Vector Preliminary

Dr. S. GangulyDepartment of Physics

Bethune College, Kolkata

E-mail: sgpresi78@gmail.com

2 S. Ganguly

0.1 Product of three vectors

1. Scalar triple product: ~A.( ~B × ~C)=(Axi+Ay j +Az k).

∣∣∣∣∣∣i j k

Bx By BzCx Cy Cz

∣∣∣∣∣∣=

∣∣∣∣∣∣Ax Ay AzBx By BzCx Cy Cz

∣∣∣∣∣∣From the properties of the determinants, it readily follows that~A.( ~B × ~C) = ~B.(~C × ~A) = ~C.( ~A× ~B)Scalar triple product remains unchanged by a cyclic permutation of the vectors, but it changes sign if the orderis violated.In a Scalar triple product, the dot and cross may be interchanged( ~B × ~C). ~A=(~C × ~A). ~B= ( ~A× ~B). ~CThe scalar triple product has a simple geometrical interpretation. It represents the volume of the parallelopipedhaving ~A, ~B and ~C as its edge. If ~A, ~B and ~C form the three sides of a paralleopiped, the ~B × ~C is the vectorSn of the base, n being the unit vector normal to both ~B × ~C.~A.( ~B × ~C) = ~A.Sn=S| ~A|cosθBut we have | ~A|cosθ=h, so ~A.( ~B × ~C)=Sh= Volume of a parallelopiped.

Condition of Coplanarity: ~A.( ~B × ~C)=0

2. Vector Triple Product : ~A× ( ~B × ~C)

Now ~B × ~C=

∣∣∣∣∣∣i j k

Bx By BzCx Cy Cz

∣∣∣∣∣∣=i(ByCz −BzCy)− j(BxCz −BzCx) + k(BxCy −ByCx)

Now ~A× ( ~B × ~C) =

∣∣∣∣∣∣i j k

Ax Ay Az(ByCz −BzCy) (BzCx −BxCz) (BxCy −ByCx)

∣∣∣∣∣∣i[Ay(BxCy −ByCx)−Az(BzCx−BxCz)]− j[Ax(BxCy −ByCx)−Az(ByCz −BzCy)] + k[Ax(BzCx−BxCz)−Ay(ByCz −BzCy)]

= i[Bx(AyCy +AzCz +AxCx)− Cx(AxBx +AyBy +AzBz)] + j[By(AxCx +AzCz +AyCy)− Cy(AxBx +

AyBy +AzBz)] + k[Bz(AxCx +AyCy +AzCz)− Cz(AxBx +AyBy +AzBz)]

= ~B( ~A. ~C)− ~C( ~A. ~B)

=BAC-CAB (Remembering Technique)

0.2 Derivative rules of a vector

1.d

dt( ~A+ ~B) =

d ~A

dt+d ~B

dt

2.d

dt( ~A. ~B) = ~A.

d ~B

dt+d ~B

dt. ~A

3.d

dt( ~A× ~B) = ~A× d ~B

dt+d ~A

dt× ~B (Order should be maintained)

4.d

dt[ ~A.( ~B × ~C)] = ~A.( ~B × d~C

dt) + ~A.(

d ~B

dt× ~C) + ( ~B × ~C).

d ~A

dt(order maintained)

5.d

dt[ ~A× ( ~B × ~C)] = ~A× ( ~B × d~C

dt) + ~A× (

d ~B

dt× ~C) +

d ~A

dt× ( ~B × ~C) (order maintained)

From the relation (2)

if ~A= ~Bd

dt( ~A. ~A) = 2 ~A.

d ~A

dtIf in particular ~A(t) be a vector function of constant magnitude | ~A2| = A2=constant

2 ~A.d ~A

dt=0

S. Ganguly 3

Since ~A 6= 0 ~A andd ~A

dtare perpendicular to each other

0.3 The vector differential operator ∇The differential operator ∇ is defined in cartesian coordinate by,

∇ = i∂

∂x+ j

∂

∂y+ k

∂

∂zAs the symbol ∇ denotes an invariant vector differentiation which requires an operand whether scalar or vectoron which it can operate, its expression in vectorial form should only be considered as Symbolic or formal.Invector analysis there are three fundamental operations with ∇ which are of physical interest.If φ is a scalarfunction and ~V is a vector function of space, these operators are as follows(i) ∇φ (ii) ∇.~V (iii) ∇× ~VA scalar field φ is a continuous function of position (coordinates) x,y,z of a point in a region R of space, thenφ=φ(x, y, z) is called a field scalar and a scalar field is said to exist in the region R. A typically scalar filed issaid to exist in the region R. A typical scalar filed is the distribution of temperature, density, electric potential(non directed quantities).The scalar field may be mapped out graphically by a series of surfaces, such as isothermal surfaces, equipotential surfacesetc on which the scalar function φ has a constant value. Such surfaces are called level surface. They cannotintersect each other.Let the two level surfaces by the constant value φ and φ+dφ of the scalar function. We examine a small portionof the two surfaces near a given point ~A on the surface denoted by the function φ. Let ~r and ~r + dr be theradius vectors from the origin to the points A and B respectively in Fig.1 From the figure it follows that AC

Figure 1:

is the least distance between the two surfaces in the direction of normal unit vector n at A which is of length

dn. If dr be the length of AB, the magnitude of rate of increase of φ at A in the direction AB=∂φ

