1VECTOR ALGEBRA

VECTOR ALGEBRA

1.1 Basic Review of Vectors1.1.1 Definition:

Physical quantities having both magnitude, a definite direction in space and it should follow the laws of vectoraddition.Example : Velocity, Acceleration, Momentum, Force, Electric Field, Torque, etc.

1.1.2 Various type of vectors:(1) Equal vectors : Vectors having same magnitude and same direction.(2) Null Vectors : Vectors having coincident initial and terminal point i.e. its magnitude is zero and it has anyarbitrary direction.

(3) Unit Vectors : Vectors having unit magnitude. Unit vector along a= a = aa

(4) Reciprocal Vector: Vector having same direction but reciprocal magnitude corresponding to originalvector.(5) Negative Vector : Vectors having same magnitude but opposite direction corresponding to original vector.

1.1.3 Orthogonal Resolution of Vectors:

Any vector A

in right- handed rectangular cartesian coordinate system can be represented as

x y z OP A A i A j A k

, where, i, j and k are the unit vectors in direction of x, y and z axis respectively

and Ax, Ay, Az are the rectangular components of vector A

along x, y, z axis.

Axi^

Azk^

Ay j^x

zP

yO

A

Magnitude of vector 2 2 2x y zA is A A A A

Unit vector along A

is A A / A x y z

2 2 2x y z

A i A j A k

A A A

Chapter 1

VECTOR ALGEBRA2

If A

makes angles with x, y, z axes respectively, then direction cosines of A

are :

l = cos = xAA ; m = cos =

yAA

; n = cos = zAA and l

2 + m2 + n2=1

So, an unit vector along A

can be written as A li mj nk

1.2 Products of Vectors1.2.1 Scalar Product or Dot Product:

.. cos cos a ba b a ba b

Properties:

(i) a b b a , (ii) For two mutually perpendicular vectors a and b

, a b

= 0,

(iii) i j j k k i = 0, i i j j k k =1

(iv) If a= z y za i a j a k and b

= x y zb i b j b k , then a b = x x y y z za b a b a b

(v) Projection of A

on B

= .A B

(vi) Work done by force F on an object in displacement of r = .F r

1.2.2 Vector Product or Cross Product :

a b = sina b n

where n is unit vector normal to the plane containing a and b

.

Properties:

(i) a b = ( )b a

(ii) For two collinear vectors (parallel or anti-parallel vectors) 0a b

.

(iii) i i j j k k = 0, , ,i j k j k i k i = j

(iv) If a

= x y za i a j a k and b

= x y zb i b j b k , then a b

=

x y z

x y z

i j ka a a

b b b

(v) Torque r (applied force)F

(vi) Angular momentum L r linear momentum P

(vii) Linear velocity v (angular velocity) r

1.2.3 Scalar Triple Product: ( )a b c

=

x y z

x y z

x y z

a a a

b b b

c c c = [abc]

3VECTOR ALGEBRAProperties:

(i) ( )a b c = ( )b c a

= ( )c a b i.e. [abc] = [bca] = [cab]

(ii) Volume of a parallelopiped having , ,a b c as concurrent edges is :V= ( )a b c

(iii) If , ,a b c

are coplanar vectors, then, [abc] = [bca] = [cab] = 0

1.2.4 Vector Triple Product:

( )a b c

= ( ) ( )b a c c a b

It represents a vector coplanar with b

and c

. and perpendicular to a

Example-1: The range of x for which between the vectors 2 2 4 and 7 2A x i xj k B i j xk

isobtuse, is equal to(a) x < 0 (b) x > 1/2 (c) 0 < x < 1/2 (d) None of these

Soln: Since, 0 090 180 , then 2cos 0 A.B 0 14x 7x 0 x 2x 1 0Therefore, 0 x 1 2

Example-2: A unit vector n on the xy-plane is at an angle of 120 with respect to i . The angle between

the vectors u ai bn and v an bi will be 60 if

(a) 3 / 2b a (b) 2 / 3b a (c) b = a/2 (d) b = a

Soln: 0. cos 60u v u v

2 2 2 2 0 2 2 0 1 . . 2 cos120 2 cos120 2a i n b n i ab ba a b ab a b ab

2 0 2 0 2 2 1cos120 cos120 2a b ab ba a b ab 2 22 5 2 0

2 2b aa ab b a or b

Example-3: The value of m for which 2 , 3 5 6 , 4A i j k B i j k C i j mk

will be co-planaris

(a) 13/2 (b) 7/2 (c) 5/2 (d) 11/2

Soln: Since, , ,A B C

are co-planar, 1 1 2

13. 0 3 5 6 02

1 4A B C m

m

Example-4: If b i a i j a j k a k

, then b

can be simplified to

(a) 0 (b) a (c) 2a (d) None of these

Soln:

