,
  
A, B and C are coplanar and concurrent vectors. 
A

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

Note: Division of vector by a vector is not defined.

The addition or subtraction of vectors can be done by following two methods:
(i) Analytical method (ii) Geometrical method

ADDITION OF TWO VECTORS

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, then their resultant is represented by third side of the triangle taken in
 
the opposite order. 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
   
 O 
A 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, then their resultant is represented by the diagonal of the parallelogram
 
passing though the point of intersection of the vectors. Suppose two vectors A and B are as shown
in figure
COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 3
 Vectors Rg-Phy. XI-2020-22


B



A

     
   
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
 +B
B R=
A

 
O  P
A
 
Analytical method: Finding A  B:


R  Bsin 
B

 
O 
A P S
A B cos 

 
It is clear from the geometry of the figure that resultant of A and B is equal to the resultant of
  
 
A  B cos  and Bsin  . By Pythagoras theorem, we have

2 2 22 2 2 2 2
 R   A  B cos     B sin    A  B cos   2AB cos   B sin 

or R  A 2  B2  2AB cos 

 
If  is the angle which resultant R makes with A , then:

Bsin 
tan  
A  Bcos 

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 4
 Vectors Rg-Phy. XI-2020-22
Special cases :
(i) For  00; Rmax  A2  B2  2AB  A B

(ii) For   1800 ; R min  A 2  B2  2AB  A  B

Thus, A  B  R  A  B

(iii) If A  B, R  A 2  A 2  2AA cos 

= 2A 2 1  cos   = 2A 2  2 cos2  / 2

= 2A cos  / 2 and  
2
 
Illustration 1: Find A  B in the diagram shown. Given A = 4 units and B = 3 units and angle
between them is 60 degrees.


Solution: R = A 2  B2  2ABcos  B   
R AB
= 16  9  2.4.3.cos 60 0

= 37

Bsin 3sin600
 
tan     A
A+Bcos 4+3cos 600
 0.472
  tan 1 (0.472)  25.30
  
Thus, the magnitude of resultant of A and B is 37 units at angle 25.3º from A in the direction
shown in figure.

Illustration2 : 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.
Solution: Given R =A =B. ; Using R 2  A 2  B2  2AB cos 
1
A 2  A 2  A 2  2AA cos  ;  cos       120o
2
SUBTRACTION OF TWO VECTORS
      
   
The subtraction of B PQ from A OP means addition of -B PQ to A  
   
A  B = A  B  
   
   
First draw vector A OP in the given direction. Then draw vector  B PQ starting from head of
     
 
the vector A . Then close the triangle. R OQ will be equal to A - B . The angle between A and
 o
 B will be equal to (180 ) .

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 5
 Vectors Rg-Phy. XI-2020-22
Therefore, the resultant will be given by:
Q
R  A2  B2  2ABcos1800  2 2
 A  B  2ABcos 

B
Bsin 1800    Bsin  
and tan    P
A  Bcos 180   0
A  B cos  A 
O

 
 is the angle made by R with A (180o  )


R
   

=A
0 B
Note: For   90 , A  B  A  B

B
ADDITION OF MORE THAN TWO VECTORS Q'
   
To find the resultant R  A  B  C , the polygon law of addition is used.
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 resultant is represented by the closing side of the polygon taken in opposite order. Here in the
   
figure, R (closing side of polygon) represents the resultant of vectors A, B and C


C
2

B

1

A


C
 1   2 

2

C
    2
 R  ABC
B 
B

1  A 1

A b 
a 
 
D C
Note: If n number of vectors makes a closed polygon, their resultant will
    
be zero. Here vectors A, B, C, D and E make closed polygon.  
      E B
A  B  C  D  E  0 
A

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 6
 Vectors Rg-Phy. XI-2020-22
Inchapter Exercise
1. At what angle should the two force vectors 2F and 2F act so that the resultant force is 10F ?
2. Two forces while acting on a particle in opposite directions, have the resultant of 10N. If they
act at right angles to each other, the resultant is found to be 50 N. Find the two forces.

Answer key
1.   450 2. 30 N, 40 N

RESOLUTION OF VECTORS INTO RECTANGULAR COMPONENTS

When a vector is split into components which are at right angle to each other then the components are
called rectangular or orthogonal components of that vectors.
 
(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 a x . Similarly OC is the projection of OA on Y-axis and is represented by a y .
   
According to law of vector addition a  OA  OB  OC
 Y (iˆ)
Thus a has been resolved into two parts, one along OX and
the other OY, which are mutually perpendicular.
C A
OB 
In OAB,  cos   OB  OA cos   a x  a cos 
OA a sin 
a

AB
 sin   AB  OA sin   OC  a y  a sin   (iˆ)
OA
O a cos 
B X
If î and ĵ denote unit vectors along OX and OY respectively then
 
OB  a cos  ˆi and OC  a sin  ˆj
    
So according to rule of vector addition OA  OB  OC  a  a x iˆ  a y ˆj  a  a cos  ˆi  a sin  ˆj

RECTANGULAR COMPONENTS OF A VECTOR IN THREE DIMENSIONS

 
1. Consider a vector a represented by OA , as shown in figure. Consider O as origin and draw a
rectangular parallelopiped with its three edges along the X, Y and Z axes.
   