∂r. In the

direction of the unit normal n this rate of increase is maximum i.e∂φ

∂nis maximum. The vector n

∂φ

∂ngive the

maximum rate of increase in φ at any point on the level surface. This vector is called the gradient φat any point and is written as

grad φ=n∂φ

∂n

Thus we see that the gradient of a scalar field is a vector field, the magnitude of which at any point is equal tothe maximum rate of increase of φ and the direction is perpendicular to the level surface at that point.(In case of equipotential surface the electric field intensity E at any point is in the direction of the maximumrate of decrease of potential φ i.e normal to the equipotential surface and has a magnitude equal to the rate ofdecrease of potential)

4 S. Ganguly

We consider the vector ~∇φ= i∂φ

∂x+ j

∂φ

∂y+ k

∂φ

∂zwhere the three terms are the vector rate of increase of φ in

the direction of X,Y,Z.

Now, (grad φ). ~dr=(n∂φ

∂n). ~dr =(

∂φ

∂n)drcosθ= dφ

Now φ=φ(x,y,z)

dφ=∂φ

∂xdx+

∂φ

∂ydy +

∂φ

∂zdz

=(i∂

∂xφ+ j

∂

∂yφ+ k

∂

∂zφ).(idx+ jdy + kdz)

=(~∇.φ).d~r=(grad φ).(d~r)

gradφ = ~∇φ = i∂φ

∂x+ j

∂φ

∂y+ k

∂φ

∂z

∗ If we consider a surface given by the equation

φ(x, y, z)= constant

then,dφ

dt= 0 and from the relation (~∇φ).

d~r

dt=dφ

dt, we get (~∇φ).

d~r

dt=0

Therefore ~∇φ must be perpendicular tod~r

dt, that is ~∇φ is a vector normal to the surface φ(x, y, z)=constant

at every point.∗ If φ1 and φ2 be two differentiable scalar functions in the region R, then we have,

(i) ~∇(φ1 + φ2)=~∇φ1+ ~∇φ2

(ii) ~∇(φ1φ2) = φ1~∇φ2 + φ2~∇φ1

(iii) ~∇(cφ1)=c~∇φ1

0.3.1 Divergence of a field vector

Let ~A(x, y, z) be a continuously differentiable vector point function. Then the scalar product of ~∇. ~A is known

as divergence of ~A

~∇. ~A= (i∂

∂x+ j

∂

∂y+ k

∂

∂z).(iAx + jAy + kAz)

=∂Ax∂x

+∂Ay∂y

+∂Az∂z

~∇. ~A is a scalar function unlike ~∇φ

Physical meaning of divergence of a vector div ~A =

limdV→0

∮~A.d~s

dV

That means the divergence is net flux per unit volume. If at a point P, it happens that~∇.~V= 0 means ~V is said to be divergenceless or Solenoidal and there no inflow or outflow of flux occurs.

~∇.~V > 0 → then ~V has a source at P and outflow of flow occurs~∇.~V < 0 → then ~V has a sink at P and inflow of flux occursWe can establish a physical meaning for the divergence of a vector by an illustration from hydrodynamics. Con-sider the flow of a fluid of density ρ as indicated in Fig.2 Consider the flow of a fluid of density ρ, as indicatedby Fig. 1.2. If ~v be the velocity of flow, then the vector ~A = ρ~v represents the mass of the fluid flowing throughunit area per second normal to the surface ABCO.The y component of ~A through ABCO per unit time is Ay(y)dxdz i.e Aydxdz The fluid flow through the oppo-site face FEDG per second is given by,

Ay(y + dy)dxdz = (Ay(y) +∂Ay∂y

dy + ......)dxdz,

expanding by Taylor’s series and neglecting the higher-order terms. So, the net increase in the mass of fluid inthe volume element dτ(=dxdydz) per second to flow through the two faces is given by,

S. Ganguly 5

Figure 2:

Aydxdz − (Ay +∂Ay∂y

)dxdz

=-∂Ay∂y

dτ

Similarly, there will be two more such terms as:

-∂Ax∂x

dτ and -∂Az∂z

dτ

due to the flow through the two other pairs of opposite faces,namely BEFA-OGDC and FGOA-EDCB. Thetotal increase mass of fluid per unit volume per unit time due to excess of inward flow over the outward is:

−

∂Ax∂x− ∂Ay

∂y− ∂Az

∂z

dτdτ

=-~∇. ~A

which is the rate of increase of the density of the fluid in the volume element dτ ,

∂ρ

∂t= −~∇. ~A

which is the well known equation of continuity of hydrodynamics.

In case the fluid is incompressible,∂ρ

∂t=0, therefore ~∇. ~A =0 (i.e. ~A is solenoidal)

0.4 Curl of a Vector:

~∇× ~A =

∣∣∣∣∣∣∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

Ax Ay Az

∣∣∣∣∣∣∣∣∣∣∣∣If ~∇× ~A=0, then the vector ~A is classified as irrotational. The physical meaning of

the curl of a vector is related to the rotation or circulation of a vector field.

