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TEXT BOOK OF TRIGONOMETRY, VECTOR CALCULUS

AND ANALYTICAL GEOMETRY
Contents
Unit 1: Lesson
1. Expansions
2. Hyperbolic functions
3. Logarithm of Complex numbers
4. summation of series
Unit 2:
5. vector columns – an Introduction
6. Differential operators
7. Integration of vectors
8. Theorems of Gauss, Stokes and Green
Unit 3:
9. Fourier series
10. Polar coordinates
Unit 4: 11. Analytical Geometry of Three Dimensions
12. Sphere
Unit 5:
13. Cone, Cylinder
14. Coincoids
1
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Unit I
Lesson - 1
Trigonometry
Contents
1.0 Aims and Objectives
1.1 Expansions
1.2 Examples
1.3 Let us sum up
1.4 Check your progress
1.5 Lesson End activities
1.6 Points for discussion
1.7 References

We shall study the expansions of cosnx and sinnx by using the concept of Demoivre’s
Theorem, the concept of combinations and the concept of the Binomial expansion.
1.1 Expansions
1.1.1. Expansions of cos n q and sin n q
We know that
(cos q +i sin q )n = cos n q + i sin n q
cos n q + i sin n q = (cos q +i sin q )n
= cosn q +nc1 cosn-1 q sin q +nc2 cosn-2 q (i sin q )2 +
nc3 cosn-3 q (i sin q )3 +nc4 cosn-4 q (i sin q )4 +………….
= cosn q +i nc1 cosn-1 q sin q - nc2 cosn-2 q sin2 q -i nc3 cosn-3 q
sin3 q + nc4 cosn-4 q sin4 q + i nc5 cosn-5 q sin5 q ………..
= cosn q - nc2 cosn-2 q sin2 q + nc4 cosn-4 q sin4 q ……………
+ i (nc1 cosn-1 q sin q - nc3 cosn-3 q sin3 q + nc5 cosn-5 q
sin5 q ….)
2
Equate real and imaginary parts

Cos n q = cosn q - nc2 cosn-2 q sin2 q + nc4 cosn-4 q sin4 q …………….
Sin n q = nc1 cosn-1 q sin q - nc3 cosn-3 q sin3 q + nc5 cosn-5 q sin5 q ……….
1.2 Examples
(1) Expand cos 6 q inpowers of cos q

Or
Prove that
Cos 6 q = 32 cos6 q +48 cos4 q + 18 cos2 q -1
Proof: we know that
Cos n q = cosn q - nc2 cosn-2 q sin2 q + nc4 cosn-4 q sin4 q ……..
Put n = 6,
Cos 6 q = cos 6 q - 6c2 cos4 q sin2 q + 6c4 cos2 q sin4 q -6c6 sin6 q
6.5 6.5.4.3
= cos 6 q - cos 4 q sin 2 q + cos 2 q sin 4 q - sin 6 q
1.2 1 .2.3.4
= cos 6 q -15 cos 4 q sin 2 q +15cos 2 q sin 4 q - sin 6 q
= cos6 q - 15 cos 4 q (1- cos 2 q ) + 15 cos 2 q ( 1-cos 2 q )2 – (1-cos 2 q )3
= cos6 q - 15 cos 4 q +15 cos6 q + 15 cos 2 q (1+ cos 4 q -2 cos 2 q ) –
(1-cos 6 q - 3 cos 2 q + 3 cos 4 q )
= cos6 q - 15 cos 4 q +15 cos6 q + 15 cos 2 q + 15 cos 6 q - 30 cos 4 q -
1+cos 6 q + 3 cos 2 q - 3 cos 4 q
= 32 cos6 q - 48 cos 4 q +18 cos 2 q -1
sin 6q
2. Expand in powers of cos q
sin q
sin 6q
Prove that = 32 cos5 q - 32 cos 3 q +6 cos q
sin q
Proof : We know that
Sin n q = nc1 cosn-1 q sin q - nc3 cosn-3 q sin3 q + nc5 cosn-5 q sin5 q ……….
Put n = 6
3
Sin 6 q = 6c1 cos5 q sin q - 6c3 cos3 q sin3 q + 6c5 cos5 q sin5 q
6 .5 .4 .
= 6 cos 5 q sin q - cos3 q sin3 q + 6 cos q sin5 q
1 . 2 .3 .
= 6 cos 5 q sin q - 20 cos3 q sin3 q + 6 cos q sin5 q
sin 6q
= 6 cos 5 q - 20 cos3 q sin2 q + 6 cos q sin4 q
sin q
= 6 cos 5 q - 20 cos3 q + 20 cos 5 q + 6 cos q (1+cos4 q -2cos2 q )
= 6 cos 5 q - 20 cos3 q + 20 cos 5 q + 6 cos q +6 cos5 q -12 cos3 q
= 32 cos 5 q - 32 cos3 q 6 cos q
1.3 Let us sum up

So far we have seen the expansion of cosn q and sinn q using Binomial theorem,
Demoivre’s theorem and concept of i2 =-1
1. find sin 2x and cos2x
sin 7q
Prove that = 64 cos6 q - 80 cos4 q + 24 cos4 q - 1
sin q
1. Prove that cos 4 q = 8 cos 4 q - 8 cos 2 q + 1
2. Prove that cos 7 q = cos 7 q - 21 cos5 q sin2 q +35 cos 3 q sin 2 q - 7cos q sin 6 q
sin 6q
3. Prove that = 32 cos 5 q - 32 cos 3 q +6 cos q
sin q
1.6 Point for discussion
1. Prove that
cos 7 q sec q = 64 cos 6 q +114 cos4 q + 56 cos2 q -7
1.7 References
1. Trigonometry by S. Narayanan
4
Lesson – 2
HYPERBOLIC FUNCTIONS
CONTENTS
2.2. Examples
2.3 Let us sum up
2.7 References

Our aim is to define the hyperbolic cosines of x and series of x using
exponentiation.
e x + e-x
Definition 1 : The hyperbolic cosine of x is defined as coshx =
2
e x - e-x
Definition 2: The hyperbolic sine of x is defined as sinhx =
2
e x + e-x
Definition 3: tanhx =
e x + e -x
More results
1. sin (i x) = i sin hx
2. cos (i x) = cos hx
3. tan (i x) = i tan hx
2.2. Examples
Separate into real and imaginary parts
a) sin (α+iβ)
sin (α+iβ) = sin α cos (iβ) + cos α sin (iβ)
= sin α cos hβ + cos α i sin hβ
5
= sin α cos hβ + i cos α sin hβ

Real part = sin α cos hβ
Imaginary part = cos α sin hβ
b) sin (α- iβ) = sin α cos (iβ) - cos α sin (iβ)
= sin α cos hβ - cos α i sin hβ
= sin α cos hβ - i cos α sin hβ
Real part = sin α cos hβ
Imaginary part = - cos α sin hβ
c) cos (α+iβ)
cos (α+iβ) = cos α cos (iβ) - sin α sin (iβ)
= cos α cos hβ - sin α (i sin hβ )
= cos α cos hβ - i sin α sin hβ
Real part = cos α cos hβ

Imaginary part = - sin α sin hβ
d) cos (α- iβ)
cos (α- iβ) = cos α cos (iβ) + sin α sin (iβ)
= cos α cos hβ + sin α (i sin hβ)
= cos α cos hβ + i sin α sin hβ
Real part = cos α cos hβ
Imaginary part = sin α sin hβ
sin(a + ib )
e) tan (α+iβ) =
cos(a + ib )
multiply the numerator and denominator by 2 cos (α-iβ)
2 sin(a + ib ) cos(a - ib )
tan (α+iβ) = ---------(1)
2 cos(a + ib ) cos(a - ib )
numerator = 2 sin (α+iβ) cos (α- iβ)
= 2 sin A cos B; A = α+iβ
B = α- iβ
= sin (A+B) + sin (A-B)
= sin 2 α + sin (2iβ)
6
= sin 2 α + i sinh2β
Denominator = 2 cos (α+iβ) cos (α- iβ)
= 2 cos A cos B; A = α+iβ
B = α- iβ
= cos (A+B) + cos (A-B)
= cos 2 α + cos (2iβ)
= cos 2 α + cos h2β
Using in 1
sin 2a + i sinh 2 b
tan (α+iβ) =
cos 2a + cosh 2 b
sin 2a sinh 2 b
= +i
cos 2a + cosh 2 b cos 2a + cosh 2 b
sin 2a
Real part =
cos 2a + cosh 2 b
sinh 2 b
Imaginary part =
cos 2a + cosh 2 b
f) sin h(α+iβ)
1
sin h(α+iβ) = [i sinh(a + ib )]
i
1
= [sin i (a + ib )]
i
1
= [sin(ia - b )]
i
= - i[sin(ia ) cos b - sin b cos(ia )]
= - i[i sinh a cos b - cosh a sin b ]
= sinh a cos b + i cosh a sin b
Real part = sin hα cos b ;
Imaginary part = cos hα sin b
g) cosh (α+iβ) = cos [i(α+iβ)]
= cos (iα-β)
7
= cos (iα) cos β + sin (iα) sin β

= cosh α cos β + i sinh α sin β
= cos α cos hβ - i sin α sin hβ
Real part = cosh α cos β
Imaginary part = sinh α sin β
1
h) tanh (α+iβ ) = [i tanh(a + ib )]
i
1
= [tan i(a + ib )]
i
1
= [tan(ia - b )]
i
= -i tan (iα-β)
sin(ia - b )
= -i
cos(ia - b )
2 cos( b + ia ) sin( b - ia )
=i ------- (1)
2 cos( b + ia ) cos( b - ia )
Numerator = 2 cos( b + ia ) sin( b - ia )
= 2 cos A sin B; A = β +iα

B = β- iα
= sin (A+B) - sin (A-B)
= sin 2 β - sin (2i α)
= sin 2 β - i sinh2 α
Denominator = 2 cos( b + ia ) cos( b - ia )
= 2 cos A cos B; A = β+iα
B = β- iα
= cos (A+B) + cos (A-B)

= cos 2 β + cos (i2 α)
= cos 2 β + cos h2 α
Using in (1)
8
sin 2 b - i sinh 2a
tan h (α+iβ) = i
cos 2 b + cosh 2a
i sin 2 b + sinh 2a
=
cos 2 b + cosh 2a
sinh 2a sin 2 b
= +i
cos 2 b + cosh 2a cos 2 b + cosh 2a
sinh 2a
Real part =
cos 2 b + cosh 2a
sin 2 b
Imaginary part =
cos 2 b + cosh 2a
Examples:
1. Prove that sinh3x = 3 sin hx + 4 sin h3 x
Proof : sin 3 q = 3 sin q - 4 sin3 q
Put q = ix
Sin (3ix) = 3 sin(ix) – 4 [sin (ix)]3
i sinh3x = i3 sinhx – 4 [i3 sinh3 x]
= i 3sinhx – 4 i3 sinh3 x
i sinh3x = 3 i sinhx + 4 i sinh3 x
/ i; sinh3x = 3sinhx + 4 sinh3 x
2. Express sinh7 q interms of hyperbolic sines of multiples of q
Solution
eq - e -q
Sin h q =
2
2 sinh q = eq - e -q
( 2 sinh q ) 7 = ( eq - e -q )7
27 sinh7 q = e7 q - 7c1 e6 q e- q +7c2 e5 q e-2 q - 7c3 e4 q e-3 q +7c4 e3 q e-4 q -7c5
e2 q e-5 q +7c6 e q e- q 6-7c7 e-7 q
= e7 q - e-7 q -7 e5 q + 7.6 e 3q - 7.6.5 eq + 7.6.5.4 e -q - 7.6 e -3q + 7e -5q

1.2 1.2.3 1.2 .3 .4 1 .2
=( e7 - e-7 -)- 7 e5 + 21 e3 - 35 e +35 e- -21 e-3 + 7 e-5 q
q q q q q q q
=( e7 q - e-7 q )- 7 (e5 q - e-5 q ) + 21 (e3 q - e-3 q ) +– 35 (e q - e- q )
9
÷ 2;
6 e 7q - e -7q æ e 5q - e -5q ö æ e 3q - e -3q ö æ eq - e -q ö

2 sinh7 q = 7çç ÷÷ + 21çç ÷÷ + -35çç ÷÷
2 è 2 ø è 2 ø è 2 ø
= sinh7 q - 7 sinh5 q + 21 sinh 3 q - 35 sinh q
1
Sinh7 q = [sinh 7q - 7 sinh5q + 21 sinh 3q - 35 sinhq ]
26
3. If sin ( q + i f ) = tan α+i sec α , prove that cos2 q cosh2 f = 3
Solution
Sin ( q +i f ) = sin q cos (i q ) + cos q sin (i q )
= sin q cosh q + cos q (i sinh q )

= sin q cosh q + i cos q - sinh q
Sin q cosh q + i cos sinh q = tan α + i sec α
Equate real and imaginary parts
Tan α = sin q cosh q
Sec α = cos q sinh q
We know
sec2 α – tan2 α = 1
cos2 q sinh2 q – sin2 q cosh2 q = 1
é1 + cos 2q ù é cosh 2f - 1ù é1 - cos 2q ù é cosh 2f + 1ù