2. Vector a is the diagonal of the parallelopiped whose projections on x, y and z axis are a x ,a y and a z
respectively.

These are the three rectangular components of A . Y
   C
Using triangle law of vector addition, OA  OE  EA
   ay
Using parallelogram law of vector addition, OE  (OB  OD) A
    
 OA  (OB  OD)  EA a
O
ax B
      X
 EA  OC  OA  OB  OD  OC
D az
     E
Now, ˆ
OA  a,OB  a x i,OC  a y ˆj and OD  a z K
ˆ Z

 a  a x ˆi  a y ˆj  a z kˆ Also (OA) 2  (OE)2  (EA) 2
COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 7
 Vectors Rg-Phy. XI-2020-22
but (OE) 2  (OB) 2  (OD) 2

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

or a 2  a 2x  a 2y  a 2z  a  a 2x  a 2y  a 2z
ADDITION AND SUBTRACTION OF VECTORS IN COMPONENT FORM
 
Let A  A x i  A y j  A z k and B  B x i  B y j  Bz k be two vectors whose addition (or
subtraction) has to be done. Then the components in the same direction will be added and new
vector will be obtained. For example, in this case,
 
A  B  (A x  Bx )i  (A y  By )j  (A z  Bz ) k

Illustration 3: Resolve horizontally and vertically a force F  8N which makes an angle of 45º
with the horizontal.
 FV
Solution: Horizontal component of F is
1
FH = F cos 45º = 8 
2 F

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

 1 
= (8)   = 4 2N
 2
Illustration 4: Resolve a weight of 10 N in two directions which are parallel and perpendicular
to a slope inclined at 30º to the horizontal.
Solution: Component perpendicular to the plane
W  Wcos300 W||
30 º
3 W
= (10)
2 30 º W  10 N
= 5 3N
and component parallel to the plane
 1
W|| = W sin 30º = (10)   = 5 N
 2
   
Illustration 5: Obtain the magnitude of 2A  3B if A  ˆi  ˆj  2kˆ and B  2iˆ  ˆj  kˆ .
 
Solution: 2A  3B  2(iˆ  ˆj  2k)
ˆ  3(2iˆ  ˆj  k)
ˆ   4iˆ  5jˆ  7kˆ
 
Magnitude of 2A  3B  (4) 2  (5) 2  (7) 2

= 16  25  49  90

Illustration 6: The magnitude of a vector A is 10 units and it makes an angle of 300 with the
X- axis. Find the components of the vector if it lies in the X-Y plane.

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 8
 Vectors Rg-Phy. XI-2020-22

Solution: Components of vector A lying in the X-Y plane are A x  A cos , A y  A sin , A z  0
10 3
Thus, A x  10 cos 30 0   8.66 units ; A y  10 sin 30 o  10 x 1/ 2 = 5
2
Az  0
 
Illustration 7: Two forces F1  1N and F2  2N act along the lines x = 0 and y = 0 respectively..
Then, find the resultant force.
Solution: 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 F  2iˆ  ˆj
Illustration 8: What vector must be added to the summation of vectors ˆi  3jˆ  2kˆ and

3iˆ  6j  7k so that the resultant vector is a unit vector along the y-axis.

Solution: ˆi  3jˆ  2kˆ  3iˆ  6ˆj  7kˆ  4iˆ  3jˆ  5kˆ

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

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

Inchapter Exercise


1. Let AB be a vector in two dimensional plane with magnitude 4 units, and making an angle of 600 with

x-axis and lying in first quadrant. Find the components of AB along x axis and y axis.Hence represent

AB in terms of unit vectors ˆi and ˆj .

2. A 1000 N block is placed on an inclined plane with angle of 300. Find the components of the weight
(i) parallel (ii) perpendicular to the inclined plane.

   
3. Let A  2iˆ  ˆj, B  3jˆ  kˆ , find 2 A -3B .

Answer Key
1. Component on x-axis = 2 ; Component along y-axis = 2 3

Hence AB  2iˆ  2 3jˆ

2. (i) 500 N (ii) 866 N

 
3. 2A  3B  4i  7j  3k

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

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 9
 Vectors Rg-Phy. XI-2020-22

Illustration 9: A body is displaced from position vector r1  (2iˆ  3jˆ  k)
ˆ m to the position vector

r2  (iˆ  ˆj  k)
ˆ m.. Find the displacement vector..
 