. . . . . . 3 2

b i a i j a j k a k

a i i i i a a j j j j a a k k k k a a a a

VECTOR ALGEBRA4

Example-5: Three unit vectors , ,a b c ( andb c

are not parallel) are such that 32a b c c

. The

angles which a makes with b

and c , respectively are(a) 0 030 ,90 (b) 0 0150 ,90 (c) 0 060 ,90 (d) 0 090 ,30

Soln. 3 3 3. . . 0 and .2 2 2a b c c b a c c a b c a c a b

The angle which a makes with b

and c are 0 0150 ,90 respectively..

1.3 Gradient, Divergence and Curl

1.3.1 Gradient of a Scalar Field :Gradient of a continuously differentiable scalar function (x, y, z) is mathematically defined as:

grad=

= i j kx y z

= i j kx y z

where,

= 'del' or 'grad' or 'nabla' operatori j kx y z

Physical interpretation : Gradient of scalar function at any point P(x, y, z) is a vector, whose magnitudeis equal to the rate of change of scalar function with distance along the normal to level surface and itsdirection is along the normal to the level surface at that point.

Level surface : It has same value of scalar function at each point. Example: Equipotential surface.

Directional Derivative : Directional derivative of in the direction of A

is defined as the component of

in the direction of vector A

and is given by, , A

= AA

Tangent Plane and Normal to the level surface:

Consider , ,x y z c be the equation of a level surface. and r xi yj zk be the position vector ofany point P (x,y,z) on this surface.

Tangent plane at P:

is a vector normal to the surface i.e. it is perpendicular to the tangent plane at P..

Let, R Xi Yj Zk

be the position vector of any point on the tangent plane at P to the surface. Therefore,

R r X x i Y y j Z z k lies in the tangent plane at P and it will be perpendicular to

i.e. . 0R r

0X x Y y Z zx y z

, which is the tangent plane at point P..

Normal at P: Let, R Xi Yj Zk

be the position vector of any point on the normal at P to the surface.

Therefore, R r X x i Y y j Z z k lies along the normal at P and it will be parellel to

i.e. 0R r

, which is the vector equation of the normal at point P to the given surface.

5VECTOR ALGEBRA1.3.2 Divergence of a vector field:

Divergence of a continuous differentiable vector point function V specified in a vector field is given by,,

V = x y zi j k V i V j V kx y z =

yx zVV V

x y z = Scalar quantity

Physical Interpetation: Divergence of A

at pt P(x, y, z) is defined as the outward flux of the vector fieldA

per unit volume enclosed by an infinitesimal closed surface surrounding point P.

Properties:

(i) If V = 0, then V

is known as solenoidal Vector.

(ii) If V = negative, then V

is a sink field i.e. vector lines are going inward.

(iii) If V = positive, then V

is a source field i.e. vector lines are the going outward.

(iv) (u v) = u v

(v) (uv) = ( u) u v( u)

1.3.3 Curl of a vector field:

Curl or rotation of a continuous differentiable vector point function V is given by,,

V

= x y zi j k v i v j v kx y z =

x y z

i j k

x y zV V V

It is the measurement of rotation of vector field and the direction of curl of vector is along axis of rotation.

Properties:

(i) If 0V

, then V is an irrotational vector and we can write V

(ii) If 0V

, then V is a rotational vector..

(iii) ( )U V

= U V

(iv) ( )uV

= ( )u V u V

1.3.4 Important Vector Identities:

If are scalar point functions and ,A B

are vector point functions in certain region then.