ê 2 úê 2 ú-ê 2 úê 2 ú =1
ë ûë û ë ûë û
(1+cos2 q ) (cosh2 f -1) - (1-cos2 q ) (cosh2 f +1) = 4
On simplification
Cos2 q cosh2 f = 3
4. If sin (x+iy) = cos q + i sin q , prove that
cos2 x = sinh2 y
Proof : sin (x+iy) = cos q + i sin q
sin x cos (iy) + cos x sin (iy) = cos q + i sin q
10
sin x coshy + cos x (i sinhy) = cos q + i sin q

sin x coshy + i cosx (i sinhy) = cos q + i sin q
equate real and imaginary parts
sin x coshy = cos q
cos x sinhy = sin q
we know cos2 x + sin2 q = 1
sin 2 x cosh2 y + cos2 x sinh2 y = 1
(1 – cos2 x) ( 1+sinh2 y) + cos2 x sinh2 y = 1
1 + sinh2 y – cos2 x – cos2 x sinh2 y + cos2 x sinh2 y = 1
cos2 x = sinh2 y
5. If sin (x+iy) = tan (α+iβ), show that
Sin2α tanhy = sin h2β tan x
Solution :
sin (x+iy) = sin x coshy + i cosx sinhy
Sin2a Sinh2b
tan (α+iβ) =
cos2a+cosh2b cos2a+cosh2b
Sin2a Sinh2b
sin x coshy = -------------- (1)
cos2a+cosh2b cos2a+cosh2b
Sin2a Sinh2b
cosx sinhy = cos2a+cosh2b
cos2a+cosh2b -------------- (2)
(1) ¸ (2)
Sin2a
tanx cothy =
Sinh2b
tanx sinh2b = sin2a tanhy =
2.3. Let us Sum up

So far we have studied the concept on finding the real and imaginary parts of
trigonometric fuction using cos(ix)=coshx: sin(ix)=isinhx
11

(i) Prove that cosh2 x=1+sinh2 x
(ii) Prove that sinh2x=2sinhx coshx
2.5 Lesson end activities
1. Prove that cosh 3x = 4 cosh3x – 3coshx
2. Express cosh6 q in a series of hyperbolic cosines of multiples of q
3. If tan ( q +i q ) = Sin (x+iy), prove that cothy sinh2y = cotx sin2 q
4. If sin ( q +i q ) = R (cos α + i sin α),
Prove that
2R2 = cosh2 f - cos2 q
And tan α tan q = tanh f
5. If (cos (a+ib) = α+iβ, prove that
i) (1 + α)2 + β2 = (cosh b + cos a)2
ii) (1 - α)2 + β2 = (cosh b - cos a)2
6. If cos (A+iB) = cos q + i sin q , prove that sin q = ± sin2 A
7. If cos (A+iB) = cos q + i sin q , prove that sin q = ± sin2 B
8. If sin (A+iB) = l +iy, prove that
l2 y2
(i) + =1
cosh 2 B sinh 2 A
l2 y2
(ii) + =1
sin 2 A cos 2 A
1
Answer : 2 : [cosh 6q + 6 cosh 4q + 15 cosh 2q + 10]
32
2.7 References
Trigonometry by Rasinghamic and Aggarval
12
Lesson - 3
Logarithm of a complex number
Contents
3.0 Aim and Objectives
3.1 To find log e (x+iy)
3.2 Examples
3.3 Let us sum up
3.6 References
In this lesson we are going to see the definition of logarithm of a complex number
using the fundamental concepts of logarithm of a function.
Definition 1: Let Z = x+iy, if log z = u, then
Z = eu
In general the logarithm of a complex number is also a complex number
3.1 To find log e (x+iy)
Let log e (x+iy) = α+iβ

x+iy = e α +iβ
= e α e iβ
= e α (cos β + i sin β)
x+iy = e α cos β + i eα sin β
equate real, imaginary parts
x = e α cos β (1)
y = eα sin β (2)
(1)2 + (2)2 gives

x2 + y2 = e 2α cos2 β + e2α sin2 β
13
= e 2α (cos2 β + sin2 β)
x2 + y2 = e 2α
2 α = loge (x2 + y2 )
Α = ½ log (x2 + y2 )
y
2 ¸ 1 gives = tan b
x
b = tan -1 æç y ö÷
è xø
1
\ l +oe xg (=i ) ly o+ xg2 ( y+)2 t i a1 -n y
2 x ( )
To find general logarithm of a complex number
Let log e ( x + iy ) = a + ib
x+iy = e α +iβ
= e α e iβ
= e α cos β + i sin β
x+iy = e α (cos (2nπ+β) +i sin (2nπ+β)) n=1,2,3,..
= e α e i(2nπ+β)
= e α +2nπi+iβ
x+iy = e α + iβ +2nπi
\ + L o ex= g i +( ya ) i+b 2n p i
L o exg +(i ) y =l xo +gi ( y+) n2p i

This is called the general logarithm of x+iy.
1
\ L o +gx ( i)= l y o 2+ gx
2
2
( y+) 1 -
( ) 2 pi
t i ay+ n
x
1
= l ox+2 g (y +)é
2 1
t i (ay+ )nùn 2 p
-
2 êë x úû
14
Note :
1. log Z is infinitely many valued function. This is called the general logarithm of Z.
2. If n = 0, we get the principal value of log Z
Important
1
1. log( x + iy ) = log( x 2 + y 2 ) + i tan -1 æç y ö÷
2 è xø
2. log e ( x + iy ) = log( x + iy ) + 2npi
3.2 Examples
1. find log (1+i)
1 + I = x + iy \ x = 1, y = 1
x2 +y2 = 2
amplitude = tan -1 æç y ö÷ = tan -1 (1) =i p

è xø 4
1
\ log( x + iy ) = log 2 + i p
2 4
2. Find Log (1+i)
Log (1+i) = log (1+i) + 2 n p i
1
= log 2 + i p + i 2np
2 4
1
= l o+ gi p 2+ n( p
2 )
2 4
3. Log i
i=0+1i \ x = 0, y = 1
è xø 0
( )
q = tan -1 æç y ö÷ = tan -1 1 = tan -1 a = p
2
x2 + y2 = 1 = 1
1
Log i = log( x 2 + y 2 ) + iq
2
1
= log 1 + i p
2 2
15
Log i = i p {\ log 1 = 0}
2
4. Log i = log i + 2nπi
= ip + 2npi
2
5. Prove that log (cos q + i sin q ) = i q , - p < q < p
Solution :
log (cos q + i sin q ) = log (x+iy)
x = cos q ; y = sin q
x 2 + y 2 = 1; q = tan -1 æç y ö÷
è xø
1
\ log(cosq + i sin q ) = log 1 + iq
2
= iq
6. Find a power series for tan-1 x using logarithm of complex number
Proof: if x is real
1
log ( 1+ix) = log(1 + x 2 ) + i tan -1 x
2
tan-1 x = Imaginary part of log (1+ix)
but log (1 + Z) = Z - Z 3 + Z 3 ...........

2 3
-1 i 2 x2 i3 x3
tan x = IP of ix - + + ......
2 3
x2 i3 x3
= IP of ix + - + ......
2 3
éæ x 2 x 4 ö æ x3 x5 öù
= IP of êçç + ..... ÷÷ + içç x - + + ...... ÷÷ú
ëè 2 4 ø è 3 5 øû
x3 x5
\ t= a-1 xn - . x . +. . . . . .
3 5
7. Obtain the general value of L o ii g
Solution : Let L o ii g= a + ib
i = i a+ib
16
= e(a+ib) log i
= e(a+ib) [i(2nπ+π/2)]
i = e -b(2nπ+π/2) e ia(2nπ+π/2)
taking modulus on both sides
1 = e -b(2nπ+π/2)
b=0
i = e ia(2nπ+π/2)
i = cos [a(2nπ+π/2)] + i sin [a(2nπ+π/2)]
= cos [a(2nπ+π/2)] = 0
a(2nπ+π/2) = 2m+π/2
æ 4 np + p ö 4 np + p
aç ÷=
è 2 ø 2
4m + 1
a= ; m, n ¬ Z
4n + 1
4m + 1
\ log ii =
4n + 1
Method 2
L o ieg
L o ii g= (1)
L o ieg
L o ieg= lnie o + g ip
2
= ip + 2npi
2
=ip ( 2
+ 2np )
æ p + 4npö
= iç n÷Î ; Z
è 2 ø
i æ 4np +öp
Similarly L o e = i çg m÷Î, Z
è 2 ø
Using in (1)
17
æ 4np + p ö
iç ÷
i è 2 ø
log i =
æ 4mp + p ö
iç ÷
è 2 ø
(4n + 1)p
= ; n, m Î Z
(4m + 1)p
8. If i x+iy = x+iy, prove that

x 2 + y 2 = e - ( 4 n +1)py
Proof : i x+iy = x+iy
x+iy = ix+iy
x + iy
= e log i
= e ( x + iy ) log i
p
( x + iy )( 2 npi + i
2
x+iy = e
( 4 np + p )
i ( x + iy )
2
x+iy = e
1
( ix - y ) 4 np +p
= e2
y
- ( 4 np + p ) ix ( 4 n p + p )
2 2
= e e
y
-
2
( 4 n +1)p é æx ö æx öù
=e êcosç 2 (4n + 1)p ÷ + i sin ç 2 (4n + 1)p ÷ú
ë è ø è øû
y
-
2
( 4 n +1)p æx ö
x= e cosç (4n + 1)p ÷
è2 ø
y
-
2
( 4 n +1)p æx ö
y= e sin ç (4n + 1)p ÷
è2 ø
\ x 2 + y 2 = e - y ( 4 n +1)p
3.3 Let us sum up.
So far we have studied the concept of finding the logarithm of a complex number
and also a here concept of the general logarithm a complex number.
18
3.4 Check you progress

(a) Find log (1- i)
(b) Find log (1+ i tan2)
3.5 Lesson End Activities
- ( 4 n +1)p
1. Prove that i i = e 2
; n is any integer
2. Prove that Log (-1) = i( 2n+1) π
3. Prove that Log (1- i) =

1
2
[ (
log 2 + i 2np + - p
4
)]
4. Prove that Log (-5) = log 5 + I (2n π+ π)
5. If = A+iB , prove that
æ pA ö B
a. tan ç ÷ = and
è 2 ø A
b. A 2 + B 2 = e -pB
x-i
6. Prove that i log = p - 2 tan -1 x
x+i
7. Show that log

a + ib
a - ib
= 2i tan -1 b
a
( )
p
8. Prove that log(1 + i tan a ) = log sec a + ia ;0 < a <
2
9. Prove that log(1 + cos 2q + i sin 2q ) = log( 2 cos q ) + iq * * - p < q < p
3.7 References
1. Trigonometry by S. Narayanan
19
Lesson - 4
Summation of Series
Contents
4.1 Summation of Series
4.2 Examples
4.3 Let us sum up
4.6 References
In this lesson, we are going to study trigonometric series using the concept of
arithmetic series using the concept of arithmetic progression, Geometric progression,
Binominal theorem, exponential Theorem and logarithmic theorem.
Model 1 :
Summation of series when angles are in Arithmetic progression
Sine series
1.4.1 Find the sum to n terms of the series
sin a + sin(a + b ) + sin(a + 2 b ) + sin(a + 3b ) + ......
Proof :
Let s n = sin a + sin(a + b ) + sin(a + 2 b ) + .... + sin(a + n - 1b )
Multiply both sides by 2 sin æç b ö÷

è 2ø
2 sin æç b ö÷ s n = 2 sin a sin b + 2 sin b sin(a + b ) +

è 2ø 2 2
------(1)
2 sin(a + 2 b ) sin b + ... + 2 sin(a + n - 1b ) sin b
2 2
We know 2 sin A sin B = cos (A-B) – cos (A+B)
20
\ 2 sin a sin b = cosæç a - b ö÷ - cosæç a + b ö÷

2 è 2ø è 2ø
2 sin(a + b ) sin b = cosæç a + b ö÷ - cosæç a + 3b ö÷

2 è 2ø è 2ø
2 sin(a + 2 b ) sin b = cosæç a + 3b ö÷ - cosæç a + 5b ö÷

2 è 2ø è 2ø
-----------------------------------------------------
----------------------------------------------------
2 sin(a + n - 12 b ) sin b
2
( ( )) ( (
= cos a + n - 3 b - cos a + n - 1 b
2 2
))
Adding the above we get
2 sin b
2
S n = cosæça - b ö÷ - cos a + n - 1 b
è 2ø 2
( ( )) ---(2)
But we know that

A+ B A- B
cos A – cos B= - 2 sin sin
2 2
A =a - b
2
;B =a + n - 1 b
2
( )
\ A+ B =a - b
2
(
+a + n - 1 b
2
)
= 2a - b + nb - b
2 2
= 2a + nb - b
= 2a + ( n - 1)
A+ B n -1
\ =a + b
2 2
A- B =a - b
2
(
-a - n - 1 b
2
)
=a - b - a - nb + b
2 2
= - nb
A- B b
\ = -n
2 2
21
(2) becomes
æ æ n - 1 ö ö æ - nb ö
2 sin b S n = -2 sin çça + ç ÷ b ÷÷ sin ç ÷
2 è è 2 ø ø è 2 ø
æ æ n - 1 ö ö æ nb ö
= 2 sin çça + ç ÷ b ÷÷ sin ç ÷
è è 2 ø ø è 2 ø
æ æ n - 1 ö ö æ nb ö
sin çça + ç ÷ b ÷÷ sin ç ÷
\ Sn =
sin æç b ö÷
è 2ø
Results
1. But b =a
\ sin a + sin 2a + sin 3a + ....... + sin na
æ æ n + 1 ö ö æ na ö
sin çç ç ÷a ÷÷ sin ç ÷
=
( )
sin a
2
4.2. Examples
Find the sum to n terms of the series
cos a + cos(a + b ) + cos(a + 2 b ) + cos(a + 3b ) + .......... .......... ..
Solution
S n = cos a + cos(a + b ) + cos(a + 2 b ) + ....... + cos(a + n - 1b )
Multiply both sides by 2 sin b

2
\ 2 sin b S n = 2 cos a sin b + 2 cos(a + b ) sin b + 2 cos(a + 2 b ) sin b

2 2 2 2 --(1)
+ .................... + 2 cos(a + n - 1b ) sin b
2
We know that
2 cos A sin B=sin (A+B) – sin (A- B)
2 cos a sin b = sin(a + b ) - sin(a - b )