Solution : The body is displaced from r1 to r2 . Therefore, displacement of the body is
  
S  r2  r1  (iˆ  ˆj  k)
ˆ  (2iˆ  3jˆ  k)
ˆ  ( iˆ  2ˆj)m

PRODUCT OF TWO VECTORS

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

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 and B is the product of the magnitudes of A and B and cosine of the
 
angle between them. Thus A.B  ABcos  . As A, B and cos  all are scalars, so their product is a
scalar quantity. Dot product is also called scalar product.
Geometrical interpretation of scalar product :
 
We have, A.B  AB cos  = A  Bcos   or   A cos   B

B


B A co s 

 
O  O 
B cos  A A

   
(a) A .B  A  B co s   (b) A.B   A cos   B
= ABcos  = ABcos 

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 B are parallel having same direction, then A.B  ABcos 00  AB
   
(v) If A and B are antiparallel, then A.B = A B cos1800   AB .
 
(vi) The scalar product of two identical vectors A.A  AA cos 00  A2
  
(vii) If i , j and k are mutually perpendicular unit vectors, then ˆi.iˆ  11 cos 0 0  1

 ˆi.iˆ  ˆj.jˆ  k.k

ˆ ˆ 1
ˆi.jˆ  11 cos 90 0  0 ˆi.jˆ  ˆj.kˆ  k.i
ˆˆ0
   
(viii) If A  A1iˆ  A 2ˆj  A 3kˆ and B  B1ˆi  B2ˆj  B3k, ˆ then A.B  A B  A B  A B .
1 1 2 2 3 3

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 10
 Vectors Rg-Phy. XI-2020-22
Applications of scalar product :
  
(i) Work done: If a force F causes displacement s, then work done w  F.s  Fs cos 
   
(ii) Angle between the vectors : For two vectors A and B , we have A .B  AB cos 
 
A.B  
 cos   or, If A  A1ˆi  A 2ˆj  A3kˆ and B  B1ˆi  B2ˆj  B3kˆ
AB
A1B1  A 2 B2  A 3 B3
then cos  
A  A 22  A32 B12  B22  B32
2
1

(iii) Component or projection of one vector along other vector:

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

  A co s 
ˆ   ABcos   B
A cos  B ˆ  A.B 
  =  B  B̂
 B    

A
 
(b) Component of vector B along vector A is given by


B
ˆ   ABcos   A
Bcos A ˆ
 
 A 
 
 A.B 
=  A  Â 
   
B co s  A

   
Illustration 10: Work done by a force F on a body is W  F . S , wheree S is the displacement of

body. Given that under a force F  (2iˆ  3jˆ  4k)N
ˆ a body is displaced from
 
position vector r1  (2iˆ  3jˆ  k)m
ˆ to the position vector r2  (iˆ  ˆj  k)m
ˆ , find
the work done by this force.
 
Solution: The body is displaced from r1 to r2 . Therefore, displacement of the body is
  
S  r2  r1  ( î  ĵ  k̂)  (2 î  3 ĵ  k̂) = ( î  2 ĵ) m
 
Now, work done by the force is W = F  S  (2 î  3 ĵ  4k̂)  ( î  2 ĵ)
= (2)(-1) + (3)(- 2) = - 8J

 
Illustration 11: Prove that the vectors A  2iˆ  3jˆ  kˆ and B  ˆi  ˆj  kˆ are mutually perpendicular..
 
Solution: A  B  (2 î  3 ĵ  k̂)  ( î  ĵ  k̂) = (2)(1) + (-3)(1) + (1)(1) = 0 = AB cos 
cos  = 0 or  = 90º
 
or the vectors A and B are mutually perpendicular..

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 11
 Vectors Rg-Phy. XI-2020-22
   
Illustration 12: Let for two vectors A and B , 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
    2 2
A.B  B.A  A – B = 0 or A = B
Thus A/B = 1 i.e. the ratio of magnitudes is 1.

Illustration 13: Find the angle between the vectors 3iˆ  2jˆ  1kˆ and 5iˆ  2jˆ  3kˆ
  2 2 2
Solution: A  32 22 12  14 B   5   2    3  38
 
A  B = AxBx + AyBy + AzBz = 3 x 5 + 2(-2) + (1) (-3) = 8
 
A.B 8
cos       0.35   cos1 (0.35)
| A || B | 14  38

Inchapter Exercise
 
1. Find the angle between two vectors A  2 î  ĵ  k̂ and B  î  k̂ .
2. If the vectors 4iˆ  ˆj  3kˆ and 2miˆ  6mjˆ  kˆ are mutually perpendicular, find the value of m.
3. What is the angle between ˆi  ˆj  kˆ and î
4. Find the magnitude of component of 3iˆ  2ˆj  kˆ along the vector 12iˆ  3jˆ  4kˆ
 1 
Answer Key: 1.  = 30º 2. m  3/14 3. θ  cos 1   4. 2
 3

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 n̂ is a unit vector perpendicular
 
to the plane of vector A and B .The direction of unit vector n̂ is given by the right hand thumb rule.
 