(1) ( )

(2) ( )

(3) ( )A B A B

(4) ( )A B A B

(5) ( ) ( ) ( ) ( ) ( )A B A B A B B A B A

(6) ( ) ( )A A A

(7) ( ) ( ) ( )A A A

VECTOR ALGEBRA6

(8) ( ) 0A

(9) ( ) 0

(10) ( ) ( ) ( )A B B A A B

(11) ( ) ( ) ( ) ( ) ( )A B B A A B A B B A

(12) 2( ) ( )A A A

Example-6: The unit vector perpendicular to the surface 2 2 1 x y z at point P (1,1,1) is equal to

(a) 2 2

3 i j k

(b) 2 2

3 i j k

(c) 2 2

3 i j k

(d) 2 2

3 i j k

Soln:2 2

2 2Given : : 1 unit normal to the surface3

i j kx y z n

Example-7: Find the unit normal to the surface of ellipsoid 2 2 2

2 2 2 1 x y za b c

at the point , ,3 3 3

a b c.

Soln. n

=

2 2 2

2 2 2 2 2 2 2 2 2

2 2 24 4, ,

3 3 3

1 1 12 2 2 3 3 31 1 14 4 4

3 3 3

a b c

x y z i j ki j k bci acj abka b ca b cx y z b c c a a b

a b ca b c

Calculation of

if is given as a function of r:

Example-8. ln r First take normal derivative of w.r.t. r and just put rrr

beside it i.e.

21 rrr r

Example-9: Show that, 2 21n nr n n r .

Soln:

2 1 2 2 2

3 2 2 2 2

. . . . .

2 . 3 2 3 1

n n n n n n

n n n n n

r r nr r nr r nr r nr r

n n r r r nr n n r nr n n r

Example-10: The vector nr r will solenoidal for 0r (a) n = 3 (b) n = -3 (c) n = 2 (d) n = -2

Soln: 1 . . . . 3 3n n n n n nr r r r r r nr r r r n r

; So this will solenoidal for n = -3

Example-11: If 4rgrad & r 0 at r=1r

, then find

Soln: Given that grad 4 5r rr r

5

xi yj zk i j kx y z r

;

7VECTOR ALGEBRA

Therefore, 5

25 2 2 2;

x x xx r x y z

5

22 2 2,

x x f y zx y z

3

2

2 2 2 2

2 2 2

1Putting , we get , constant of integration3

r x y z t f y z

x y z

31 ,

3 r f y z

r ; Similarly, 3 3

1 1, and ,3 3

r f x z r f x yr r

.

So, 31

3 r C

r ; Since, at r =1,

10 03

r C 1C3

Putting this in equation (1), we get 31 1

3 3r

r

Example-12: If F

is a constant vector and r is the position vector then .F r would be

(a) .r F

(b) r F (c) .F r

(d) F

Soln: 1 2 3 1 2 3 1 2 3 .F Fi F j F k F r F x F y F z F i F j F k F

Example-13: For the function 2 2,xx y

x y

, find the magnitude of the directional derivative along a line

making an angle 30 with the positive x-axis at (0, 2)

Soln.

2 22 2 2 2 2 2 2 22 21 x x x yxi j k i j

x y z x y x y x y x y

2 2

2 22 2 2 2

2 y x xyi jx y x y

at the point (0, 2)

j

C30 i^

A

B

2 2

2 0 24 0 40 4 0 4ii j

Directional derivative at the point (0, 2) in the direction CA

i.e. 3 1 2 2

i j

3 1 3 .4 2 2 8i i j

cos30 sin 30

3 1 2 2

CA CB BA i j

i j

Example-14: Find the directional derivative of 2V

, where 2 2 2 V xy zy j xz k

, at the point (2, 0, 3) in thedirection of the outward normal to the sphere 2 2 2 14x y z at the point (3, 2, 1).

Soln. 2 .V V V = 2 2 2 2 2 2 2 2 4 2 4 2 4 . .V V V xy i zy j xz k xy i zy j xz k x y z y x z

VECTOR ALGEBRA8

42 2 4 2 4 2 V i j k x y z y x zx y z

4 4 2 3 3 2 4 2 3 2 2 4 4 2 4xy xz i x y y z j y z x z k 2V at (2, 0, 3) = 0 2 2 81 0 0 0 4 4 27i j k

324 432 108 3 4i k i k

Normal to 2 2 2 14x y z 2 2 2 14i j k x y zx y z

2 2 2xi yj zk Normal vector at (3, 2, 1) 6 4 2i j k

Unit normal vector 2 3 2 6 4 2 3 2

36 16 4 2 14 14

i j ki j k i j k

Direction derivative along the normal 108 9 43 2 1404108 3 4 .