2 2 2
2 cos(a + b ) sin b = sin(a + 3b ) - sin(a + b )

2 2 2 2
22
2 cos(a + 2 b ) sin b = sin(a + 5 b ) - sin(a + 3b )

2 2 2
-------------------------------------------------------
-------------------------------------------------------
2 cos(a + n - 1b ) sin b = sin(a + (n - 1 / 2) b ) - sin(a + (n - 3 / 2) b )

2
\ using in (1)
é æ 1ö ù
2 sin b S n = sin êa + ç n - ÷ b ú - sin æça - b ö÷ ----(2)
2 ë è 2ø û è 2ø
But we know that

A+ B A- B
sin A - sin B = 2 cos sin
2 2
( )
Here A = a + n - 1 b ; B = a - b
2 2
A + B = a + (n - 1 )b + a - b
2 2
= a + nb - b +a - b
2 2
= 2a + nb - b
= 2a + ( n - 1) b
A+ B n -1
\ =a + b
2 2
1
A - B = a + nb - b -a + b
2 2
= nb
A - B nb
\ =
2 2
\ using in (2)
é æ n - 1 ö ù æ nb ö
2 sin b S n = 2 cos êa + ç ÷ b ú sin ç ÷
2 ë è 2 ø û è 2 ø
23
cos êa + ç ÷ b ú sin ç ÷
ë è 2 ø û è 2 ø
\ Sn =
sin æç b ö÷
è 2ø
Cor 1: Put b = a
æ n + 1ö æ na ö
cosç ÷a sin ç ÷
è 2 ø è 2 ø
\ cos a + cos 2a + cos 3a + ... + cos na =
sin a
2
( )
1. First the sum to n terms of the series
Sin2 a +sin2 2 a +sin2 3 a +……………..
Solution
Sn = Sin2 a +sin2 2 a +sin2 3 a +…………+sin2 n a
= ½ [2 Sin2 a +2 sin2 2 a +2 sin2 3 a +…………+2 sin2 n a ]
But we know 2sin2 a = 1 – cos 2 a
1
\ S n= [(1 - cos 2a ) + (1 - cos 4a ) + (1 - cos 6a ) + ...... + (1 - cos 2na )]
2
1
= [n - (cos 2a + cos 4a + cos 6a + ...... + cos 2na )]
2
We know
cos êa + ç ÷ b ú sin ç ÷
ë è 2 ø û è 2 ø
cosa + cos(a + 2 b ) + ..... + cos(a + n - 1b ) =
sin æç b ö÷
è 2ø
Hence ‘ a ’=2 a ; b = 2a
24
é é n - 1 ù æ n - 2a ö ù
ê cos ê2a + 2a ú sin ç ÷ú
1 ë 2 û è 2 øú
\ S n = ên -
2ê sin 2a ú
ê 2 ú
ë û
1é cos( n + 1)a sin( na ) ù
= ê n- úû
2ë sin a
2. Find the sum to n terms of the series

cos2 a +cos2 2 a +cos2 2 a +…………+cos2 n a +…… a
Proof:
Let
Sn = cos2 a +cos2 2 a +cos2 2 a +…………+cos2 n a
= ½ [2 cos2 a +2 cos2 2 a +2 cos2 2 a +…………+2 cos2 n a ]
But 1 + cos 2 q =2cos2 q

1
\ S n= [(1 + cos 2a ) + (1 + cos 4a ) + (1 + cos 6a ) + ...... + (1 + cos 2na )]
2
1
= [n + (cos 2a + cos 4a + cos 6a + ...... + cos 2na )]
2
é é n - 1 ù æ n - 2a ö ù
ê cos ê2a + 2a ú sin ç ÷ú
1 ë 2 û è 2 øú
= ên +
2ê sin 2a ú
êë 2 úû
1é cos( n + 1)a sin( na ) ù

= ê n+ úû
2ë sin a
3. Find the sum of the series
Sin3 a +sin3 2 a +sin3 3 a +……………..+sin3 n a
25
Proof:
We know sin3 a = 3 sin a - 4 sin3 a
\ 4sin3 a =3 sin a - sin3 a
Let Sn = Sin3 a +sin3 2 a +sin3 3 a +…………+sin3 n a
= ¼ [4 Sin3 a +4 sin3 2 a +4 sin3 3 a +…………+4 sin3 n a ]
1 é(3 sin a - sin 3a ) + (3 sin 2a - sin 6a ) + (3 sin 3a - sin 9a )ù
=
4 êë +......... + (3 sin na - sin 3na ú
û
1
= [3(sin a + sin 2a + sin 3a + ......... + sin na )]
4
-[sin 3a + sin 6a + sin 9a + ......... + sin 3na ]
3
= [sin a + sin 2a + sin 3a + ......... + sin na ]
4
1
- [sin 3a + sin 6a + sin 9a + ......... + sin 3na ]
4
é n -1 ù æ n ö é n - 1 ù æ 3na ö
sin êa + a ú sin ç a ÷ sin ê3a + 3a ú sin ç ÷
3 ë 2 û è 2 ø -1 ë 2 û è 2 ø
= -
4 sin a
2
( ) 4 sin 3a
2
( )
æ n + 1ö æ na ö - 1 é (n + 1)3a ù æ 3na ö
sin ç ÷a sin ç ÷ sin ê úû sin çè 2 ÷ø
3 è 2 ø è 2 ø 4 ë 2
= -
4 ( )
sin 3a
2
sin 3a
2
( )
4. Find the sum to n terms of

Cos3 a +cos3 2 a +cos2 3 a +…………+cos3 n a
Proof:
We know cos3 q = 4cos3 q -3cos q
\ 4cos3 q =cos3 q +3cos a
Let Sn = cos3 a +cos3 2 a +cos3 3 a +…………+cos3 n a
= ¼ [4 cos3 a +4cos3 2 a +4cos3 3 a +…………+4 cos3 n a ]
1 é(cos 3a + 3 cos a ) + (cos 6a + 3 cos 2a ) + (cos 9a - 3 cos 3a )ù
=
4 êë +......... + (cos 3na - 3 cos na úû
26
1
= [cos 3a + cos 6a + cos 9a + ......... + cos 3na ] +
4
3[cos a + cos 2a + cos 3a + ......... + cos a ]
1
= [cos 3a + cos 6a + cos 9a + ......... + cos 3na ]
4
3
+ [cos a + cos 2a + cos 3a + ......... + cos na ]
4
é n - 1 ù æ 3na ö é n -1 ù æ n ö
cos ê3a + 3a ú sin ç ÷ cos êa + a ú sin ç a ÷
1 ë 2 û è 2 ø 3 ë 2 û è2 ø
= +
4 sin 3a
2
( ) 4 sin a
2
( )
1
cos
(n + 3)a sinæ 3na ö cosæ (n + 1)a ö sinæ na ö
ç ÷ ç ÷ ç ÷
=
4 2 è 2 ø+3 è 2 ø è 2 ø
sin 3a
2
( ) 4 sin a
2
( )
Model 3 C+is method)
Type 1: Problems based on ex and e-x

x x2 x3
1. e x = 1 + + + + ......................¥
1! 2! 3!
-x x x2 x3
2. e = 1- + - + ......................¥
1! 2! 3!
Problems
1. Find the sum to infinity of the series
x2
sin a + x sin(a + b ) + sin(a + 2 b ) + ......................¥
2!
Proof:
x2
Let S = sin a + x sin(a + ib ) + sin(a + 2 b ) + ......................¥
2!
x2
C = cos a + x cos(a + ib ) + cos(a + 2 b ) + ......................¥
2!
27
x2
C + iS = (cos a + i sin a ) + x cos(a + b ) + ix sin(a + b ) + cos(a + 2 b ) +
2!
x2
i cos(a + 2 b )......................¥
2!
ia i (a + b ) x 2 i (a + 2 b )
=e + xe + e + ......................¥
2!
x 2 ia i 2 b
= e ia + x e ia e ib + e e + ......................¥
2!
é xe ib x 2 i 2 b ù
= e ia ê1 + + e + ......................¥ ú
ë 1! 2! û
é xe ib 1 ù
= e ia ê1 +
1!
+ xe ib
2!
( )2
+ ......................¥ ú
ë û
[ ]
= e ia e xe
ib
= e ia e x (cos b + i sin b )
= e ia e x cos b +ix sin b
= e x cos b e ia e ix sin b
= e x cos b e i (a + x sin b )
= e x cos b [cos(a + x sin b ) + i sin(a + x sin b )]
C + is = e x cos b cos(a + x sin b ) + ie x cos b sin(a + x sin b )
Equate imaginary part

S = e x cos b sin(a + x sin b )
2. Find the sum of the series

1 1
sin a + sin 2a + + sin 3a + .......... ........ ¥
2! 3!
Proof
1 1
S = sin a + sin 2a + + sin 3a + ..................¥
2! 3!
28
Let
1 1
C = 1 + cos a + cos 2a + + cos 3a + ..................¥
2! 3!
1 1
\ C + is = 1 + (cos a + i sin a ) + (cos 2a + i sin 2a ) + + (cos 3a + i sin 3a ) + ..................¥
2! 3!
1 i 2a 1 i 3a
= 1 + e ia + e + e + ..................¥
2! 3!
1 ia 1 ia
= 1 + e ia + ( )
2!
e
2
+ ( )
3!
e
3
+ ..................¥
1 2 1 3
= 1+ y + y + y + ..................¥ where y = e ia
2! 3!
= ey
ia
= ee
= e cos a + i sin a
= e cos a e i sin a
= e cos a [cos(sin a ) + i sin(sin a )]
C + is = e cos a cos(sin a ) + ie cos a sin(sin a )
Equate imaginary parts

S = e cos a sin(sin a )
3. Sum to infinity the series

sin a cos 2a sin 2 a cos 3a
cos a + + + .........................¥
1! 2!
Solution
sin a cos 2a sin 2 a cos 3a
C = cos a + + + .........................¥
1! 2!
sin a sin 2 a sin 2 a sin 3a
S = sin a + + + .........................¥
1! 2!
29
sin a
\ C + is = (cos a + i sin a ) + (cos 2a + i sin 2a )
1!
sin 2 a
+ (cos 3a + i sin 3a ) + .........................¥
2!
sin a i 2a sin 2 a i 3a
ia
=e + e + e + .........................¥
1! 2!
é sin a ia sin 2 a i 2a ù
= e ia ê1 + e + e + .........................¥ ú
ë 1! 2! û
é sin a ia
ia
= e ê1 + e +
sin a e ia ( )2
ù
+ .........................¥ ú
ëê 1! 2! ûú
é y y2 ù
= e ia ê1 + + + .........................¥ ú where y = sin ae ia
ë 1! 2! û
= e ia e y
ia
= e ia e sin ae
= e ia e sin a (cos a +i sin a )
sin a cos a + i sin 2 a
= e ia +
2
= e sin a cos a e i (a +sin a)
[
= e sin a cos a cos(a + sin 2 a ) + i sin(a + sin 2 a ) ]
= e sin a cos a cos(a + sin 2 a ) + ie sin a cos a sin(a + sin 2 a )
Equate real part

C = e sin a cos a cos(a + sin 2 a )
4. Sum the series to infinity

cos 2q cos 4q
1+ + + ..................¥
2! 2!
cos 2q cos 4q
Let C = 1+ + + ..................¥
2! 2!
30
sin 2q sin 4q
S= + + ..................¥
2! 2!
1
C + is = 1 + [cos 2q + i sin 2q ] + 1 [cos 4q + sin 4q ] + ..................¥
2! 4!
1 i 2q 1 i 4q
= 1+ e + e + ..................¥
2! 4!
1 iq 1 iq
= 1+
2!
( )
e
2
+ ( )
4!
e
4
+ ..................¥
1 2 1 4
= 1+ x + x + ..................¥ where x = e iq
2! 4!
e x + e-x
= = cosh x
2
=
2
[
1 e iq iq
]
e + e -e = cosh(e iq )
= cos(cos q + i sin q )
= cos[i (cosq + i sin q )]
= cos[i cosq - sin q )]
= cos(i cos q ) cos(sin q ) + sin(i cos q ) sin(sin q )
= cosh(cos q ) cos(sin q ) + i sinh(cos q ) sin(sin q )
Equate real part
C = cosh(cos q ) cos(sin q )
Type 2 Summation of series based on logarithmic series
31
FORMULA
x2 x3 x4
1. log(1 + x) = x - + - + ........¥
2 3 4
x2 x3 x4
2. log(1 - x) = - x - - - - ........¥
2 3 4
(1) sum the series to infinity

a 2 cos 2q a 3 cos 3q
a cosq + + + ............¥
2 3
a 2 cos 2q a 3 cos 3q
Let C = a cosq + + + ............¥
2 3
a 2 sin 2q a 3 sin 3q
S = a sin q + + + ............¥
2 3
a2
C + is = a (cos q + i sin q ) + (cos 2q + i sin 2q ) +
2
a3
(cos 3q + i sin 3q ) + ............¥
3
iqa 2 i 2q a 3 i 3q
= ae + ae + ae + ...........¥
2 3
x2 x3
= x+ + + ...........¥
2 3
Where x = ae iq
= - log(1 - x )
= - log(1 - ae iq )
= - log[1 - a (cosq + i sin q )]
= - log[1 - a cosq - isainq ]
é1 æ - a sin q öù
[ ]
= - ê log 1 - a cosq ) 2 + a 2 sin 2 q + i tan -1 ç ÷ú
ë2 è 1 - a cosq øû
32
Equal to real part