Angular velocity of a rotor: Meaning of Curl

Let a rigid body rotate with a constant angular velocity ω about an axis through a point O in the body.Also let P be a point in the body such that ~OP=~r. Linear velocity of P is given by ~V = ~ω × ~r.If along with rotation the body moves as a whole with a linear velocity ~v0, the total linear velocity of P is givenby,(see Fig. 3)

6 S. Ganguly

~V = ~v0 + ω × ~r

Figure 3:

~∇× ~V = ~v0 + ~∇× (ω × ~r)

=~∇× (ω × ~r) [~v0 is constant]

Now ~V = ~ω × ~r

=

∣∣∣∣∣∣∣∣∣∣i j k

ωx ωy ωz

x y z

∣∣∣∣∣∣∣∣∣∣= i(zωy − yωz)− j(zωx − xωz) + k(yωx − xωy)

~∇× ~V =

∣∣∣∣∣∣∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

(zωy − yωz) (zωx − xωz) (yωx − xωy)

∣∣∣∣∣∣∣∣∣∣∣∣= i(2ωx) + j(2ωy) + k(2ωz)

= 2~ω

So, ~ω =1

2(~∇× ~V )

If ~∇× ~V = 0 , the motion is irrotational

Important Relations1. ~∇.( ~A+ ~B) = ~∇. ~A+ ~∇. ~B2. ~∇× ( ~A+ ~B) = ~∇× ~A+ ~∇× ~B

Laplacian Operator (∇2):~∇2φ=~∇. ~∇φ

= (i∂

∂x+ j

∂

∂y+ k

∂

∂z).(i

∂φ

∂x+ j

∂φ

∂y+ k

∂φ

∂z)

S. Ganguly 7

=∂2φ

∂x2+∂2φ

∂y2+∂2φ

∂z2

So, ~∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2

Prob:1 Prove that ~∇.(~∇× ~A) = 0

=(i∂

∂x+ j

∂

∂y+ k

∂

∂z).

∣∣∣∣∣∣∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

Ax Ay Az

∣∣∣∣∣∣∣∣∣∣∣∣=0

Important Relation:

~∇× (~∇× ~A) = ~∇(~∇. ~A)− ~∇2 ~A

This is using the relation ~A× ( ~B × ~C) = ~B( ~A. ~C)− ~C( ~A. ~B)

Prove that (~∇× ~∇φ = 0)

~∇× ~∇φ =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

∂φ

∂x

∂φ

∂y

∂φ

∂z

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣=0

Physics importance of the above formulaif ∇× ~E=0; ~E = −~∇VWhere ~E is the electric field and V is the potential.

Some Important Identities:

1. ~∇.(φ ~A) = φ(~∇. ~A) + ~A.~∇φ2. ~∇× (φ ~A) = φ(∇× ~A) + (~∇φ)× ~A

3. ~∇.( ~A× ~B) = ~B.(~∇× ~A)− ~A.(~∇× ~B)

4. ~∇× ( ~A× ~B) = ~A(~∇. ~B)− ~B(~∇. ~A) + ( ~B.~∇) ~A− ( ~A.~∇). ~B

5.~∇× (~∇× ~A) = ~∇(~∇. ~A)− ~∇2 ~AProof of relation 2:

~∇× (φ ~A) =

∣∣∣∣∣∣∣∣i j k

∂

∂x

∂

∂y

∂

∂zφAx φAy φAz

∣∣∣∣∣∣∣∣=i(φ

∂Az∂y−φ∂Ay

∂z)+ i(Az

∂φ

∂y−Ay

∂φ

∂z)+ j(φ

∂Ax∂z−φ∂Az

∂x)+ j(Ax

∂φ

∂z−Az

∂φ

∂x)+k(φ

∂Ay∂x−φ∂Ax

∂y)+k(Ay

∂φ

∂x−

Ax∂φ

∂y)

=φi(∂Az∂y−∂Ay∂z

)+φj(∂Ax∂z−∂Az∂x

)+φk(∂Ay∂x−∂Ax∂y

)+i(Az∂φ

∂y−Ay

∂φ

∂z)+j(Ax

∂φ

∂z−Az

∂φ

∂x)+k(Ay

∂φ

∂x−Ax

∂φ

∂y)

=φ(~∇× ~A) + (~∇φ)× ~AProof of Relation 4

~∇× ( ~A× ~B) =

∣∣∣∣∣∣∣∣i j k

∂

∂x

∂

∂y

∂

∂z( ~A× ~B)x ( ~A× ~B)y ( ~A× ~B)z

∣∣∣∣∣∣∣∣

8 S. Ganguly

x component of ~∇× ( ~A× ~B)

=∂

∂y( ~A× ~B)z −

∂

∂z( ~A× ~B)y

=∂

∂y(AxBy −AyBx)− ∂

∂z(AzBx −AxBz)

=Ax(∂By∂y

+∂Bz∂z

)−Bx(∂Ay∂y

+∂Az∂z

) + (By∂

∂y+Bz

∂

∂z)Ax − (Ay

∂

∂y+Az

∂

∂z)Bx

=Ax(∂By∂y

+∂Bz∂z

+∂Bx∂x

)−(Ay∂

∂y+Az

∂

∂z+Ax

∂

∂x)Bx−Bx(

∂Ay∂y

+∂Az∂z

+∂Ax∂x

)+(By∂

∂y+Bz

∂

∂z+Bx

∂

∂x)Ax

=Ax(~∇. ~B)− ( ~A.~∇)Bx −Bx(~∇. ~A) + ( ~B.~∇)AxSimilarly we get other two components, adding them up we get,