The fingers of right hand should be oriented along the direction of A such that they curl towards B .
The direction of outstretched thumb gives direction of n̂
n̂  
AB n̂  
AB

 
B
 
A 
 B
A
 
BA

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 12
 Vectors Rg-Phy. XI-2020-22
Geometrical interpretation of vector product :
S Q T R


B
Bsin 



O A P
 
Suppose two vectors A and B are represented by the sides OP and OQ of a parallelogram, as
   
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.
1
 Area of triangle OPQ = (area of parallelogram OPRQ).
2
1  
= | A B |.
2
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 vectorsis zero.

  0
  AB sin 0 0 nˆ  0
A  B

A  B  ABsin180 nˆ  0

(iv) Vector product of two identical vectors is also zero.

 
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 |  ABsin 90 o = AB.
(vi) For unit vectors ˆi, ˆj and kˆ
ˆi  ˆi  (1)(1)sin 0o nˆ  0

 ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  0
and ˆi  ˆj  (1)(1) sin 90o kˆ = k̂

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 13
 Vectors Rg-Phy. XI-2020-22
Similarly ˆj  kˆ  ˆi
and kˆ  ˆi  ˆj

 ˆj  ˆi   k,
ˆ kˆ  ˆj  ˆi, ˆi  kˆ  ˆj

(vii) Cross product can be used to find angle between two vectors.
 
According to definition of vector product of two vectors A  B  ABsin nˆ
 
   AB 
So, A  B  ABsin  i.e.   sin 1    
 A B 
 
However, dot product method is more convenient to find angle between vectors.
Note: To find the cross product of two vectors, the following methods can be used:

Cross Product Method 1:

 
A  B  (A x ˆi  A y ˆj  A z k)
ˆ  (B ˆi  B ˆj  B k)
x y z
ˆ

 A x B x (iˆ  ˆi)  A y Bx (ˆj  i)

ˆ  A B (kˆ  ˆi)
z x

 A x B y (iˆ  ˆj)  A y B y (ˆj  ˆj)  A z B y (kˆ  ˆj)

 A x Bz (iˆ  k)
ˆ  A B (ˆj  k)
y z
ˆ  A B (kˆ  k)
z z
ˆ

[As ˆi  ˆi  0, ˆj  ˆj  0, kˆ  kˆ  0 and iˆ  ˆj  k,
ˆ ˆj  ˆi   k,k
ˆ ˆ  ˆi  ˆj, ˆi  kˆ  ˆj, kˆ  ˆj  ˆi]
 
ˆ  A B ˆj  A B kˆ  A B (ˆi)  A B (ˆj)  A B (i)
So, we have A  B  A y Bx ( k) ˆ
z x x y z y x z y z

 (A y Bz  A z By )iˆ  (A x Bz  A z Bx )jˆ  (A x B y  A y Bx )kˆ

Cross Product Method 2:

 
Cross product of two vectors A and B can be obtained easily by using the determinant method.
 
A  B  (A x ˆi  A y ˆj  A z k)
ˆ  (B iˆ  B ˆj  B k)
x y z
ˆ

iˆ ĵ kˆ
ˆi ˆj kˆ Ax Ay Az

 Ax Ay Az
B x B y Bz
Bx By Bz

Here, we will use ˆi, ˆj,kˆ one by one. When î is chosen, its corresponding row and column become
bound and remaining elements are subtracted after cross multiplication.

So, (A y Bz  By A z ), is the component along î .

Similarly, in the case of ĵ , the row and column in which it is present become bound and remaining
elements are subtracted after cross multiplication.
So, ˆi(A y Bz  By A z )  ˆj(A x Bz  Bx A z ) is the component along î and ĵ

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 14
 Vectors Rg-Phy. XI-2020-22

iˆ ĵ k̂
Ax Ay Az

Bx By Bz

Same argument will follow for k̂ as is for î and ĵ

 
 A  B  ˆi(A y Bz  By A z )  ˆj(A x Bz  Bx A z )  k̂(A x By  Bx A y )

iˆ ĵ k̂
Az
Ax Ay

Bx By Bz

 
Illustration: 14: If A  3iˆ  ˆj  2kˆ and B  2iˆ  2jˆ  4kˆ
 
(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 A and B

iˆ ˆj kˆ
 
Solution: (i) A  B  3 1 2  8iˆ  8jˆ  8kˆ
2 2 4
   
Magnitude of A  B  | A  B |  (8) 2  ( 8) 2  ( 8) 2  8 3

 
A  B 8iˆ  8jˆ  8kˆ 1 ˆ ˆ ˆ
(ii) n̂      (i  j  k)
|AB| 8 3 3
  1 ˆ ˆ ˆ
There are two unit vectors perpendicular to both A and B , they are  n̂   (i  j  k)
3
   
8 3 AB 2 A.B 3
(iii) sin     cos  
AB 14 24 7 AB 7
 
Illustration15: Let a force F be acting on a body free to rotate about a point O and let r be the
position vector of any point P on the line of action of the force. Then torque of
  
this force about point O is defined as :   r  F
 
Given, F  (2iˆ  3jˆ  k)N
ˆ and r  (iˆ  ˆj  6k)m
ˆ , find the torque of this force.