1
[
C = - log (1 - a cos q ) + a 2 sin 2 q
2
2
]
1
[
= - log 1 + a 2 cos 2 q + a 2 sin 2 q - 2a cos q
2
]
1
[
= - log 1 + a 2 - 2a cos q
2
]
(2) sum the series to infinity
1 1
c sin a - c 2 sin(a + b ) + c 3 sin(a + 2 b )..........¥
2 3
Solution
1 1
S = c sin a - c 2 sin(a + b ) + c 3 sin(a + 2 b )..........¥
2 3
1 1
C = c cos a - c 2 cos(a + b ) + c 3 cos(a + 2 b )..........¥
2 3
1
\ C + is = c(cos a + i sin a ) - c 2 [cos(a + b ) + i sin(a + b )]
2
1
+ c 3 [cos(a + 2 b ) + i sin(a + 2 b )]
3
1 2 i (a + b ) 1 3 i (a + 2 b )
= c e ia - c e + c e ..............¥
2 3
1 2 ia ib 1 3 ia i 2 b
= c e ia - c e e + c e e ..............¥
2 3
é 1 1 ù
= e ia êc - c 2 e ib + c 3 e i 2 b .................¥ ú
ë 2 3 û
e ia é ib 1 2 i 2 b 1 3 i 3 b ù
= êce + 2 c e + 3 c e ................¥ ú
e ib ë û
é 1 1 ù
= e i (a - b ) ê x - x 2 + x 3 ................¥ ú
ë 2 3 û
Where x = ce ib
= e i (a -b ) ´ log(1 + x)
= e i (a - b ) log(1 + ce ib )
= e i (a - b ) log[1 + c(cos b + i sin b )]
33
= e i (a - b ) log[1 + c cos b ) + ic sin b )]
é1 ù
( 2 2 2
ê 2 log (1 + c cos b ) + c sin b + ú )
= [cos(a - b ) + i sin(a - b )]ê ú
ê -1 æ c sin b öú
ê i tan çç ÷÷ú
ë è 1 + c cos b øû
Equate imaginary part

1 æ c sin b ö
S= [ ]
sin(a - b ) log (1 + c cos b ) 2 + c 2 sin 2 b + cos(a - b ) tan -1 çç ÷÷
2 è 1 + c cos b ø
Type 3: Summation of series – using Binomial series

1. Sum the series
1 1 .3 1.3 .5
1 - cos q + cos 2q - cos 3q + ........................¥
2 2 .4 2.4 .6
Solution :
1 1 .3 1 .3 .5
C = 1 - cos q + cos 2q - cos 3q + ........................¥
2 2 .4 2 .4 .6
1 1.3 1.3.5
S = - sin q + sin 2q - sin 3q + ........................¥
2 2.4 2.4.6
1 1.3
C + is = 1 - (cos q + i sin 2q ) + (cos 2q + i sin 2q )
2 2.4
1.3.5
- (cos 3q + i sin 3q ) + ........................¥
2.4.6
1 iq 1.3 i 2q 1.3.5
=1- e + e - e i 3q + ........................¥
2 1.2.2.2 1.2.2.2.3.2
1 iq 1.3 e i 2q 1.3.5 e i 3q
=1- e + - + ........................¥
2 2! 2 2 3! 2 3
2 3
1 æ e iq ö 1.3 æ e iq ö 1.3.5 æ e iq ö
= 1 - çç ÷÷ + çç ÷÷ - ç ÷÷ + ........................¥
1! è 2 ø 2! è 2 ø 3! çè 2 ø
This is of the form
3
p p ( p + 8) æ x ö 2 p ( p + q )( p + 2q ) æ x ö
1 - æç x ö÷ + ç q÷ - çç ÷÷ + ........................¥
1! è q ø 2! è ø 3! èqø
34
p = 1; p+q =3 \ q=2
x e iq
=
q 2
x e iq
=
2 2
x = e iq
-p
q
\ C + is = (1 + x)
-1
= (1 + e iq ) 2
-1
2
C + is = (1 + cos q + i sin q )
[
= 2 cos 2 q
2
+ i 2 sin q cos q
2 2
]
-1
2
(
= 2 cos q
2
) [cos(q 2 ) + i sin (q 2 )]
-1
2
-1
2
-1
[ ( 2 )]
= 2 cos q
-1
2 æç e iq 2 ö÷
è ø
2
[ ( 2 )] æçè e ö÷ø
= 2 cos q
-1
2 iq
4
C + is = ( 2 cos q )[cos q - i sin q ]

4 4
Equate real part on both sides
C = (2 cos q )
-1
2
cos q ( 4)
4.3 Let us sum up
So far we have studied the concept of finding a trigonometric series using ap, gp,
etc. Also we have seen the sum to arrive a certain trigonometric series using the
fundamental forms of sin3x, cos3x, sin3 x, cos3 x, sin2 x, cos2 x
4.4. Check you progress
(a) Find sum for cos2a+cos4a+cosba+….a
(b) Find sum for sin2a+sin4a+sinba+….a
35
4.5. Lesson end Activities

sum the series to infinity
2
1. cos a + cos b cos(a + b ) + cos b cos(a + 2 b ) + ............¥
1! 2!
2
b
Ans: e cos cos(a + sin b + cos b )
sin(a + 2 b ) sin(a + 4 b )
2. sin a + + ............¥
2! 4!
Ans: [sin a cos(cos b ). cosh(sin b ) - cos a sin b sinh(sin b )]
sin 3q sin 5q
3. sin q cos q + cos 3q + cos 5q + ...........
3! 5!
Ans: sinh(sin q cosq ) cos(sin 2 q )
1 1 1.3
4. 1 + cos 2a - cos 4a + cos 6a ........................¥
2 2.4 2.4.6
Ans: 2 cos a cos a ( 2)
a2 a3
5. a sin q + sin 2q + sin 3q + ...........¥
2 3
é a sin q ù
Ans: tan -1 ê
ë1 - a cos q úû
1 1
6. cos a sin a - cos 2 a sin 2a + cos 3 a sin 3a
2 3
é sin a cos a ù
Ans: tan -1 ê
ë 1 + cos a úû
2
4.7 Sources
1) Trigonometry : M.L. Khanna

2) Trigonometry : S. Narayanan
36
Unit II
Lesson - 5
Vector Calculus
Contents
5.1 Scalar and Vector point functions – differential vectors
5.2 Examples
5.3 Let us sum up
5.6 References

We are going to study the concept of Scalar and Vector point functions,
differential vectors in detail.
5.1 Scalar and Vector point functions – differential vectors

5.1.1. Scalar point function
Let A be any subset of the set of real numbers. If to each element a of A, we
associate by some rule a unique real number f(a), then this rule defines a scalar function
of the scalar variable a. Here f(a) is a scalar quantity and therefore f is a scalar function.
5.1.2. Vector point function
Let A be any subset of the set of real numbers. If to each element a of A, we
associate by some rule a unique vector f (a ) , then this rule defines a vector function of
the scalar variable ‘a’. Here f (a ) is a vector quantity and f is a vector function.
5.1.3. Derivative of a vector function with respect to a scalar
Let r = f (t ) be a vector function of the scalar variable ‘t’.
37
dr
If exists, there r is said to be differentiable.
dt
dr d 2r d 3r
is denoted by r . Similarly , ........ are denoted by r , r …….
dt dt 2 dt 3
respectively.
5.1.4. Some results on differentiation of vectors.
Let a, b, c be differentiable vector function of a scalar t. let f be a differentiable
scalar point function of the same variable ‘t’ , then
d d a db
a. ( a + b) = +
dt dt dt
d db da
b. (a.b) = a. + .b
dt dt dt
d db d a
c. ( a ´ b) = a ´ + ´b
dt dt dt
d d a df
d. (f a ) = f + a
dt dt dt
éd a ù é db ù é dc ù
e.
d
dt
[ ]
abc = ê bc ú + ê a
dt dt
c ú + ê ab ú
ëê ûú ëê ûú ëê dt ûú
æ db ö æ dc ö
f.
d
dt
[ ]
a ´ (b ´ c) =
da
dt
´ (b ´ c) + a ´ ç
ç dt
´ c÷ + a ´ çb ´ ÷
÷ ç dt ÷ø
è ø è
5.1.5. Derivative of a constant vector.
A vector is said to be constant only if both its magnitude and direction are fixed.
Let r be a constant vector function of the scalar variable t
Let r = c ; c is a constant vector
\ r +dr = c
\ r +dr - r = 0
dr =0
38
dr
\ =0
dt
dr
\ l im =0
dt ® 0 dt
dr
\ = 0 (zero vector)
dt
5.1.6 Derivative of a vector point function in terms of its components
Let r = xi + y j + z k where x, y, z are scalar functions of the sector variable ‘t’.
d r dx dy dz
Then = i+ j+ k
dt dt dt dt
Results
(1) The necessary and sufficient condition for the vector point function a (t ) to be
da
constant is that =0
dt
(2) If a is a differentiable vector point function of the scalar variable ‘t’ and if a = a, than
then
d æ 2ö da
a. ç a ÷ = 2a
dt è ø dt
da da
b. a. =a
dt dt
da da
c. if a has constant length, then a. ^ r , if =0
dt dt
d. The necessary and sufficient condition for the vector a (t ) to have constant
da
magnitude is a. =0
dt
39
5.2 Examples
1. If a is a differentiable vector function of the scalar variable ‘t’ , then
d æç d a ö÷ d2a
a´ = a´ 2
dt çè dt ÷ø dt
Proof :
d æç d a ö÷ d a d a d æ d a ö÷
a´ = ´ + ç
+a´
dt çè dt ÷ø dt dt dt çè dt ÷ø
d2a
= 0+ a´
dt 2
d2a
= a´ 2
dt
{\ a ´ a = 0}
2. The necessary and sufficient condition for the vector a (t) to have a constant direction
da
is a ´ = 0.
dt
Proof: Let a be the unit vector in the direction of a
\ a = a aˆ
r
d a r d ˆ a rd
\ = a + a aˆ
d t d t d t
r
r d ra d ér a ˆ rd ù
\ a ´ a = .aˆ ´ê a + aˆ ú a
d t d ë t d ût
2
daˆ r
a aˆ ´ +0
dt
2 daˆ
= a aˆ ´ (1)
dt
The condition is necessary
Suppose a has a constant direction then â is a constant vector
40
d
\ (aˆ ) = 0
dt
da
\ using in (1) a ´ =0
dt
\ The condition is necessary
The condition is sufficient
da
Suppose a ´ =0
dt
2 daˆ
From (1), a aˆ ´ =0
dt
daˆ
\ aˆ ´ =0 ---(2)
dt
Since â is of constant length,
daˆ
\ aˆ. =0 ---(3)
dt
\ â is a constant vector is the direction of a is constant
3. Find a unit tangent vector to the curve x = 3t+2; y=5t2 , z=2t-1 at t=1
dr dr
Solution : Unit tangent vector is defined as ¸ï ï
dt dt
r = xi + y j + z k
= (3t + 2)i + 5t 2 j + (2t - 1)k
dr
= 3i + 10t j + 2k
dt
dr ù
\ ú = 3i + 10t j + 2k
dt ûú
t =1
dr
= 9 + 100 + 4 = 113
dt
41
3i + 10t j + 2k
\ unit tangent vector =
113
4. A particle moves along a curve whose parametric equations are x = e-t ; y = 2 cos3t, z =
2sin3t, where t is the time. Find its velocity and acceleration at t = 0.
Solution :
r = xi + y j + z k
= e - t i + 2 cos 3t + 2 sin 3t k
dr
= -e -t i - 6 sin 3t j + 6 cos 3t k
dt
d2r
= e -t i - 18 cos 3t j - 18 sin 3t k
dt
dr
\ At t = 0, = -i + 6k
dt
Which is the required velocity vector
d2r
\ At t = 0, = i - 18 j
dt 2
Which is the required acceleration vector.
5.3. Let us sum up
So far we have studied in finding the velocity, acceleration of a particle using
fundamental of calculus.
5.4. Check your progress
dr d2r 2
r=
(1) Find ,= n a2 ´+bnifr=0
dt dt
r r r r
r = x +i y + 2i xwhere
x=t, y=t2 , z=t3 at t =1
r
d x r r r
(2) Find i f= rc ao+ s i s bi nior t=0
d t
1. If r = a cos nt + b sin nt where a,b ,n are constants, prove that
42
dr
a. r = = na ´ b
dt
d2r
b. 2
+ n2 r = 0
dt
é dr d 2r ù
c. êr 2 ú
=0
ëê dt dt ûú
2. Find the unit tangent vector to the curve x = a cost; y = a sint; z = ct
a sin t i + a cos t y + c k
Ans
a2 + c2
3. Find the velocity vector, the speed and the acceleration vector for the particle whose
path is given by
x = 3 cos2t; z = 2 sin3t
Ans:
i) Velocity = - 6 sin 2t i + 6 cos 3t k
ii) Speed = 36(cos 2 3t + sin 2 2t
iii) Acceleration = - 12 cos 2t i - 18 sin 3t k

4. The position vector of a moving point as given by r = cos at i + sin at j . Show that the
velocity v is perpendicular to r
5. If r = cos at i + sin at j , show that r ´ v is a constant vector.
5.7 References
1) Vector Calculus by - Namasivayam
2) Vector Calculus by - Rasinghmia aggarval
3) Vector calculus by - P. Durai Pandian
4) Vector calculus by - Chatterjee
43
Lesson - 6
Contents
6.1 Differential operators and directional derivative
6.2 Example
Let us sum up
Check your progress
Lesson End activities
References