~∇× ( ~A× ~B) = ~A(~∇. ~B)− ~B(~∇. ~A) + ( ~B.~∇) ~A− ( ~A.~∇). ~BProb 2 Find a unit vector perpendicular to the surface x2 + y2 − z = 1 at the point P(1,1,1)Ans: Here φ(x, y, z) = x2 + y2 − z~∇φ=2xi+ 2yj − kso, (~∇φ)P=(2i+ 2j − k).Unit vector normal to P is given by,

n =~∇φ|~∇φ|

=2i+ 2j − k

3

Prob 3 Prove that ~∇2(1

r) = 0 where r2 = x2 + y2 + z2

Ans:∂

∂x(1

r) = − x

(x2 + y2 + z2)3/2

∂2

∂x2(

−x(x2 + y2 + z2)3/2

) =3x2

(x2 + y2 + z2)5/2− 1

(x2 + y2 + z2)3/2

∂2

∂y2(1

r) =

3y2

(x2 + y2 + z2)5/2− 1

(x2 + y2 + z2)3/2

∂2

∂z2(1

r) =

3z2

(x2 + y2 + z2)5/2− 1

(x2 + y2 + z2)3/2

∂2

∂x2(1

r) +

∂2

∂y2(1

r) +

∂2

∂z2(1

r) =

3(x2 + y2 + z2)

(x2 + y2 + z2)5/2− 3

(x2 + y2 + z2)3/2

=0

Prob 4: Show that rn~r is an irrotational vector for any value of n but it is solenoidal if n=-3. ~r being theposition vector of a point.

Ans: ~∇× (rn~r) = rn(~∇× ~r)− ~r × ~∇(rn)

~∇× ~r =

∣∣∣∣∣∣∣∣i j k

∂

∂x

∂

∂y

∂

∂zx y z

∣∣∣∣∣∣∣∣= 0

~∇(rn) = (i∂

∂x+ j

∂

∂y+ k

∂

∂z)(x2 + y2 + z2)n/2

=in

2(x2 + y2 + z2)

n

2−1

.2x+ jn

2(x2 + y2 + z2)

n

2−1

.2y + kn

2(x2 + y2 + z2)

n

2−1

.2z

=n(x2 + y2 + z2)

n

2−1

(xi+ yj + zk)

=nrn−2~r

S. Ganguly 9

Now ~r × ~∇(rn) = ~r × nrn−2~r=0so, rn~r is an irrotational vector for any value of n.~∇(rn~r) = rn(~∇.~r) + ~r.(~∇rn)

= 3rn + ~r.nrn−2~r=3rn + nrn−2r2

=(3+n)rn

The expression would thus become zero for n=-3. For n=-3 the vector rn~r is solenoidalProblem 5: Find the directional derivatives of φ = x2yz + 4xz2 at the point (1,-2,-1) in the direction of the

vector 2i− j − 2k.Ans: φ(x, y, z) = x2yz + 4xz2

~∇φ = (i∂

∂x+ j

∂

∂y+ j

∂

∂z)(x2yz + 4xz2)

=i(2xyz + 4z2) + j(x2z) + k(x2y + 8xz)

~∇φ|x=1,y=−2,z=−1=8i− j − 10k

Let ~r=2i− j − 2k, then the unit vector along ~r = n=~r

r=

2i− j − 2k

3

The directional derivative of φ(x, y, z) along ~r is given by,dφ

dr= ~∇φ.n

= (8i− j − 10k).(2i− j − 2k

3)

=37

3

Prob 6: The surfaces x2 + y2 + z2 = 9 and z = x2 + y2 = 3 intersect at (2,-1,2). Find the angle betweenthe surfaces at the point of intersection.Ans: We know that the angle between the surfaces is equal to the angle between their normal. A normal tox2 + y2 + z2 = 9 at (2,1,-2) is

~∇φ1=~∇(x2 + y2 + z2 − 9) = 2xi+ 2yj + 2zk~∇φ1|x=2,y=−1,z=2 = 4i− 2j + 4k

Similarly, a normal to the surface ~∇φ2=~∇(x2 + y2 − z − 3) is ~∇φ2|x=2,y=−1,z=2 = (4i− 2j − k)If θ be the angle between the surfaces φ1=constant and φ2= constant then,~∇φ1.~∇φ2=|~∇φ1||~∇φ2|cosθ at (2,1,-2)

cosθ=16

6√

21= 0.5819

θ = 54o25′

Prob 7: If ~A is a constant vector, then prove that ~∇.(~r. ~A) = ~A

Ans: We have ~r = xi+ yj + zk)

~A = A1i+A2j +A3k

~∇(~r. ~A) = (i∂

∂x+ j

∂

∂y+ k

∂

∂z)(xA1 + yA2 + zA3)

= iA1 + jA2 + kA3

= ~A (proved)

Prob 8: Prove that ∇2(lnr) =1

r2

Ans: We have ~r = xi+ yj + zkr2 = ~r.~r = x2 + y2 + z2

r = (x2 + y2 + z2)

1

2

lnr = ln(x2 + y2 + z2)

1

2

10 S. Ganguly

=1

2ln(x2 + y2 + z2)

Now, ∇(lnr) =1

2∇(x2 + y2 + z2) =

1

2(i∂

∂x+ j

∂

∂y+ k

∂

∂z)ln(x2 + y2 + z2)

=1

2[i

2x

x2 + y2 + z2+ j

2y

x2 + y2 + z2+ k

2z

x2 + y2 + z2]

=1

2(2xi+ 2yj + 2zk

x2 + y2 + z2)

=~r

r2

Prob 9: Given that ~B = i(x2 + 2yz) + j2yz − k(z2 + 2zx). Find a vector ~A such that ~∇× ~A = ~B. Is the

Solution of ~A is unique?