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 15
 Vectors Rg-Phy. XI-2020-22

ˆi ˆj kˆ
  
Solution:   r  F  1 1 6  ˆi(1  18)  ˆj(1  12)  k(3
ˆ  2)
2 3 1

 17iˆ  11jˆ  5kˆ

 
Illustration 16: Two vectors, A  2iˆ  2jˆ  pkˆ and B  ˆi  ˆj  kˆ are given. Find the value of p if
(i) the two vectors are perpendicular (ii) the two vectors are parallel.

   
Solution: (i) A and B will be perpendicular if A  B = 0
 
Now A  B = AxBx + AyBy + AzBz = 2 x 1 + 2 x 1 + p x 1 = 4 + p
 
For A.B = 0, we must have 0 = 4 + p or p = –4

^ ^ ^

    i j k
 
(ii) A and B will be parallel if A  B = 0 ; Now A  B = 2 2 p
1 1 1

= ˆi(2  p)  ˆj(p  2)  k(2

ˆ  2) = ˆi(2  p)  ˆj(p  2)

 
For A  B = 0, we must have each component to be zero. That is 0 = 2 – p, and 0 = p – 2
(both conditions are similar). Thus p = 2

Inchapter exercise

 
1. Find a unit vector perpendicular to both A  2iˆ  ˆj and B  ˆi  2ˆj
 
2. The torque of a force F   3iˆ  ˆj  5kˆ acting at the point r  7iˆ  3jˆ  kˆ is:
(A) 14iˆ  38jˆ  16kˆ (B) 4iˆ  4ˆj  6kˆ
(C) 21iˆ  4ˆj  4kˆ (D) 14iˆ  34ˆj  16kˆ
3. The magnitude of the vector porduct of two vectors is 3 times their scalar porduct. What is the
angle between the two vectors?

  
4. The linear velocity of a rotating body is given by v =  × r .
  
If  = ˆi - 2ˆj + 2 kˆ a n d r = 4 ˆj -3 kˆ , then what is the magnitude of v ?

Answer key

1. k̂ or - k̂ 2.(A) 3. θ = 6 0 o 4. 29 units

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 16
 Vectors Rg-Phy. XI-2020-22

JEE MAIN EXERCISE


1. The value of a unit vector in the direction of vector A  5iˆ  12ˆj , is –

(A) î (B) ˆj  
(C) ˆi  ˆj / 13  
(D) 5iˆ  12jˆ /13

2. For the figure –

C B

           
(A) A  B  C (B) B  C  A (C) C  A  B (D) A  B  C  0

3. Two forces of 4 dyne and 3 dyne act upon a body. The resultant force on the body can only be –
(A) more than 3 dynes (B) more than 4 dynes
(C) between 3 and 4 dynes (D) between 1 and 7 dynes

4. A force of 6 N and another of 8 N can be applied together to produce the effect of a single force of-
(A) 1 N (B) 11 N (C) 15 N (D) 20 N

5. Which of the sets given below may represent the magnitudes of three vectors adding to zero ?
(A) 2, 4, 8 (B) 4, 8, 16 (C) 1, 2, 1 (D) 0.5, 1, 2

6. A blind person after walking each 10 steps in one direction, each of length 80 cm, turns randomly to
the left or to right by 900 . After walking a total of 40 steps the maximum possible displacement of
the person from his starting position could be –
(A) 320 m (B) 32 m (C) 16 2m (D) 16 2m

7. If the angle between vector a and b is an acute angle, then the difference a – b is –
(A) the main diagonal of the parallelogram (B) the minor diagonal of the parallelogram
(C) any of the above (D) none of the above

   
8.   
Angle between P  Q and P  Q will be – 
(A) 00 only
(B) 900 only
(C) 1800only
(D) between 00 and 1800 (both the values inclusive)

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 17
 Vectors Rg-Phy. XI-2020-22
9. The three vectors OA, OB and OC have the same magnitude R. Then the sum of these vectors
have magnitude –
B
C

450
450
A
O

(A) R (B) 2R (C) 3R 

(D) 1  2 R 
 
10. Two vectors P  2iˆ  bjˆ  2kˆ and Q  ˆi  ˆj  kˆ are perpendicular to each other if -
(A) b = 0 (B) b = 1 (C) b = 2 (D) b = –4

11. A vector is not changed if –

(A) It is rotated through an arbitrary angle (B) It is multiplied by an arbitrary scale
(C) It is cross multiplied by a unit vector (D) It is a slide parallel to it self

      
12. If A  B  C and the magnitudes of A , B and C are 5, 4 and 3 units, the angle between A and

C is–
3 1  4   1  3 
(A) cos 1   (B) cos   (C) (D) sin  
5 5 2 4
 
13. The vector A and B are such that –
   
AB AB
     
(A) A  B  0 (B) A  B  0 (C) A  0 (D) B  0

14. Two forces each of magnitude F have a result of the same magnitude F. The angle between the two
forces is –
(A) 450 (B) 1200 (C) 1500 (D) 600