We are going to study the concept of new ideas on del operator. We also study the
u r u r
concept of operators line dir f , curl and grade f . Also our aim is to study the concept of
solenoidal vectors and irrotational vectors.
6.1 Differential operators and directional derivative
6.1.1 The vector differential operator Ñ (del) is defined as
¶ ¶ ¶
Ñ=i + j +k
¶x ¶y ¶z
6.1.2 The gradient
Let f ( x, y, z ) be a differentiable scalar field. Then gradient of f is defined as
¶f ¶f ¶f
i + j +k and is denoted by grad f
¶x ¶y ¶z
¶f ¶f ¶f
\ gradf = i + j +k
¶x ¶y ¶z
æ ¶ ¶ ¶f ö
= çç i + j + k ÷f
è ¶x ¶y ¶z ÷ø
= Ñf
Ñf is a vector field.
44
6.1.3. Divergence of a vector point function Let f ( x, y, z ) be a vector point function

differentiable at each point ( x, y, z ) in a certain region of space. The divergence of
f is defined as
¶f ¶f ¶f
i. + j. + k. and is
¶x ¶y ¶z
Written as div f
u r u r u r
u r ¶ fr r¶ f u ¶rf
\ d =i v. +f . i +j . k
¶x ¶y ¶z
æ r ¶ r ¶ rö ¶u r
= ç i + j +÷ k .f
è ¶x ¶y ø ¶ z
= Ñ. f
6.1.4. Solenoidal vector
A vector point function is called solenoidal if Ñ. f = 0
6.1.5 Curl of a vector point function.
Let f ( x, y, z ) be a differentiable vector point function in a certain region of space. Then
¶f ¶f ¶f
the curl of f is defined as i ´ + j´ +k´ and is written as curl f .
¶x ¶y ¶z
¶f ¶f ¶f
\ curl f = i ´ + j´ +k´
¶x ¶y ¶z
æ r ¶ r ¶ rö ¶ u r
= ç i + j +÷ k ´ f
è ¶x ¶y ø ¶ z
= Ñ´ f
6.1.6 A vector point function f is called irrotational or rotation of f if Ñ ´ f = 0
6.1.7 Ñ. f and Ñ ´ f in terms of component’s
1. Let f = f1 i + f 2 j + f 3 k
¶f1 ¶f 2 ¶f 3
Ñ. f = + +
¶x ¶y ¶z
45
2. f = f1 i + f 2 j + f 3 k
i j k
Ñ´ f = ¶ ¶ ¶
¶x ¶y ¶z
f1 f2 f3
é ¶f ¶f ù é ¶f ¶f ù é ¶f ¶f ù
= ê 3 - 2 úi - jê 3 - 1 ú + k ê 2 - 1 ú
ë ¶y ¶y û ë ¶x ¶z û ë ¶x ¶y û
6.1.8 Directional derivative
1. Let f be a scalar point function of the variable t.
Let f(A), f(B) be the functional values of
f ( B ) - f ( A)
the scalar point function f at A and B respectively. Then if lim exists, it is
B® A AB
called the directional derivative of the scalar functional A along AB.
2. Let f be a vector point function.
f ( B) - f ( A)
Then lim if it exists, is called the directional derivative of f at A
B® A AB
along AB.
¶f ¶f ¶f
Note (1) : , , are the directional derivatives of f at A in the directions of the
¶x ¶y ¶z
coordinate axes
¶f ¶f ¶f
(2) , , are the directional derivatives of f at A in the directional derivatives of
¶x ¶y ¶z
f at A in the direction of the coordinate axes.
6.1.9 Directional derivative of f along any line

Let f be a scalar point function consider the line AB
46
Let be A (x,y,z)
Let the direction cosines of AB be l , m, n then the directional derivative of f along AB is
¶f ¶f ¶f
defined as l +m +n
¶x ¶y ¶z
6.1.10 Let f be a vector point function consider the line AB. Let A be (x,y,z)
¶f ¶f ¶f
Let the direction derivative of f along AB is defined as l +m +n
¶x ¶y ¶z
Ñf .n
6.1.11. The directional derivative of f in the direction of n is defined as
n
a. If f is a constant, then the directional derivative is zero.
b. Ñf is a vector normal to the level surface f ( x, y , z ) = c, c is a constant.
6.1.12. Level surface. Let f(x, y, z) be a scalar point function in a certain region of
space. The set of all points of the region for which f becomes a constant is called a
level surface and is written as f(x, y, z) = c, c is a constant.
a. The angle between the surfaces f1 ( x, y, z ) = c1 and f ( x, y, z ) = c 2 is defined
as the angle between their normal.
6.2 Example
Type 1
1. Find the directional derivative of Z 2 + 2 xy at (1,-1,3) in the direction of i + 2 j + 2k

Solution :
n = i + 2 j + 2k
n = 1+ 4 + 4 = 3
f ( x, y, z ) = Z 2 + 2 xy
¶f ¶f ¶f
= 2 y; = 2 x; = 2z
¶x ¶y ¶z
¶f ¶f ¶f
Ñf = i + j +k
¶x ¶y ¶z
= 2 yi + 2 x j + 2 z k
47
(1,-1,3) = (-2)i + 2 j + 6k
= -2i + 2 j + 6k
( )(
Ñf .n = - 2i + 2 j + 6k . i + 2 j + 2k )
= -2 + 4 + 12
= 14
The directional derivative of f along n
Ñf .n
=
n
14
=
3
2. Find the maximum directional derivative of f = xyz 2 at (1,0,3)
Solution:
f = xyz 2
¶f ¶f ¶f
= yz 2 ; = xz 2 ; = 2 xyz
¶x ¶y ¶z
¶f ¶f ¶f
Ñf = i + j +k
¶x ¶y ¶z
= yz 2 i + xz 2 j + 2 xyz 2k
Ñf )(1, 0,3) = 9 j
Ñf = 81 = 9
\ The maximum directional derivative of f is Ñf = 9
3. Find the magnitude and the direction of the greatest directional derivative of
x 2 yz 3 at (2,1,-1)
Solution :
The direction of the greatest directional derivative is along Ñf and magnitude is
Ñf
f = x 2 yz 3
48
¶f ¶f ¶f
= 2 xyz 3 ; = x2 z3; = 3 x 2 yz 2
¶x ¶y ¶z
¶f ¶f ¶f
Ñf = i + j +k
¶x ¶y ¶z
= 2 xyz 3 i + x 2 z 3 j + 3 x 2 yz 2 k
Ñf )( 2,1, -1) = -4i + 4 j + 12k
Ñf = 16 + 16 + 144 = 176 = 11 ´ 16 = 4 11
The direction of the greatest directional derivative is along Ñf = -4i + 4 j + 12k
4. Find the unit normal to the surface Z = x 2 + y 2 at the point (-1,-2,5)
f = xyz = x 2 + y 2 - z
¶f ¶f ¶f
= 2 x; = 2 y; = -1
¶x ¶y ¶z
¶f ¶f ¶f
Ñf = i + j +k
¶x ¶y ¶z
= 2 xi + 2 y j - k
Ñf )( -1, 2,5) = -2i - 4 j - k
Ñf = 4 + 16 + 1 = 21
Ñf
FORMULA unit normal vector to the surface f ( xyz ) = c is
Ñf
= -2i - 4 j - k
\ unit normal vector
21
4. Find the equation of the tangent plane to the surface yz – zx + xy + 5 = 0 at (1,-1,2)
Solution
f = yz
yz –- zx
zn + xy
¶f
= -z + y
¶x
¶f
= z+x
¶y
49
¶f
= y-x
¶y
¶f ¶f ¶f
Ñf = i + j +k
¶x ¶y ¶z
= ( y - z )i + ( x + z ) j + ( y - x)k
Ñf )(1, -1, 2,) = -3i + 3 j - 2k
\ The direction ratios of the normal to the tangent plane at (1,-1,2) are -3, 3, -2
The tangent plane passes through (1,-1,2) we know that equation of any plane passing
through (x1 ,y1 ,z1 ) is
a ( x - x1 ) + b( y - y1 ) + c( z - z1 ) = 0 ------(1)
Here a,b,c = -3,3,-2; (x1 ,y1 ,z1 ) = (1,-1,2)
Using in (1)
Equation of the tangent at (1, -1, 2) is
-3(x-1) + 3(y+1) – d(z-2) = 0.
-3x+3+3y+3-2z+4 = 0
-3x+3y-2z+10 = 0
or 3x – 3y + 2z – 10 = 0
5. Find the angle between the surfaces Z = x2 +y2 -3; and x2 +y2 +z2 = 9 at the point
(2, -1, 2)
Proof:
f1 = x 2 + y 2 - z ; f 2 = x 2 + y 2 + z 2 ;
The angle between f1 and f 2 is
Ñf1. Ñf2
---(1)
Ñf1 Ñf2
f1 = x 2 + y 2 - z f2 = x 2 + y 2 + z 2
¶f1 ¶f 2
= 2x = 2x
¶x ¶x
¶f1 ¶f 2
= 2y = 2y
¶y ¶y
50
¶ f 1 ¶f 2
= - 1 = 2z
¶ ¶z
¶f1 ¶f ¶f
Ñf1 = i + j 1 +k 1
¶x ¶y ¶z
= 2 xi + 2 y j - k
Ñf1 )( 2, -1, 2,) = 4i - 2 j - k
¶f 2 ¶f ¶f
Ñf 2 = i + j 2 +k 2
¶x ¶y ¶z
= 2 xi + 2 y j - 2 zk
Ñf 2 )( 2, -1, 2,) = 4i - 2 j - 4k
Ñf1 = 16 + 4 + 1 = 21
Ñf 2 = 16 + 4 + 16 = 36 = 6
Ñf1 .Ñf 2 = 16
Using in (1)
16
cos q =
3 21
é 16 ù
q = cos -1 ê ú
ë 3 21 û
6. Find the function f if
Ñf = ( y 2 - 2 xyz 3 )i + (3 + 2 xy - x 2 z 3 ) j + (6 z 3 - 3 x 2 yz 2 )k
Solution
By definition
¶f ¶f ¶f
Ñf = i + j +k
¶x ¶y ¶z
¶f
\ = y 2 - 2 xyz 3 ; ---(1)
¶x
¶f
= 3 + 2 xy - x 2 z 3 ---(2)
¶y
51
¶f
= 6 z 3 + 3 x 2 yz 2
¶z
Integrating both side of (1), (2), (3) w.r.t. x,y,z resply.
f = xy 2 - x 2 yz 3 + a function not containing x
f = 3 y + xy 2 - x 2 yz 3 + a function not containing y

3 4
f= z - x 2 yz 3 + a function not containing z
2
The general form of f is
3 4
f = 3y + z + xy 2 - x 2 yz 3 + c
2
7. Find Ñ.r if r = xi + y j + z k
r = xi + y j + z k
= f1 i + f 2 j + f 3 k
f1 = x; f 2 = y; f3 = z
¶f 1 ¶f 2 ¶f 3
=1 =1 =1
¶x ¶y ¶z
¶f1 ¶f 2 ¶f 3
Ñ. f = + +
¶x ¶y ¶z
= 1+1+1 = 3.
9. Find Ñ.r if r = xi + y j + z k
r = xi + y j + z k
= f1 i + f 2 j + f 3 k
f1 = x; f 2 = y; f3 = z
¶f 1 ¶f 2 ¶f 3
=0 =0 =0
¶y ¶z ¶x
¶f 1 ¶f 2 ¶f 3
=0 =0 =0
¶z ¶x ¶y
52
i j k
Ñ´r = ¶ ¶ ¶
¶x ¶y ¶z
f1 f2 f3
= ê 3 - 2 úi - ê 3 - 1 ú j + ê 2 - 1 ú k
ë ¶y ¶z û ë ¶x ¶z û ë ¶x ¶y û
=¶
9. If f = x 2 yi - 2 xz j + 2 yz k , find (i) Ñ. f (ii) Ñ ´ f
Solution (i) f = x 2 yi - 2 xz j + 2 yz k
= f1 i + f 2 j + f 3 k
f 1 = x 2 y; f 2 = -2 xz; f 3 = 2 yz
¶f1 ¶f 2 ¶f 3
= 2 xy =0 = 2y
¶y ¶z ¶x
¶f1 ¶f 2 ¶f 3
Ñ. f = + +
¶x ¶y ¶z
= 2 xy + 0 + 2 y
= 2 xy + 2 y
= 2 y ( x + 1)
(ii) f 1 = x 2 y; f 2 = -2 xz; f 3 = 2 yz
¶f 1 ¶f 2 ¶f 3
= x2 = -2 x =0
¶y ¶z ¶x
¶f1 ¶f 2 ¶f 3
=0 = -2 z = 2z
¶z ¶x ¶y
i j k
Ñ´ f = ¶ ¶ ¶
¶x ¶y ¶z
f1 f2 f3
53
= ê 3 - 2 úi - ê 3 - 1 ú j + ê 2 - 1 ú k
[
= [2 z + 2 x ]i - [0 - 0] j + - 2 z - x 2 k ]
[
= [2 x + 2 z ]i - x 2 + 2 z k ]
10. Find the constants a,b,c so that the vector
f = ( x + 2 y + az )i + (bx - 3 y - z ) j + (4 x + cy + 2 z )k is irrotational.
Solution: f = ( x + 2 y + az )i + (bx - 3 y - z ) j + (4 x + cy + 2 z )k
= f1 i + f 2 j + f 3 k
f1 = x + 2 y + az; f 2 = bx - 3 y - z; f 3 = 4 x + cy + 2 z
¶f 1 ¶f 2 ¶f 3
=z = -1 =4
¶y ¶z ¶x
¶f 1 ¶f 2 ¶f 3
=a =b =c
¶z ¶x ¶y
i j k
Ñ´ f = ¶ ¶ ¶
¶x ¶y ¶z
f1 f2 f3
= ê 3 - 2 úi - ê 3 - 1 ú j + ê 2 - 1 ú k
= [c + 1]i - [4 - a ] j + [b - 2]k
f is irrotational Ñ ´ f = 0
\ [c + 1]i - [4 - a ] j + [b - 2]k = 0
\ c + 1 = 0; 4 - a = 0; b-2=0
c = -1; a = 4; b=2
Hence a = 4, b = 2, c = -1;
11. Prove that r n r is irrotational for any value of n.
Proof : r = xi + y j + z k
54
r = r = x2 + y2 + z2
r 2 = x2 + y2 + z2
¶r ¶r ¶r
\ 2r = 2x 2r = 2y 2r = 2z
¶x ¶y ¶z
¶r x ¶r y ¶r z
= = =
¶x r ¶y r ¶z r
f = rnr
= r n ( xi + y j + z k )
= r n xi + r n y j + r n z k
= f1 i + f 2 j + f 3 k
f1 = r n x
¶f 1 ¶r y
= xnr n -1 = x.nr n -1 .
¶y ¶y r
= nxyr n - 2
¶f1 ¶z
= xnr n -1 = nxzr n - 2
¶z ¶r
f2 = r n y
¶f 2 ¶r
= ynr n -1 = nyzr n - 2
¶z ¶z
¶f 2 ¶r
= ynr n -1 = yxnr n - 2
¶x ¶x
f2 = r n z
¶f 3 ¶r
= zxnr n -1 = nxzr n -1
¶x ¶n
¶f 3 ¶r
= znr n -1 = yznr n - 2
¶y ¶y
55
i j k
Ñ´ f = ¶ ¶ ¶
¶x ¶y ¶z
f1 f2 f3
= ê 3 - 2 úi - ê 3 - 1 ú j + ê 2 - 1 ú k
[ ] [ ] [
= nyzr n - 2 - nyzr n - 2 i - nxzr n - 2 - nyzr n - 2 j + xynr n - 2 - xynr n - 2 k]
=0
\ f is irrotational for any value of
6.2.1. Important vector identities
1. Prove that, if g be a differentiable scalar point function and f be a differentiable vector
point function Ñ.( f g ) = (Ñg ). f + g (Ñ. f )