Ans: We have to find a vector ~A whose components satisfy the equation

∣∣∣∣∣∣∣∣i j k

∂

∂x

∂

∂y

∂

∂zAx Ay Az

∣∣∣∣∣∣∣∣ = ~B

∂Az∂y− ∂Ay

∂z= x2 + 2yz

∂Ax∂z− ∂Az

∂x= 2yz

∂Ay∂x− ∂Ax

∂y= −(z2 + 2zx)

Choose Az=0, Then the equation becomes,

−∂Ay∂z

= x2 + 2yz (1)

∂Ax∂z

= 2yz (2)

∂Ay∂x− ∂Ax

∂y= −(z2 + 2zx) (3)

From Equation (1) integrating with respect to z we get,

Ay = −x2z − yz2 + f1(x, y) (4)

Integrating equation(2) with respect to z

Ax = yz2 + f2(x, y) (5)

Differentiating Eqn.4 with respect to ”x”

∂Ay∂x

= −2xz +∂f1∂x

(6)

Differentiate eqn. 5 with respect to y we get,

∂Ax∂y

= z2 +∂fz∂y

(7)

Substituting Eqn.6, 7 in Eqn. 3

= −2xz +∂f1∂x− z2 − ∂f2

∂y= −z2 − 2zx

∂f1∂x

=∂f2∂y

let f1=0, then f2(x, y)=c (constant)

Ax = yz2 + cAy = −x2z − yz2Az = 0The choice is not unique.

S. Ganguly 11

0.5 Properties of vectors under rotations:

The coordinate axes is rotated about Z axis. From the Fig.4x′ = OC ′ + C ′B′

As C ′D = DBsinθ

DB′ = ADsinθ

x′ = xcosθ + C ′D +DB′

=xcosθ + ysinθ (As DB+BD=AD)

Similarly, from the Figure, y′ = −xsinθ + ycosθ

So, A′x = Axcosθ +Aysinθ

B′x = Bxcosθ +Bysinθ

A′y = −Axsinθ +Aycosθ

B′y = −Bxsinθ +Bycosθ

Az′=Az ; B′z = BzThe dot product ~A′. ~B′ = A′xB

′x +A′yB

′y +A′zB

′z

Figure 4:

=(Axcosθ +Aysinθ)(Bxcosθ +Bysinθ) + (−Axsinθ +Aycosθ)(−Bxsinθ +Bycosθ) +AzBz

=AxBxcos2θ+AxBycosθsinθ+AyBxsinθcosθ+AyBysin

2θ+AxBxsin2θ−AxBysinθcosθ−AyBxcosθsinθ+

AyBycos2θ +AzBz

=AxBx +AyBy +AzBz

So, the scalar product is invariant under coordinate rotation.

0.6 Idea of linear independence:

A set of vectors αi, i=1,2,,....n in a vector space V is said to be linearly dependent if there exists acorresponding set of scalarsai, i=1,2,....,n not all zero such that,

n∑i=1

aiαi = 0

On the other hand a set of vectors αi, i=1,2,3,....n in a vector space V is said to be linearly independent, if

n∑i=1

aiαi = 0, whereai = 0 for all i′s

A set of non-zero vectors in a vector field V is linearly dependent if and only if someelements of that set is a linear combination of others.

12 S. Ganguly

Let αi, i=1,2,3,......n be a set of vectors in the vector space V over that field F which are linearly dependent.Then there exists a set of scalars ai, i=1,2,.....n in F such that

∑ni=1 aiαi = 0

As the set of vectors is linearly dependent, at least one of the scalars ai, say ak 6= 0 (k < n). Then from theabove equation

akαk = −n∑i=1

aiαi(i 6= k)

So, αk = −∑ni=1(aia

−1k )αi (i 6= k)

This shows that the vector αk can be expressed as a linear combination of others.Conversely, let αm of the set of vectors αi, i=1,2,....,n which can be expressed as a linear combination of theother elements of αi, i.e.,

αm =

n∑i=1

biαi (i 6= m)

where the scalar bsi are in F.Now, from the postulate of vector space,αm + (−1)αm = 0

or,∑biαi + (−1)αm=0

Hence, the set of vectors αi, i=1,2,3,...n is linearly dependent.Prob 10: Find whether the vectors (i+ 2j + 3k), (2i− j)and(3i− 2k) are linearly dependent or not?