  
15. Three vectors A  2iˆ  ˆj  kˆ , B  ˆi  3jˆ  5kˆ and C  3iˆ  4ˆj  4kˆ are sides of an –
(A) equilateral triangle (B) right angled triangle
(C) isosceles triangle (D) none of the above

  
16. If A, B and C are three vectors, then the wrong relation is –
           
  
(A) A  B  C  A  B  C  (B) A. B  C  A.B  A.C    
            
    
(C) A  B  C  A  B  A  C  (D) A  B .C  A  B.C   
 
17. Two vector A and B have equal magnitude and perpendicular to each other. Then the vector
 
A  B is perpendicular to –
     
(A) A  B (B) A  B (C) 3A  3B (D) all of these

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 18
 Vectors Rg-Phy. XI-2020-22
    
18. The angle between two vector A and B is  . Then the magnitude of the product A. B  A is – 
(A) A 2 B (B) A 2 Bsin  (C) A 2 Bsin  cos  (D) Zero

19. Two force are such that the sum of their magnitudes is 18 N, magnitude of the resultant is 12 N and
their resultant is perpendicular to the smaller force. Then the magnitude of the forces are –
(A) 12 N, 6 N (B) 13 N, 5 N (C) 10 N, 8 N (D) 16 N, 2 N

     
20. If A  B  B  A , then the angle between A and B is –
(A)  (B)  3 (C)  2 (D)  4

     
21. If A.B  magnitude of A  B , then the angle between vectors A and B is
(A) 300 (B) 450 (C) 600 (D) 750

   
22.   
The magnitude of the resultant of A  B and A  B is 
(A) 2A (B) 2B (C) A 2  B2 (D) A 2  B2

23. If ˆi, ˆj and k̂ are unit vectors along x, y and z-axes respectively, the angle  between the vector i
and ˆi  ˆj  kˆ vector is given by

 1  1  1  1
 3  3
(A)   cos 1   (B)   sin   (C)   cos   (D)   sin 1  
 3  3  2   2 

24. Given that 0.2iˆ  0.6ˆj  akˆ is a unit vector. What is the value of a ?
(A) 0.3 (B) 0.4 (C) 0.6 (D) 0.8

   
25. Given A  2iˆ  3jˆ and B  iˆ  ˆj . The component of vector A along vector B is
1 ˆ ˆ 3 ˆ ˆ 5 ˆ ˆ 7 ˆ ˆ
(A)
2
 
i j (B)
2
 
i j (C)
2
 
i j (D)
2
 
j i

EXERCISE - 1
OBJECTIVE PROBLEMS WITH ONE CORRECT ANSWER

1. 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

2. 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

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 19
 Vectors Rg-Phy. XI-2020-22
3. 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

4. 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

5. 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

6. 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

7. 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


8. A body is constrained to move only in Y-direction under the action of a force F   2iˆ  15ˆj  6kˆ
N. If the body moves a distance of 10m then the work done is :
(A) 150J (B) 190J (C) 163J (D) 185J

    
9. If A  i  j & B  i  j , then a vector C perpendicular to both A & B and having a magnitude
equal to 3 is :
(A) 3kˆ 
(B) 3 ˆi  ˆj  
(C) 3 ˆi  2kˆ  
(D) 3 ˆi  ˆj 
   
10. A vector A points vertically downward & B points towards east, then the vector product A  B
is :
(A) along west (B) along east (C) along north (D) along south

11. A particle is moving eastward with a velocity of 5m/s. In 10 seconds the velocity changes to

v
5m/s northwards. The average acceleration in this time is :
t
(A) Zero (B) 1 / 2m / s2 towards north - west
(C) 1 / 2m / s2 towards north - east (D) 1 / 2m / s2 towards north

12. 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

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 20
 Vectors Rg-Phy. XI-2020-22
13. At t = 0, a particle at (1,0,0) starts moving towards (4,4,12) with a constant speed of 65 m/s. The
position of the particle is measured in metres and time in secs. Assuming constant velocity, the
position of the particle at t=2 sec is :
   
    

(A)  13 i  120 j  40 k m (B)  40 i  31 j  120 k m

   
    

(C)  13 i  40 j  12 k m (D)  31 i  40 j  120 k m

14. 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

15. 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 :
o o o o
(A) 30 (B) 60 (C) 120 (D) 150

     
16. Three forces P , Q and R are acting at a point in the plane, the angle between P and Q & Q and
 o o   
R are 150 and 120 respectively. then for equilibrium forces P , Q and R are in the ratio :
(A) 1:2:3 (B) 1:2: 3 (C) 3:2:1 (D) 3 : 2:1

  
17. The resultant of two vectors u and v is perpendicular to the vector u and its magnitude is equal to
  
half of the magnitude of vector v , The angle between u and v in degrees is
(A) 120 (B) 60 (C) 90 (D) 150

18. Choose the wrong statement

(A) Three vectors of different magnitudes may be combined to give zero resultant.
(B) Two vectors of different magnitudes can be combined to give a zero resultant.
(C) The product of a scalar and a vector is a vector quantity.
(D) All of the above statements are wrong.