Proof:
æ ¶ ¶ö
¶
Ñ.( f g ) = çç i + j + k ÷÷. f g ( )
è ¶x ¶y ¶z ø
= i.
¶
¶x
( ) ¶
fg + j . fg +k
¶y
¶
( )
¶z
fg ( )
é¶ f ¶g ù é¶ f ¶g ù é¶ f ¶g ù
= i.ê g+ f ú+ j.ê g+ f ú + k .ê g+ f ú
êë ¶x ¶x úû êë ¶y ¶y úû êë ¶z ¶z úû
é ¶f ¶f ¶f ù é ¶g ¶g ¶g ù
= êi. + j. + k. ú g + êi + j + k ú. f
êë ¶x ¶y ¶z úû ë ¶x ¶y ¶z û
é¶ f ¶g ù é¶ f ¶g ù é¶ f ¶g ù
= i.ê ´g- ´ fú+ j.ê ´g- ´ f ú + k .ê ´g- ´ fú
êë ¶x ¶x úû êë ¶y ¶y úû êë ¶z ¶z úû
æ¶ f ö æ¶ f ö æ¶ f ö
= i.ç ´ g÷+ j.ç ´ g ÷ + k .ç ´ g÷
ç ¶x ÷ ç ¶y ÷ ç ¶z ÷
è ø è ø è ø
é æ ¶g ö æ ¶g ö æ ¶g öù
- êi.ç ´ f ÷ + jç ´ f ÷ + kç ´ f ÷ú
êë çè ¶x ÷ ç ¶y
ø è
÷ ç ¶y
ø è
÷ú
øû
56
æ ¶f ö æ ö æ ö
= çi ´ ÷.g + ç j ´ ¶ f ÷.g + ç k ´ ¶ f ÷.g -
ç ¶x ÷ ç ¶y ÷ ç ¶z ÷ø
è ø è ø è
éæ ¶ g ö æ ö æ ö ù
êç i ´ ÷. f + ç j ´ ¶ g ÷. f + ç k ´ ¶ g ÷. f ú
êëçè ¶x ÷ø ç
è ¶y ÷ø ç
è ¶y ÷ø úû
é ¶f ¶f ¶f ù é ¶g ¶g ¶g ù
= êi ´ + j´ +k´ ú.g - êi ´ + j´ +k´ ú. f
ëê ¶x ¶y ¶z úû ëê ¶x ¶y ¶y ûú
( ) (
= Ñ ´ f .g - Ñ ´ g . f )
4. Prove that Ñ.( f ´ g ) = ( g.Ñ). f - g (Ñ. f ) - ( f .Ñ) g + f (Ñ.g )

Proof:
¶ ¶ ¶
Ñ ´ ( f ´ g) = i ´ ( f ´ g) + j ´ ( f ´ g) + k ´ ( f ´ g)
¶x ¶y ¶z
æ¶ f ¶ g ö÷ æ¶ f ¶ g ö÷ æ¶ f ¶ g ö÷
= i´ç ´g+ f ´ + j´ç ´g+ f ´ + k ´ç ´g+ f ´
ç ¶x ¶x ÷ø ç ¶y ¶y ÷ø ç ¶z ¶z ÷ø
è è è
æ¶ f ö æ ¶ g ö÷ æ¶ f ö
= i´ç ´ g÷ + i´ç f ´ + j´ç ´ g÷+
ç ¶x ÷ ç ¶x ÷ø ç ¶y ÷
è ø è è ø
æ ¶ g ö÷ æ¶ f ö æ ¶ g ö÷
j´ç f ´ + k ´ç ´ g÷ + k ´ç f ´
ç ¶y ÷ø ç ¶z ÷ ç ¶z ÷ø
è è ø è
( )¶¶xf - æçç i. ¶¶xf ö÷÷ g + æçç i. ¶¶xg ö÷÷ f - (i. f )¶¶xf + ( j.g )¶¶yf - æçç j. ¶¶yf ö÷÷ g
= i.g
è ø è ø è ø
æ ¶g ö ¶ f æç ¶ f ö æ ö
+ ç j. ÷ f - j. f
ç ¶y ÷
¶g
¶y
( )
+ k .g
¶ z
- k.
ç ¶z
( ) ÷ ç ¶z ÷
( )
÷ g + ç k. ¶ g ÷ f - k. f ¶ g
¶z
è ø è ø è ø
( )¶¶xf + (g. j )¶¶yf + (g.k )¶¶zf - éêæçç i. ¶¶xf ö÷÷ g + æçç j. ¶¶yf ö÷÷ g + æçç k. ¶¶zf ö÷÷ g ùú
= g .i
ëêè ø è ø è ø ûú
é ¶ g ù æç ¶ g ö÷ æ ¶g ö æ ¶g ö
( )
- ê f .i
¶g
¶x
+ f.j
¶g
¶y
( )
+ f .k ( )
ú + ç i. ÷ f + çç j. ÷÷ f
¶z úû è ¶x ø
+ ç k. ÷ f
ç ¶z ÷
êë è ¶y ø è ø
57
æ ¶ ¶ ¶ö é ¶f ¶f ¶f ù
= çç g.i + j + k ÷÷ f - êi. + j. + k. ú g -
è ¶x ¶y ¶z ø êë ¶x ¶y ¶z úû
æ ¶ ¶ ¶ö é ¶g ¶g ¶g ù
çç f .i + j + k ÷÷ g - êi. + j. + k. ú f
è ¶x ¶y ¶z ø ëê ¶x ¶y ¶z ûú
( ) [ ] ( ) [ ]
= g .Ñ f - Ñ. f g - f .Ñ g + Ñ.g f
5. Ñ.( f .g ) = ( g.Ñ). f + ( f .Ñ) g + g ´ (Ñ ´ f .) + f ´ (Ñ ´ g )

Proof:
¶ ¶ ¶
Ñ ( f .g ) = i ( f .g ) + j ( f .g ) + k ( f .g )
¶x ¶y ¶z
æ¶ f ¶g ö æ¶ f ¶g ö æ ¶ f ¶g ö
= i.ç .g + f . ÷ + jç .g + f . ÷ + k ç .g + f . ÷
ç ¶x ¶x ÷ø ç ¶y ¶y ÷ø çè ¶z ¶z ÷ø
è è
æ¶ f ö æ ¶ f ö æ ¶ f ö æ ¶g ö
= iç .g ÷ + jç .g ÷ + k ç .g ÷ + iç f . ÷
ç ¶x ÷ ç ¶y ÷ ç ¶z ÷ ç ¶x ÷
è ø è ø è ø è ø
----(1)
æ ¶g ö æ ¶g ö
+ jç f . ÷ + k ç f . ÷
ç ¶y ÷ ç ¶z ÷
è ø è ø
æ ¶g ö ¶ g æç ¶ g ö÷
But f ´ ç ´ i ÷ = f .i
ç ¶x ÷ ¶
( )
x
- f.
ç ¶x ÷
i
è ø è ø
æ ¶g ö æ ¶g ö
\ ç f . ÷i = f .i
ç ¶x ÷
¶g
¶x
( )
- f ´ç ´i÷
ç ¶x ÷
è ø è ø
( ) ¶¶xg + f ´ æçç i ´ ¶¶ xg ö÷÷

= f .i
è ø
æ ¶g ö æ ¶g ö
å ç f . ÷i = å
ç ¶x ÷
( f .i )¶¶xg + å f ´ çi ´ ÷
ç ¶x ÷
è ø è ø
58
æ ¶f ö æ ¶f ö
Also g ´ ç i. ´
ç ¶x
÷ = ç g.
÷ ç ¶x ÷
( )
÷i - g.i ¶ f
¶x
è ø è ø
æ ¶f ö æ ö
\ ç g.
ç ¶x ÷ ç ¶x ÷
( )
÷i = g ´ ç i. ¶ f ÷ + g.i ¶ f
¶x
è ø è ø
æ ¶f ö æ ö
\ å ç g.
ç ¶x
÷i = å g ´ ç i. ¶ f ÷ + å g.i ¶ f
÷ ç ¶x ÷ ¶x
( )
è ø è ø
Using in (1)
æ ¶f ö æ ¶g ö
Ñ.( f .g ) = å g ´ ç i ´
ç ¶x ÷
( )
÷ + å g.i ¶ f + å
¶x
( f .i )¶¶xg + å f ´ çi ´ ÷
ç ¶x ÷
è ø è ø
= g ´ (Ñ ´ f ) + ( g.Ñ) f + ( f .Ñ) g + f ´ (Ñ ´ g )
6.2.2 Second order differential operation

The operator Ñ 2 is defined as
¶2 ¶2 ¶2
Ñ2 = + +
¶x 2 ¶y 2 ¶z 2
Ñ 2 is called Laplacian operator

æ ¶j ¶j ¶j ö
Ñ.(Ñf ) = Ñ.çç i + j +k ÷
è ¶x ¶y ¶z ÷ø
¶ æ ¶j ö ¶ æ ¶j ö ¶ æ ¶j ö
= ç ÷+ ç ÷+ ç ÷
¶x è ¶x ø ¶y çè ¶y ÷ø ¶z è ¶z ø
¶ 2f ¶ 2f ¶ 2f
= + +
¶x 2 ¶y 2 ¶z 2
Example 1 : Prove that Ñ.(Ñf ) = 0 (or) Curl (grad f ) = 0

Proof:
¶j ¶j ¶j
Ñf = i +j +k
¶x ¶y ¶z
59
i j k
¶ ¶ ¶
Ñ ´ Ñf) =
¶x ¶y ¶z
¶j ¶j ¶j
¶x ¶y ¶z
é ¶ 2j ¶ 2j ù é ¶ 2j ¶ 2j ù é ¶ 2j ¶ 2j ù
=ê - i
ú ê- - ú j + k ê - ú
ë ¶y¶z ¶y¶z û ë ¶x¶z ¶z¶x û ë ¶x¶y ¶y¶x û
=0
Ñ ´ Ñf ) = 0
2. Prove that Ñ.(Ñ ´ f ) = 0 (or) dir curl f = 0