Ans: If it is linearly independent then a(i+ 2j + 3k) + b(3i− j) + c(3i− 2k) = 0a+3b+3c= 0

2a-b=03a-2c=0

yields a=b=c=0So the vectors are linearly independent.Basis of a Vector Space:If an arbitrary vector ψ in a vector space V over the field F can be expressed as a linear combination of thevectors αi in V so that,

ψ =

n∑i=1

ai ~αi

where ai are scalar, then ~αi is said to span the vector space V.At the same time if ~αi’s are linearly independentthen the set S is a basis for ψ. Suppose another set T= ~α1, ~α2, ~α3...... ~αn, ~α1 + ~α2 are the set of vectors whichmay call the span of T in vector space V, but T is not linearly independent so T is not the basis of V.Prob: 11 Show that the vectors (2,3), (7,0) form a basis of the vector space V2 over the fields

Ans: First we have to check whether the vectors creates a span(s) of V2So, 2c1+7c2=x13c1=x2c1=

x23

c2=x17− 2

21x2

wher x1 and x2 are any two basis vectors in the space V2

checking of linear independence:

Choose x1 and x2=0 which gives c1 = c2 = 0

This proves the set of vectors are the basis of V2

Prob: 12 Let ~∇φ = 2xyz3i+ x2z3j + 3x2yz2k Find φ(x, y, z) if φ(1,−2, 2) = 4

Ans: i∂φ

∂x+ j

∂φ

∂y+ k

∂φ

∂z= 2xyz3i+ x2z3j + 3x2yz2k

i∂φ

∂x= 2xyz3i

S. Ganguly 13

φ = yz3x2 + f(y, z)Similarly for the other components,φ = x2z3y + f(x, z)φ = x2yz3 + f(x, z)φ = x2yz3 + f(x, y)Now given that φ(1,−2, 2) = 4Substitute it in any above equation we get,4=-16 + f(y,z)f(y,z)=20φ = yz3x2 + 20

Prob.13 Show that the vector field ~V = − xi− yj√x2 + y2

is a ”Sink Field”.

Ans: We know that if ~∇. ~A is negative at a point in the vector field, then the vector field ~A is called a sink field.

~V = − xi− yj√x2 + y2

~∇.~V = −~∇.[ xi+ yj√x2 + y2

]

=[i∂

∂x+ j

∂

∂y+ k

∂

∂z].[

xi+ yj√x2 + y2

]

=-[∂

∂x(

x√x2 + y2

) +∂

∂y(

y√x2 + y2

)]

=-1√

x2 + y2

So, ~∇.~V is negative, so V is a sink field.

0.7 Vector Integral

If a vector ~B(u) such that its derivative with respect to u is equal to ~A(u), then,

~A(u) =d

du~B(u)∫ ~A(u)du = ~B(u) + C

For definite integral∫ nm

~A(u)du =∫ nm

[d

du~B(u)]du

= ~B(n)− ~B(m)

Three types of vector integration can observed 1. Line integration 2. Surface Integration 3. Volume Inte-gration .

Line integralIt refers to the integration of a vector along a line that is a curve. ~A is the direction of tangent to the curve atthe point P(x,y,z).~dl is an element of displacement at P(x,y,z) directed along the tangent.

I=∫ P2

P1

~A.~dl

=∫ P2

P1

~A.~dl =∫ P2

P1Adlcosθ

=∫ P2

P1(Axdx+Aydy +Azdz)

Example:-Suppose ~F (r) is the variable force. The work done in moving a particle from P1 to P2 is defined as,

W=∫ P2

P1

~F . ~dr =∫ v2v1md~v

dt. ~dr

= m∫ v2v1

d~v

dt.~vdt

14 S. Ganguly

= m∫ v2v1d(~v.~v)

=1

2m(v22 − v21)

=∆ KE

v1 and v2 are the velocities at point P1 and P2

line Integral∮~A.~dl is a closed line integration of the vector It is also called the circulation of the vector ~A in a given closed

curve. If ~F = − ~∇φ where φ= some scalar function.∮~F . ~dr = −

∮~∇φ. ~dr =

∮dφ = 0

Which implies that final Kinetic Energy=Initial Kinetic Energy , i.e. energy is conserved.Prob 14: If ~F = (2x+ y2)i+ (3y− 4x)j evaluate

∮~F . ~dr around the right angled triangle OAB having vertices

O(0,0), A(2,0), B(2,1) and right angled at A.

Ans: ~F . ~dr = (2x+ y2)dx+ (3y − 4x)dy

Figure 5:

∮c~F . ~dr =

∮OA

~F . ~dr +∮AB

~F . ~dr +∮BO

~F . ~drALONG OA:

y=0; dy=0, x varies from 0 to 2∮OA

~F . ~dr = 2∫ 2

x=0xdx = 4

ALONG AB:

x=2, dx=0, y varies from 0 to 1∮AB

~F . ~dr =∫ 1

0(3y − 8)dy = −dfrac132

Equation of BO:

y − 1

1− 0=x− 2

2− 0

x=2y

∮BO

~F . ~dr =∫ 0

x=2(2x+

x2

4)dx+

∫ 0

x=2

3.x/2− 4x

2dx

=-13

6∮~F . ~dr = 4− 13

2− 13

6= −14

3

Prob 15: Show that if the line integral of a vector ~F vanishes over a closed path C, the vector can bewritten as the gradient of a scalar function φ called potential.