19. 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

20. Out of the following pairs of forces, the resultant of which cannot be 4 newton
(A) 2 newton and 2 newton (B) 2 newton and 4 newton
(C) 2 newton and 6 newton (D) 2 newton and 8 newton

   
21. If a  ˆi  2ˆj  2kˆ and b  2iˆ  ˆj  2kˆ . Find the projection vector of b on a .
8 ˆ ˆ 8 ˆ ˆ 9 ˆ ˆ ˆ 9 ˆ ˆ
(A)
9
i  2 j  2kˆ  (B)
9
2i  j  2kˆ  (C)
8
i  2j k  (D)
8
2i  j  2kˆ 

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 21
 Vectors Rg-Phy. XI-2020-22

EXERCISE - 2
OBJECTIVE PROBLEMS WITH MULTIPLE CORRECT ANSWERS

   
1. Given two vectors A=3i+4j ˆ ˆ and B=i+j.
ˆ ˆ θ is the angle between A and B . Which of the
following statements is/are correct?
  ˆi+jˆ   
(A) A cosθ   is the component of A along B .
 2
  ˆi-jˆ   
(B) A sinθ   is the component of A perpendicular to B .
 2
  ˆi-jˆ   
(C) A cosθ   is the component of A along B .
 2 
  ˆj-iˆ   
(D) A sinθ   is the component of A perpendicular to B .
 2
 
2. ˆ ˆ ˆ and B=i+j+k
If A=2i+j+k ˆ ˆ ˆ are two vectors, then the unit vector

  ˆj  kˆ  ˆ ˆ ˆ
  2i+j+k 
(A) perpendicular to A is  2  (B) parallel to A is  
   6 
  ˆj  kˆ   ˆi  ˆj  kˆ
(C) perpendicular to B is  2  (D) parallel to A is
  3

   
3. If (v1  v 2 ) is perpendicular to (v1  v 2 ) , then
   
(A) v1 is perpendicular to v 2 (B) v1  v 2
  
(C) v1 is null vector (D) the angle between v1 and v2 can have any value

     
4. Two vectors A and B lie in one plane. Vector C lies in a different plane. Then, A  B  C
(A) cannot be zero
(b) can be zero
 
(C) lies in the plane of A and B
(D) lies in a plane different from that of any of the three vectors.

5. Which of the following expressions are meaningful?

           
(A) u.(v  w) (B) (u.v).w (C) (u.v)w (D) u  (v.w)

6. The magnitude of scalar product of two vectors is 8 and of vector product is 8 3 . The angle
between them can be :
(A) 30o (B) 60o (C) 120o (D) 150o

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 22
 Vectors Rg-Phy. XI-2020-22
7. A situation may be described by using different sets of coordinate axes having different orientations.
Which of the following do not depend on the orientation of the axes?
(A) The value of a scalar (B) component of a vector
(C) a vector (D) the magnitude of a vector

8. The x-component of the resultant of several vectors

(A) is equal to the sum of the x-components of the vectors
(B) may be smaller than the sum of the magnitudes of the vector
(C) may be greater than the sum of the magnitudes of the vector
(D) may be equal to the sum of the magnitudes of the vector

 
9. The magnitude of the vector product of two vectors A and B may be
(A) greater than AB (B) equal to AB
(C) less than AB (D) equal to zero

EXERCISE - 3
MATCH THE FOLLOWING
 
1. A  4iˆ  4ˆj , B  4iˆ  4ˆj .Then,
Column A Column B
 
i. AB (a) 8
 
ii. AB (b) 4
 
iii. A.B (c) 32
 
iv. AB (d) 0

   
2. A  1, B  2, Angle between A and B is 900 .

Column A Column B
 
i. A.B (a) 2
 
ii. AB (b) 4
 
iii. AB (c) 5
 
iv. AB (d) 0

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 23
 Vectors Rg-Phy. XI-2020-22

EXERCISE - 4
SUBJECTIVE PROBLEMS
1. Which of the following are correct (meaningful) vector expressions? What is wrong with any incorrect
expression?
           
(A) A . B .C ; (B) A x B . C ; (C) A . B x C ; (D) A x B x C ;
       

        
(E) A   B .C ; (F) A   B x C ; (G) 5  A; (H) 5   B . C 
     
     
2. Suppose that a vector F is given by F  q(V  B) , where q is a number and V and B are

vectors. What are the directions of F in the below three situations, if q is (i) a positive quantity (ii)
a negative quantity?

y y y
  
B v B

v
x x x
 
B v
z z z

(A) (B) (C)

3. (A) In unit-vector notation, what is
      
r  a  b  c if a  5iˆ  4ˆj  6k,
ˆ b   2iˆ  2ˆj  3k,
ˆ and c  4iˆ  3jˆ  2kˆ


(B) Calculate the angle between r and the positive z axis.
 