Proof:
¶j ¶j
f = f1 i + f 2 j + f3 k
¶y ¶z
i j k
¶ ¶ ¶
Ñ´ f =
¶x ¶y ¶z
f1 f2 f3
= ê 3 - 2 úi - ê 3 - 1 ú j + ê 2 - 1 ú k
= ê 3 - 2 úi + ê 1 - 3 ú j + ê 2 - 1 ú k
ë ¶y ¶z û ë ¶z ¶x û ë ¶x ¶y û
¶ æ ¶f 3 ¶f 2 ö ¶ æ ¶f1 ¶f 3 ö ¶ æ ¶f 2 ¶f1 ö
\ Ñ.(Ñ ´ f ) = ç - ÷+ ç - ÷+ ç - ÷
¶x çè ¶y ¶z ÷ø ¶y è ¶z ¶x ø ¶z çè ¶x ¶y ÷ø
¶ 2 f 3 ¶ 2 f 2 ¶ 2 f1 ¶ 2 f 3 ¶ 2 f 2 ¶ 2 f1
= - + - + -
¶x¶y ¶x¶z ¶y¶z ¶y¶x ¶z¶x ¶z¶y
=0
Ñ.(Ñ. f ) = 0
60
3. Prove that Ñ 2 r n = n(n + 1)r n - 2
Proof: f = r n r = x2 + y2 + z2 \r 2 = x2 + y2 + z2
Ñ 2 . f = Ñ.(Ñf )
é ¶f ¶f ¶f ù
= Ñ.êi + j +k ú ---(1)
ë ¶x ¶y ¶z û
n
f =r
¶f ¶r x
= nr n -1 = nr n -1
¶x ¶x r
= nxr n - 2
¶f ¶r y
= nr n -1 = nr n -1
¶y ¶y r
= nyr n - 2
¶f ¶r z
= nr n -1 = nr n -1
¶z ¶z r
= nzr n - 2
[
Ñ.(Ñf ) = Ñ. nxr n - 2 i + nyr n - 2 j + nzr n - 2 k ]
¶ ¶ ¶
¶x
( )
nxr n - 2 +
¶y
(
nyr n - 2 +
¶z
)
nzr n - 2 ( )
é ¶r y ¶r ù
= n ê x ( n - 2) r n - 3 + r n - 2 .1 + y ( n - 2) r n - 3 . + r n - 2 + z ( n - 2) r n - 3 + r n - 2 .1ú
ë ¶n r ¶z û
é r y zù
= n ê3r n - 2 + x( n - 2)r n -3 + y (n - 2)r n -3 . + z (n - 2)r n -3
ë r r r úû
[
= n 3r n - 2 + x 2 (n - 2)r n - 4 + y 2 (n - 2)r n - 4 + z 2 (n - 2)r n - 4 ]
= n[3r n-2
+ (n - 2)r n - 4 ( x 2 + y 2 + z 2 ) ]
= n[3r n-2
+ (n - 2)r n - 4 r 2 ]
= n[3r n-2
+ (n - 2)r n - 2 ]
= n[3 + n - 2]r n - 2
Ñ.Ñr n = n(n + 1)r n - 2
\ Ñ 2 (r n ) = n(n + 1)r n - 2
61
6.3 Let us sum up

urururur u r
So far we have studied the concept of y. f , Ùy fand y f. Also the definition of
solenoidal vectors and irrotational vectors.
(1) Find f = x2 y2 z2
u r u r
fareyxfif
(2) Find r r r
f =x i +y i+ z x
6.5. Lesson End Activities
1. Find Ñf if f = 3 x 2 y - y 3 z 2 at (1,-2,-1)
2. Find a unit normal to the surface x 2 y + 2 xz = 4 at (2,-2,3)
3. If a is a constant vector and r is the position vector of any point (x,y,z), prove that
(i) Ñ(a.r ) = a (ii) Ñ.(a ´ r ) = 0 (iii) (a.Ñ)r = a (iv) Ñ ´ (a ´ r ) = 2a
4. Find the directional derivative of 4 xz 2 + x 2 yz at (1,-2,-1) in the direction of
2i - j - 2k .
5. Find the maximum directional derivative of x 2 + y 2 + z 2 at (2,-2,-2)

6. Find the magnitude and the direction of the greatest directional derivative of
xy + yz + zx at (1,1,3)
7. Find the unit vector normal to the surface x 4 - 3 xyz + z 2 + 1 = 0 at (1,1,1)
8. Find the equation of the tangent plane to the surface x 2 + y 2 + z 2 = 25 at the point
(4,0,3)
9. Find the angle of intersection of the surfaces x 2 + y 2 + z 2 = 29 and
x 2 + y 2 + z 2 + 4 x - 6 y - 8 z - 47 = 0 at (4,-3,2)
10. Find Ñ. f and Ñ ´ f if f = xy 2 i + 2 x 2 yz j - 3 yz 2 k at (1,-1, 1)
11. If f = ( x + y + 1)i + j - (- x - y )k , prove that f .(Ñ ´ f ) = 0
62
12. Determine the constant a so that the vector
f = ( x + 3 y )i + ( y - 2 z ) j + ( x + az )k , is solenoidal.
13. Show that the vector f = (sin y + z )i + ( x cos y - z ) j + ( x - y )k is irrotational.

14. Find the constants a,b,c so that the vector
f = (axy - z 3 )i + (a - z ) x 2 j + (1 - a ) xz 2 k is irrotational
15. Show that f = (2 x 2 + 8 xy 2 z )i + (3 x 3 y - 3 xy ) j + (4 y 2 z 2 + 2 x 3 z )k , is not solenoidal
but g = xyz 2 f is solenoidal.

1. Let a be a constant vector. Prove that

æ r ö 2 é æ 1 öù 3
(i) Ñ(a ´ r ) = 2a (ii) Ñ.(r 3 r ) = 6r 3 (iii) Ñ ´ ç 2 ÷= r (iv) Ñ êrÑç 3 ÷ú = 4
çr ÷ r2 ë è r øû r
è ø
2. Prove that Ñr n = nr n - 2 r where r = xi + y j + z k and r = r
3. Prove that Ñ.(r n r ) = (n + 3)r n
4. For what value of n, Ñ.(r n r ) = 0
[ ]
5. Prove that Ñ ´ f (r )r = 0 if f(r) its differentiable
6.7 Sources
63
Lesson 7
Contents
7.0Aim and Objectives
7.1. Integration of Vector’s
7.2. Surface Integral
7.3. Volume Integral
7.4 Let us sum up
7.7 References

In this lesson we are going to study about the different types of integrals in vector
calculus (viz) line integral, surface integral and volume integral using the fundamentals
of integration.
7.1. Integration of Vector’s

Line Integral
Let f ( x, y, z ) be a vector point function defined through out some region of space
then ò f .dr is defined as the line integral of

c
f along C. Where C is any curve in that
region.
B
Also ò f .d r is called the tangential line integral over C from A to B.
A
Let f = f1 i + f 2 j + f 3 k
r = xi + y j + z k
dr = dxi + dy j + dz k
\ f .dr = f1 dx + f 2 dy + f 3 dz
64
\ ò f .dr = ò ( f dx + f
C C
1 2 dy + f 3 dz )
More over ò f ´ dr and ò f dr are also line integrals.

C C
Problems:
1. Evaluate ò f .dr where f = x 2 i + y 3 j along y = x 2 from (0,0) to (1,1)

C
Solution : The limits for x are x = 0 to x = 1

y = x2
dz = 2 xdx ---(1)
r = xi + y j
dr = dxi + dy j
( )
f .dr = x 2 i + y 3 j .(dxi + dy j )
= x 2 dx + y 3 dy
= x 2 dx + x 6 .2 xdx, using (1)
= x 2 dx + 2 x 7 dx ,
1
\ ò f .dr = ò x 2 dx + 2 x 7 dx
C 0
1 1
x3 ù æ 8ö
= ú + 2ç x ÷
3 û0 è 8 ø0
1 2 14 7
= + = =
3 8 24 12
2. If f = (3 x 2 + 6 y )i - 14 yz j + 20 xz 2 k , evaluate ò f .dr where C is a straight line joining

C
(0,0,0) to (1,1,1)
Solution : Step: Equation of the line joining the points: (x1 ,y1 ,z1 ) and (x2 ,y2 ,z2 ) is
x - x1 y - x1 z - x1
= =
x 2 - x1 y 2 - y1 x 2 - x1
65
x - 0 y - 01 z - 0
= =
1 1 1
x = y = z = t (say )
\ x = t, y = t, z = t
At (0,0,0), t = 0, at (1,1,1)t = 1
\ t = 0 to t = 1
r = xi + y j + z k
(
r =ti+ j+k )
(
dr = i + j + k dt )
( )
f = 3 x 2 + 6 y i + 14 yz j + 20 xz 2 k
\ ( )
f = 3t 2 + 6t i + 14t 2 j + 20t 3 k
f .dr = 3t 2 + 6t - 14t 2 + 20t 3
= 20t 3 - 11t 2 + 6t
1
(
\ ò f .dr = ò 20t 3 - 11t 2 + 6t dt )
C 0
13
=
3
7.2. Surface Integral
In order to evaluate surface integrals it is convenient to express them as double
integrals taken over the orthogonal projection of the surface S on the line of coordinate
places. Let S be a given surface. Let f be the vector point function. Let n be the unit
outward drawn normal vector to the surface S. Then the surface integral is defined as
ò ò f .nds
S
dxdy
a. If the position is on the xy plane, then the surface integral is ò ò f .n
R n.x
where R is the orthogonal projection.
66
b. If the Orthogonal projection R is on yz plane then the surface integral is
dydz
òò f .n
R n. j
dzdx
c. If the Orthogonal projection R is on zx plane then the surface integral is ò ò f .n
R n.k
Examples:
1. Evaluate ò ò f .nds where f = z i + x j + 3 y 2 z k and S is the surface of the

S
cylinder x 2 + y 2 = 16 included in the first octant between z =0 and z = 5.

Solution :
¶j ¶j
f = x2 + y2; = 2 x; = 2y
¶x ¶y
¶j ¶j
Ñj = i + j
¶x ¶y
= 2 xi + 2 y j
Ñf 2 xi + 2 y j
n= =
Ñf 4x 2 + 4 y 2
=
(
2 xi + y j )
2 x2 + y2
xi + y j
=
16
n=
1
4
(
xi + y j )
Let R be the projection of S on xz plane
dxdz
ò ò f .nds = ò ò f .n ---(1)
S R n. j
( 1
)
f .n = z i + x j - 3 y 2 z k . ( xi + y j )
4
67
1
= ( xz + xy )
4
1
= x( y + z )
4
1 y
n. j = ( xi + y j ). j =
4 4
1
x( y + z )
4
\ using in (1) ò ò f .nds = ò ò
S R
y
3
4
5 4 xz + xy
ò ò
z =0 z =0 y
5 4 æ xz ö
=ò ò çç + x ÷÷dxdz
z =0 z =0
è y ø
5 4æ xz ö
=ò ò ç + x ÷÷dxdz
z =0 z =0 ç 2
è 16 - x ø
= 90
7.3. Volume Integral
Let V be a volume bounded by a surface suppose f(x,y,z) is a single valued function of
position defined over V.
Then òòò f ( x, y, z )dv is defined as a volume integral.

V
If we sub divide the volume V into small cuboids by drawing lines parallel to the
three co-ordinate axes, then dv = dxdydz and the volume integral becomes
òòò f ( x, y, z )dxdydz
V
If f is a vector point function, then òòò f dv

V
2
Example: Evaluate òòòfdv where f =45x y and v is the closed region bounded by the
V
planes
4 x + 2 y + 2 = 8; x = 0, y = 0, z = 0
Solution: We have
68
2 4 - 2 x8 - 4 x - 2 y
2
òòòfdv = ò ò ò 45 x
V x =0 y =0 z =0
ydxdydz
In the equation 4 x + 2 y + 2 = 8,
Put y =0, z = 0, 4x = 8
x=2
\ x = 0 to x = 2
Put z = 0,
4x + 2 y + 2 = 8
2x + y = 8
y = 8 - 2x
\ y = 0 to y = 8 - 2 x
Also from
4x + 2 y + 2 = 8
z = 8 - 4x - 2 y
\ z = 0 to y = 8 - 4 x - 2 y
2 4- 2 x
8- 4 x - 2 y
òòòfdv = 45 ò ò x y ( z)
2
dxdy
0
V x =0 y =0
2 4- 2 x
2
= 45ò òx y (8 x - 4 x - 2 y )dydx
0 0
2 4- 2 x
2
= 45ò ò (8 x y - 4 x 3 y - 2 x 2 y 2 )dydx
0 0
2 4-2 x 4-2 x
æ 2 y2 y2 y3 ö
= 45ò ò çç 8 x - 4x3 - 2x 2 ÷ dx
0 0 è 2 3 3 ÷ø 0
2
é x3 ù
= 45ò ê4 x 2 (4 - 2 x) 2 - 2 x 3 (4 - 2 x) 2 - 2 (4 - 2 x) 3 ú dx
0 ë
3 û
= 128
69
7.4 Let us sum up

We have studied so far evaluating the line integral surface integrals and volume
integrals, with the fundamentals of integration of vectors.
1. Evaluate ò f .dr where f = x 2 i + y 3 j and the curve C is the arc of the parabola y = x2
C
in the xy plane from (0,0) to (1,1)

æ 7ö
ç Ans : ÷
è 12 ø
2. Evaluate ò f .dr where f = ( x 2 - y 2 )i + xy j and the curve C is the arc of the curve
C
y=x3 from (0,0) to (2,8)

æ 821 ö
ç Ans : ÷
è 21 ø
3. Evaluate ò ( xdy - ydx ) around the circle x 2 + y 2 = 1
( Ans : 2p )
4. Evaluate ò f .dr where f = xyi + ( x 2 + y 2 ) j and the curve C is the arc of the curve
C
y=x3 -4 from (2,0) to (4, 12)

( Ans : 732)
5. Evaluate ò ò f .nds where
S
f = yzi + zx j + xy k and S is that part of the surface of the
sphere x 2 + y 2 + z 2 = 1 which lies in the first quadrant
æ 3ö
ç Ans : ÷
è 8ø
7.6. Points for discussion.
1. Evaluate ò ò f .nds where f = ( x + y 2 )i + 2 x j + 2 yz k and S is the surface of the

S
plane 2 x + y + 2 z = 6 in the first octant.
( Ans : 81)
70
2. Evaluate ò ò f .nds where

S
f = yi + 2 x j - 2k .Where S is the surface of the
plane 2 x + y = 6 in the first octant cut off by the plane z = 4.

( Ans : 108)
3. If f = (2 x 2 - 3)i + 2 xy j - 4 x k then evaluate . òòò Ñ. f dv Where V is the closed region

V
bounded by the planes x=0, y=0, z=0, and 2x+2y+z=4.

æ 8ö
ç Ans : ÷
è 3ø
7.7 References
71
Lesson – 8
Contents
8.1Theorems of Gauss, Green and Stoke’s.
8.2. Examples
8.3 Let us sum up
8.6 References
In this lesson we are going to study about the theorems of gauss, green and
stroke’s. These theorems help us in evaluating a particular type of integral.
8.1Theorems of Gauss, Green and Stoke’s.