Ans:∮~∇φ.~dl =

∫ BA~∇φ.~dl +

∫ AB~∇φ.~dl

=∫ BAdφ+

∫ ABdφ

S. Ganguly 15

=(φB − φA) + (φA − φB)

=0=∮c~F .~dl

Since ~dl is completely arbitrary, we may write

~F = ~∇φ = gradφ

Prob 16: Determine whether the integral∫

(2xyz2)dx + (x2z2 + zcosyz)dy + (2x2yz + ycosyz)dz is inde-

pendent of the path of integration. Find the corresponding potential, and evaluate it from (1,0,1) to (0,π

2,1)

Ans: ~F = 2xyz2i+ (x2z2 + zcosyz)j + (2x2yz + ycosyz)k~∇× ~F = 0Hence the line integral is independent of path.dφ = ~∇φ. ~dr = ~F . ~dr

dφ = 2xyz2dx+ (x2z2 + zcosyz)dy + (2x2yz + ycosyz)dz

=(2xdx)yz2 + x2(dy)z2 + x2y(2zdz) + [(cosyzdz)z + (cosyzdz)y]=d(x2yz2) + d(sinyz)=d(x2yz2 + sinyz)φ = x2yz2 + sinyz

[φ]BA =

[x2yz + sinyz

]0,π2 ,11,0,1

=1-0

=1

Prob 17 ABCD is a square in the plane. The x,y co-ordinate of A,B,C and D are (0,0), (a,0), (a,a) and

(0,a) respectively.Show that the line integral of ~F = (2xy + z3)i + x2j + 3xz2k is zero over ABCDA. What isthe corresponding potential?

Ans: ~dr = dxi+ dyj∮~F . ~dr =

∮(2xy + z3)dx+ x2dy

Figure 6:

A→ B y=0, dy=0∮(2.x.0 + 03)dx = 0

=0

B → C x=a, dx=0; y varies from 0 to a∫ a0a2dy = a3

16 S. Ganguly

C → D y=a, dy=0

∫ 0

a2axdx = −a3

D → A x=0∮~F . ~dr = 0

Then∮~F . ~dr = 0

So, ~F = ~∇φ

(2xy + z3)i+ x2j + 3xz2k = i∂φ

∂x+ j

∂φ

∂y+ k

∂φ

∂z

∂φ

∂x= (2xy + z3)

φ = yx2 + z3x+ f(y, z)

∂φ

∂y= x2

φ = x2y + f(x, z)

∂φ

∂z= 3xz2

φ = xz3 + f(x, y)

φ = x2y + xz3 +constantProb 18: If ~F = (2x + y)i + (3y − x)j then evaluate

∫c~F . ~dr where C is the curve in XY plane consisting of

the straight line segments from (0,0) to (2,0) and then (3,2).

Ans:∮~F . ~dr =

∫c(2x+ y)dx+ (3y − x)dy

O → A

x=0 to 2 , y=0∮~F . ~dr =

∫ 2

02xdx = 2× 2 = 4

A→ B

Straight line joining (2,0) to (3,2)

x− 2

3− 2=y − 0

2− 0= t

x = t+ 2; y = 2t

dx=dt dy=2dt∫ t=1

t=0(2(t+ 2) + 2t)dt+ (6t− t− 2)2dt∫ t=1

t=0(2t+ 4 + 2t+ 12t− 2t− 4)dt

14∫ t=1

t=0tdt

= 7∮~F . ~dr=4+7=11

S. Ganguly 17

0.8 Surface Integral: Flux of a vector

Consider a surface S of any arbitrary shape Fig.7. in the field vector ~A. Divide S into an infinite number ofelementary area ds’s. Any elementary area may be considered as a vector surface ~ds .

~A. ~ds = Acosθds is said to be the flux of the vector ~A through the area ~ds , θ being the angle between ~A and ~ds∫∫s~A. ~ds over a closed surface S is called the the flux of ~A through the surface.∫∫

s~A. ~ds =

∫∫∫s~A.nds

Figure 7:

0.9 Examples of double integral

Prob: 19 Evaluate∫∫R

(x+ y)dydx , R is the region bounded by x=0, x=2, y=x and y=x+2Ans: Let I =

∫∫R

(x+ y)dxdy

Figure 8:

18 S. Ganguly

The limits are x=0,x=2, y=x and y=x+2

I =∫ 2

0dx∫ x+2

x(x+ y)dy

=∫ 2

0[xy +

y2

2]x+2x dx

=∫ 2

0[x(x+ 2) +

(x+ 2)2

2− x2 − x2

2]dx

=2∫ 2

0(2x+ 1)dx

=12

Problem 20: Evaluate∫∫

r3drdθ over the area bounded by the circles r = 2cosθ and r = 4cosθAns: Let I =

∫∫r3drdθ

Figure 9:

=∫ π2−π

2

dθ∫ 4cosθ

2cosθr3dr

=1

4

∫ π2−π

2

dθ(256cos4θ − 16cos4θ)

=240

4

∫ π2−π

2

cos4θdθ

=120∫ π20 cos4θdθ

=45π

2[as cosθ is even function]

0.10 Volume Integral

Let φ be any scalar field, ~A any field vector and dV(=dxdydz) an infinitesimal volume in the field, then

∫∫∫V~AdV is defined as the volume integral.

S. Ganguly 19

Gauss’s Divergence TheoremThe surface integral of the normal component of a vector taken around a closed surface is equal to the integralof the divergence of the vector taken over the volume enclosed by the surface.