14 14 14i j ki k

Example-15: A fluid motion is given by

2 sin sin sin 2 cosv y z x i x z yz j xy z y k is the motion irrotational? If so, find the velocitypotential

Soln. Curl v v

2 sin sin sin 2 cosi j k y z x i x z yz j xy z y kx y z

2

sin sin sin 2 cos

i j k

x y zy z x x z yz xy z y

cos 2 cos 2 cos cos sin sin 0x z y x z y i y z y z j z z k Hence, the motion is irrotational.So, v

where is called velocity potential.

d dx dy dzx y dz

. . .i j k idx jdy kdz dr v drx y z

2 sin sin sin 2 cos .y z x i x z yz j xy z y k idx jdy kdz 2sin sin sin 2 cosy z x dx x z yz dy xy z y dz

9VECTOR ALGEBRA

2sin sin cos sin 2y zdx xdy z xy zdz xdx yzdy y dz

2sin cosd xy z d x d y z

2sin cosd xy z d x d y z 2sin cosxy z x y z c

Hence, Velocity potential 2sin cosxy z x y z c

1.4 Line Integration of VectorsThe integration of a vector a curve is known as line integral.

Let , ,F x y z

be a vector function and a curve AB.

Line integral of a vector function F

along the curve AB is defined as integral of the component of F

alongthe tangent to the curve AB.Component of F

along a tangent PT at P = Dot product of F

and unit vector along PT

= is a unit vector along tangent PTdr drFds ds

Line integral = .drFds

from A to B along the curve.

Therefore, line integral c c

drF ds F drds

BT

F

P

A

O X

r

(i) If F

represents the variable force acting on a particle along arc AB, then the total work done B

AF dr

(ii) If V

represents the velocity of a liquid then c

V dr is called the circulation of V

round closed curve c.

(iii) When the path of integration is a closed curve then notation of integration is in place of .Properties:

(i) Work done by a conservative field A

in moving a particle from point P to Q =

Q

P

F dr

=Q

P

dr

= Q

P

d = Q P = independent of path.

(ii) Work done by a conservative field A

in moving a particle around a closed path C = c

F dr

= 0

Example-16: If 2 5 6 2 4F xy x i y x j

, then evaluate the line integral .C

F dr along the curve C

in the x-y plane given by 3y x from the point (1,1) to (2,8) .

VECTOR ALGEBRA10

Soln:

3 2

2 4 2 3 2

1

2 5 4 3 2

1

3

, . 5 6 2 4 . 3

6 5 12 6 35

C

r x i x j dr dx i x dx j

So F dr x x i x x j dx i x dx j

x x x x dx

Example-17: If a force field 2 3 2A y x i x y j

is applied to a particle, then the work done bythe force field in traversing the particle around a circle C in x-y plane, with center at the origin and radius 2units is equal to ( C is traversed in the counter-clockwise direction )

(a) 2 (b) 4 (c) 8 (d) 16

Soln:

2

0

. 2 3 2 Putting 2cos and 2sin we get,

2sin 4cos 2sin 6cos 4sin 2cos 8

C C

A dr y x dx x y dy x y

d d

Example-18: Evaluate .C

F dr

, where 2 3 F x i y j and curve C is the arc of parabola 2y x in the x-y

plane from (0, 0) to (1, 1).

Soln. Now along the curve C, 2y x . Therefore, 2dy xdx

1 2 6

0. 2

C xF dr x dx x x dx

13 81 2 7

00

2 723 8 12x xx x dx

Example-19: If 2 3F x y i y x j

, evaluate .C

F dr

where C is the curve in the xy-plane consisting

of the straight lines from (0, 0) to (2, 0) and then to (3, 2).Soln. The path of integration C has been shown in the figure. It consists of the straight lines OA and AB.

O (0, 0) A(2, 0)x

yB(3, 2)

We have, .C

F dr

= 2 3 .C

x y i y x j dx i dy j 2 3

Cx y dx y x dy

Now along the straight line OA, y = 0, dy = 0 and x varies from 0 to 2. The equation of the straight line AB

is 2 00 2 . . 2 43 2

y x i e y x

Therefore, along AB, y = 2x4, dy = 2dx and x varies from 2 to 3.