(C) What is the magnitude of component of a along the direction of b ?
 
4. Two vectors are given by a  3iˆ  5jˆ and b  2iˆ  4ˆj . Find :
   
(A) a  b (B) a.b
    
 
(C) a+b .b (D) the magnitude of component of a along the direction of b .
  
5. Three vectors are given a  3iˆ  3jˆ  2k,
ˆ b   1iˆ  4ˆj  2kˆ and c  2iˆ  2jˆ  1kˆ . Find
        
   
(a) a . b  c , (b) a . b  c , and (c) a  (b  c) .


6. A vector with a magnitude of 8m, is added to a vector A , which lies along an x-axis. The sum of
these two vectors is a third vector that lies along the y-axis and has a magnitude that is twice the
 
magnitude of A . What is the magnitude of A ?

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 24
 Vectors Rg-Phy. XI-2020-22
7. A ship standing at point O sets out to sail to a point A 120 km due north of O. An unexpected
storm blows the ship to a point B 100 km due east of O. East directon is along î and north

direction is along ˆj . Find vector BA

        
8. ˆ ˆ , solve for a and b ?
If a-b=2c, a+b=4c, c=3i+4j

 
9. Let A and B be the two vectors of magnitude 10 units each. If they are inclined to the x-axis at
 
angles30º and 60º respectively, find the resultant of A & B in the form piˆ  qjˆ

10. A particle whose speed is 50 m/s moves along the line from A(2, 1) to B (9, 25) . Find its velocity
vector in the form of p ˆi  q ˆj .

11. Two vectors acting in the opposite directions have a resultant of 17 units. If they act at right angles
to each other, then the resultant is 25 units. Calculate the magnitude of two vectors .

12. Answer the following :

(A) Should a quantity having a magnitude & direction be necessarily a vector ?
(B) Can two similar vectors of different magnitude yield a zero resultant ?Can three yield ?
      
(C) R  A  B, Is it possible that | R || A | & | R || B | are both true?
     
(D) If a  b  c & a  b  c , what further information you can have about these vectors ?
     
(E) If a & b are two non zero vectors such that a  b  a  b ,then what is the angle between
 
a & b ?
(F) Time has magnitude as well as direction. Is it a vector ?
     
(G) When will a x b  a . b ?( a & b are two non - zero vectors )
(H) Do unit vectors ˆi , ˆj& kˆ have units ?

13. Two vectors have magnitude 2m and 3m. The angle between them in degrees is 60. Find
(A) the scalar product of the two vectors.
(B) the magnitude of their vector product.

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 25
 Vectors Rg-Phy. XI-2020-22

ANSWER KEY
JEE MAIN EXERCISE
1. (D) 2. (C) 3. (D) 4. (B) 5. (C)
6. (D) 7. (B) 8. (D) 9. (D) 10. (D)
11. (D) 12. (A) 13. (D) 14. (B) 15. (B)
16. (D) 17. (D) 18. (D) 19. (B) 20. (A)
21. (B) 22. (A) 23. (A) 24. (C) 25. (C)

EXERCISE-1 (OBJECTIVE PROBLEMS WITH ONE CORRECT ANSWER)

1. (A) 2. (D) 3. (D) 4. (D) 5. (B)
6. (C) 7. (A) 8. (A) 9. (A) 10. (D)
11. (B) 12. (A) 13. (D) 14. (A) 15. (C)
16. (D) 17. (D) 18. (B) 19. (B) 20. (D)
21. (A)

EXERCISE-2 (OBJECTIVE PROBLEMS WITH MULTIPLE CORRECT ANSWERS)

1. (AD) 2. (ABC) 3. (BD) 4.(AD) 5. (AC)
6. (BC) 7. (ACD) 8.(ABD) 9. (BCD)

EXERCISE-3 (MATCH THE FOLLOWING)

1. (I) A (II) A (III) D (IV) C 2. (I) D (II) A (III) C (IV) C

EXERCISE-4 (SUBJECTIVE PROBLEMS)

1. (A) WRONG (B) WRONG (C) RIGHT (D) RIGHT (E) WRONG
(F) RIGHT (G) WRONG (H) RIGHT
2. (I) (1) – k̂ (2) ZERO (3) î (II) (1) k̂ (2) ZERO (3) î
 7  20
3. (A) 11 î + 5 ĵ – 7 k̂ (B) COS-1  
 (C)
 195  17

13
4. (A) 2 k̂ (B) 26 (C) 46 (D)
5


5. (A) –21 (B) –9 (C) 5 î – 11 ĵ – 9 k̂ 6. M 7. 100iˆ  120ˆj
5
 
8. a = 9 î + 12 ĵ b = 3 î + 4 ĵ 9. 5   
3  1 iˆ  ˆj

10. 14iˆ  48ˆj 11. 24, 7

12 (A) NO (B) NO, YES (C) YES (D) VECTORS ARE COLLINEAR
(E) 90O (F) NO (G) NEVER (H) NO

13. (A) 3 (B) 3 3

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE # 26