If V be the volume bounded by a closed surface S and ~A is a vector function of position with continuousderivatives, then according to the divergence theorem of Gauss,∫∫

s~A.nds =

∫∫∫V

(~∇. ~A)dV

∫∫∫s~A.nds as the limit of sum:

We consider the outward surface S of a body in a vector field ~A . Let ∆SP be an element of surface S surround-ing point Pp where the vector field is Ap and the outward normal is np (see figure 10) The normal flux of ~Ap at

Figure 10:

Pp is∑Mp=1

~Ap.np∆Sp where p=1,2,.....M

When M→ ∞, the largest ∆Sp → 0, if the limit exists then∫∫s~A.nds is called the surface integral of the

normal component of ~A over S.

To show∫∫s~A.nds =

∫∫R~A.n

dxdy∣∣∣np.k∣∣∣From the figure 10, the projection of ∆Sp on the xy plane is

∣∣∣(nP∆S).k∣∣∣ =

∣∣∣np.k∣∣∣∆SpBut

∣∣∣(np.k)∣∣∣∆Sp = ∆xp∆yp

∴ ∆Sp =∆xp∆yp∣∣∣np.k∣∣∣

Using this value of ∆Sp in the result of limit of sum of∫∫s~A.nds we get,

∑Mp=1

~Ap.np∆xp∆yp∣∣∣np.k∣∣∣

In the limit, when M →∞ largest ∆xp and ∆yp tend to zero at the same time, we have,

∑Mp=1

~Ap.np∆xp∆yp∣∣∣np.k∣∣∣ =

∫∫R~A.n

dxdy∣∣∣np.k∣∣∣So,

∫∫s~A.nds =

∫∫R~A.n

dxdy∣∣∣np.k∣∣∣

20 S. Ganguly

Proof of Gauss’ Divergence Theorem:We consider a closed surface S of any arbitrary shape drawn in the vector field ~A enclosing a volume V (Figure

Figure 11:

11). A line drawn parallel to the z-axis cuts S in two points, say, P1(x, y, z1) and P2(x, y, z2) (z1 ≤ z2) andmeets at Q(x,y,0) on the projection R of V on the xy plane.

Now,∫∫∫

V(~∇. ~A)dV =

∫∫∫V

(∂A1

∂x+∂A2

∂y+∂A3

∂z)dxdydz

where ~A = iA1 + jA2 + kA3

or,∫∫∫

V(~∇. ~A)dV =

∫∫∫V

∂A1

∂xdxdydz +

∫∫∫V

∂A2

∂ydxdydz +

∫∫∫V

∂A3

∂zdxdydz

Now,∫∫∫

V

∂A3

∂zdxdydz =

∫∫R

[∫ z=f2(x,y)z=f1(x,y)

∂A3

∂z]dxdy

=∫∫R

[A3(x, y, f2)−A3(x, y, f1)]dxdy

Let S1 and S2 be respectively the lower and upper parts of S corresponding to P1 and P2 and n an outwardunit normal to S.

∴ ds =dxdy

n.k, for S2

and ds = −dxdyn.k

, for S1

for n makes an acute angle with k and an obtuse angle at S1 with k.

∴∫∫RA3(x, y, f2)dxdy =

∫∫S2A3(n.k)ds

and∫∫RA3(x, y, f1)dxdy = −

∫∫S1A3(n.k)ds

∴∫∫∫

V

∂A3

∂zdxdydz =

∫S2

A3(n.k)ds+

∫S1

A3(n.k)ds =

∫s

A3(n.k)ds (8)

Like wise we can show, ∫∫∫V

∂A2

∂ydxdydz =

∫s

A2(n.j)ds (9)

∫∫∫V

∂A1

∂xdxdydz =

∫s

A1(n.i)ds (10)

S. Ganguly 21

Now, adding equations (8), (9) and (10) , we get,∫∫∫V

(∂A1

∂x+∂A2

∂y+∂A3

∂z)dxdydz =

∫s(iA1 + jA2 + kA3).nds

or,

∫∫∫V

~∇. ~AdV =

∫∫s

~A.nds

Prob 21: Prove∫∫∫

V(φ ~∇2ψ − ψ ~∇2φ)dV =

∫∫s(φ~∇ψ − ψ~∇φ). ~ds where V is the volume bounded by the

surface S and φ, ψ are scalar field.Ans: From Gauss’s Divergence Theorem,∫∫∫

V~∇. ~AdV =

∫∫s~A.nds

Let ~A = φ~∇ψ

~∇.(φ~∇ψ) = φ ~∇2ψ + ~∇φ~∇ψ∫∫∫V~∇.(φ~∇ψ) =

∫∫s(φ~∇ψ).nds

∫∫∫V

(φ ~∇2ψ + ~∇φ~∇ψ)dv =

∫∫s

(φ~∇ψ).nds (11)

Interchanging φ and ψ we get, ∫∫∫V

(ψ ~∇2φ+ ~∇ψ~∇φ)dv =

∫∫s

(ψ~∇φ).nds (12)

Subtraction equation (11) and (12) we get,∫∫∫V

(φ ~∇2ψ − ψ ~∇2φ)dV =∫∫s(φ~∇ψ − ψ~∇φ). ~ds

All materials are taken from the following references:References: 1. Fundamentals of Mathematical Physics-A.B.Gupta (Publisher: Books and Allied Private

Limited; Edition: 2014)2. Principles of Mathematical Physics-S.P. Kuila (Publisher: New Central Book Agency(P) Limited; Edition2015)3. Vector Analysis- Murry R Spigel (Publisher: Schaum Outline Series)4. Mathematical Physics-H.K.Dass, Rama Verma (S.Chand Publisher Edition: 2015)