8.1.1.Green’s Theorem: Let R be a closed bounded region in the xy plane whose
boundary C consists of finitely many smooth curves.
Let M and N be continuous function of x and y having continuous partial derivatives
¶M ¶N æ ¶N ¶M ö
¶y
and
¶x
in R. Then ò ò ççè ¶x
R
-
¶y
÷÷dxdy = ò ( Mdx + Ndy )
ø C
8.1.2. Gauss Theorem: Suppose V is the volume bounded by a closed piece wise smooth
surface S. Suppose f = ( x, y, z ) be a vector point function which is continuous and has

continuous first order partial derivatives in V. then
òòò Ñ. f dv = ò ò f .n ds
V S
8.1.3.Stoke’s Theorem: Let S be a piece wise smooth open surface bounded by a piece
wise smooth simple closed curve C. Let f(x,y,z) be a continuous vector point function
which has continuous first order partial derivatives in a region of space which contains S
in its interior. Then
ò f .dr = ò ò (Ñ ´ f ).nds
C S
72
8.2. Examples
1. Gauss Theorem : Verify Gauss Theorem for
f = ( x 2 - yz )i + ( y 2 - zx) j + ( z 2 4 - xy )k taken over the rectangular parallelepiped.

0 £ x £ a , 0 £ y £ b, 0 £ z £ c ,
Solution: Step 1. To find òòò (Ñ. f )dv

V
f = ( x 2 - yz )i + ( y 2 - zx) j + ( z 2 - xy )k
f 1 = x 2 - yz; f 2 = x 2 - zx; f 3 = z 2 - xy;
¶f1 ¶f 2 ¶f 3
= 2 x; = 2 y; = 2 z;
¶x ¶y ¶z
Ñ. f = 2 x + 2 y + 2 z
= 2( x + y + z )
a b c
òòò Ñ. f dv = 2ò ò ò 2( x + y + z )d xdydz
V 0 0 0
a b c
= 2 ò ò ò ( x + y + z )d xdydz
0 0 0
a b
[
= 2 ò ò xz + yz + z 2
2
] dydz
c
0
0 0
[ ]dydz
a b
2
= 2 ò ò cx + ycy + c
2
0 0
[ ]
a b
2
= 2 ò ò cxy + cy 2 + c . y dx
2 2 0
0
[ ]
a
2
= 2 ò ò bcx + b 2 c + b c dx
2 2
0
a a
é bcx 2 b 2 cx bc 2 x ù
= 2ò ê + + ú
0 ë
2 2 2 û0
é a 2 bc ab 2 c abc 2 ù
= 2ê + + ú
ë 2 2 2 û
73
abc
=2 [a + b + c]
2
= abc[a + b + c ] -------(1)
Step 2: To evaluate ò ò f .n ds
S
Z
C
E
H
G
X
A
O
B F
Y
ò ò f .n ds = òò + òò + òò + òò + òò + òò
S S1 S2 S3 S4 S5 S6
Where S1 is OAFB, S2 : GECH; S3 : AEGF; S4 : OBHC; S5 : OCEA; S6 : HBFG
To evaluate òò Put z = 0 in f
S1
z = 0 in n = -k f = - xy k
b a
\ òò = ò ò xydxdz f .n = - xy k .(-k )
OAFB y = 0 x = 0
b a
x2 y ù
= ò ú dy = xy
y =0
2 û0
b
1 2
= a ydy
2 òy
b
1 y2 ù 1
= a 2 ú = a 2b 2
2 2 û0 4
To evaluate S2
74
òò = òò f .n ds
S2 GECH
z = c; n = k
\ f .n = (c 2 - xy )
b a
2
òò = ò ò (c
S2 y =0 x =0
- xy )dxdz
b a
é 2 x2 y ù
= ò êc x - ú dy
y =0 ë
2 û0
b
é 2 a2 y ù
= òê c a - ú dy
y =0 ë
2 û
b
2 a2 y ù
= c ay - ú
2 2 û0
a 2b 2
= abc 2 -
4
To evaluate òò = òò f .n ds
S3 AEGF
x = a; n = ±i
f .n = +( x 2 - yz )
= +(a 2 - yz ) = - yz + a 2
b c
2
\ òò = ò ò (- yz + a )dzdy
S3 y =0 z =0
b c
é yz 2 ù
= ò ê- + a 2 z ú dz
y =0 ë
2 û0
b 2
é y 2 ù
= òy =0 êë 2 + a cúû dy
- c
b
c2 y2 ù
=- + a 2 cy ú
4 û0
b 2c 2
=- + a 2 bc
4
75
S4 OBHC
x = 0; n = -i
f .n = +(- yzi + y 2 j + 3 2 k ).(-i )

= + yz
b c
\ òò = + ò ò yzdzdy
S4 y =0 z =0
b b c
é yz 2 ù yc 2
=+ò ê ú dz = + ò dy
y =0 ë 2 û 0 y =0
2
b
c2 y2 ù
=+ ú
2 2 û0
b 2c 2
=+
4
S5 OCEA
y = 0; n = - j
f .n = ( x 2 i - zx j + z 2 k ).(- j )
= + zx
c a
\ òò = ò ò zxdxdz
S5 z =0 x =0
c a
zx 2 ù
= ò ú dz
z =0
2 û0
c
1
= a 2 ò zdz
2 0
c
1 z2 ö
= a 2 ÷÷
2 2 ø0
a 2c 2
=
4
76
To Find
òò = òò f .n ds
S6 HBFG
n = j; y = b
\ f .n = b 2 - zn
c a
2
\ òò = ò ò (b - zx)dxdz
S6 z =0 x =0
c a
é zx 2 ù
= ò êb 2 x - ú dz
z =0 ë
2 û0
c
é 2 a2z ù
= ò êb a - ú dz
z =0 ë
2 û
c
a2 z2 ù
2
= b az - ú
2 2 û0
a 2c 2
= b 2 ac -
4
\ òò + òò + òò + òò + òò + òò =
S1 S2 S3 S4 S5 S6
2 2
a b a 2b 2 b 2 c 2 b 2c 2 a 2c 2 a 2c 2
+ abc 2 - + + a 2 bc - + + b 2 ac -
4 4 4 4 4 4
= abc 2 + a 2 bc + ab 2 c
= abc ( a + b + c )
\ Gauss Theorem is verified.
2. Stoke’s theorem:
ò f .dr = òò (Ñ ´ f ).nds
S
Problem verify stroke’s theorem for
f = (2 x - y )i - yz 2 j - y 2 z k where S is the upper half surface of the sphere
x 2 + y 2 + z 2 = 1 and C is its boundary

Solution: Step 1: The boundary C of it’s a circle in the xy plane of radius = 1 and centre
at the origin.
x2 + y2 = 1
77
\ x = cos t ; y = sin t ; z = 0,
\ x = cos t ; y = sin t ; z = 0,
t = 0 to 2p
f = (2 x - y )i - yz 2 j - y 2 z k
f .dr = (2 x - y )dx - yz 2 dy - y 2 zdz

x = cos t y = sin t z =0
dx = - sin t dt dy = cos t dt dz = 0
2p
ò f .dr = ò (2 cos t - sin t )(- sin t )dt

0
2p
= ò (-2 sin t cos t + sin 2 t )dt
0
2p 2p
= ò (2 sin t cos t dt + ò (sin 2 t dt
0 0
2p 2p
1
= - ò (sin 2t dt + (1 - cos 2t )dt
0
2 ò0
2p 2p 2p
é - cos 2t ù 1é ù
= -ê + dt - cos 2t dt
2 ë ò0 ò0
ê ú
ë 2 úû 0 û
2p
1 1 æ sin 2t ö
= (1 - 1) + ç t - ÷
2 2è 2 ø0
1
= [(25 - 0) - (0)]
2
ò f .dr = p
Step 2: To evaluate òò (Ñ ´ f ).nds
S
f = (2 x - y )i - yz 2 j - y 2 z k
= f1 i + f 2 j + f 3 k
f1 = 2 x - y f 2 = - yz 2 f3 = - y 2 z
78
¶f 1 ¶f 2 ¶f 3
= -1 = -2 yz =0
¶y ¶z ¶x
¶f 1 ¶f 2 ¶f 3
=0 =0 = -2 yz
¶z ¶x ¶y
i j k
¶ ¶ ¶
Ñ´ f =
¶x ¶y ¶z
f1 f2 f3
æ ¶f ¶f ö æ ¶f ¶f ö æ ¶f ¶f ö
= çç 3 - 2 ÷÷i - ç 3 - 3 ÷ j + çç 2 - 1 ÷÷k .
è ¶y ¶z ø è ¶x ¶z ø è ¶x ¶y ø
= (- xyz + xyz )i - (0 - 0 ) j + (0 + 1)k .
=k
\ Ñ´ f = k
Here n = k
\ òò (Ñ ´ f ).n ds = òò (Ñ ´ f ).k ds
S S
= òò k .k ds
S
= ò ò ds = p
S S
\ Stoke’s theorem is verified.
Green’s Theorem:
æ ¶N ¶M ö
òòR ççè ¶x - ¶y ÷÷dxdy = ò Mdx + Ndy
ø C
Verify Green’s theorem in the plane for ò ( xy + y 2 )dx + x 2 dy where C is the closed curve
C
of the region bounded by y = x and y = x2
79
Solution : Step 1
M = xy = y 2 N = x2
¶M ¶N
= 2y + x = 2x
¶y ¶x
y=x ---(1) and y = x 2 ---(2)
(1) and (2) intersect at (0,0) and (1,1)
\ x = 0 to x = 1
y = x 2 to x
1 x
æ ¶N ¶M ö
òòR ççè ¶x - ¶y ÷÷dxdy =
ø
ò ò [2 x - x - 2 y ]dydx
x =0 y = x 2
1 x
= ò ò [x - 2 y ]dydx
x =0 y = x 2
1 x
ò [xy - y ]
2
= dx
x =0 x2
1
[(
= ò x 2 - x 2 - ( x 3 - x 4 ) dx ) ]
0
1
(
= ò x 4 - x 3 dx )
0
1
x5 x4 ù 1 1 1
= - ú = - =-
5 4 û0 5 4 20
Step 2: To evaluate ò Mdx + Ndy

C
2
y=x
dy = 2 xdx
y
y = x2
y=x
(1,1)
0 x
80
Along y = x2
y = x, dy = dx
x = 0 to x = 1
1
[ ]
= ò x.x + x 2 dx + x 2 dx
0
1
= ò ( x 2 + x 2 + x 2 )dx
0
1
= ò 3 x 2 dx
0
0
3ù
= 3 x ú = -1
3 û1
19 1
\ The required integral = -1 = -
20 20
\ Green’s Theorem is verified
8.3. Let us sum up

So far we have seen the verification of gauss, stoke’s and green’s theorem. For
example if we want to evaluate.
f mdn+nly, we can use green’s theorem depending on the problem given.

1. State Green’s theorem
2. State Gauss theorem
3. State Stoke’s theorem
1. Verify Green’s theorem in a plane with respect to ò ( x 2 - y 2 )dx + 2 xydy Where C
is the boundary of the rectangle in the xoy plane bounded by the lines x = o, x =
a, y = o, y = b (Ans: 2ab2 )
2
2. Verify Green’s Theorem for òò (3x - 8 y 2 )dx + (4 y - 6 xy )dy Where C is the
boundary of the region defined by the lines x = 0, y = 0, x + y = 1. (Ans: 5/3)
81
3. Use Green’s theorem in a plane to evaluate ò (2 x - y )dx + ( x - y )dy where C is
the boundary of the circle x2 +y2 = a2 in xoy plane.
2
4. Use Green’s theorem in a plane to evaluate òx (1 + y )dx + ( x 3 + y 3 )dy Where C
is the square formed by x = ± 1 and y = ± 1.
5. Verify Green’s theorem in a plane with respect to ò ( x 2 dx - xydy ) Where C is the
boundary of the square formed by x = 0, y = 0, x = a , y = a
6. Verify Gauss divergence theorem for f = ( x 2 - yz )i + ( y 2 - zx) j + ( z 2 - xy )k and

the closed surface of the rectangular parallel piped formed by x = 0, x = 1, y = 0,
y = 2, z = 0, z = 3.
7. Verify divergence theorem for f = 4 xzi - y j + yz k when S is a closed surface of

the cube formed by x = o, x = 1, y = o, y = 1, z = 0, z = 1.
2
8. Use divergence theorem to evaluate òò ( yz i + 2 x 2 j + 2 z 2 k ).ds Where S is the
S
closed surface boundary by the xoy plane and the upper half of the sphere
x 2 + y 2 + z 2 = a 2 above this plane
2
9. Use divergence theorem to evaluate òò (4 xi - 2 y j + z 2 k ).ds Where S is the
S
closed surface bounded by the cylinder x 2 + y 2 = 4 and the planes z = 0 and z = 3.

(April 2004)
10. Verify gauss’s theorem for f = 2 xzi + yz j + z 2 k over the upper half of the
sphere x 2 + y 2 + z 2 = a 2 (April 2005)
8.6. Points for discussion
1. Verify divergence theorem for f = ( x 2 - yz )i + ( y 2 - zx) j + ( z 2 - xy )k taken over

the rectangular parallellopiped 0 £ x £ a, 0 £ y £ b (November 2006)
2. Using gauss theorem find the value of ò f .n ds where f = xy 2 i + yz j + zx 2 k and

S is the surface bounded by x = 0,x = 1, y = 0, y =2, z = 0, z = 3 (November 2005)
82
3. Verify gauss theorem for f = 4 xzi + y 2 j - yz k over the cube x = 0, x = 1, y = 0, y

= 1, z = 0, z = 1 (November 2000)
4. Verify Stoke’s theorem for f = xi + z 2 j + y 2 k over the surface x + y + z = 1

lying in the first octant. (November 2000)
5. Verify Stoke’s theorem for f = (2 x - y )i - yz 2 j + y 2 z k Where S is the upper half

of the sphere x 2 + y 2 + z 2 = 1 and C its boundary (April 2004)
6. Verify stoke’s theorem for the function f = - yi + 2 yz j + z 2 k = a 2 and C its

boundary (April 2005)
7. Verify stoke’s theorem for the function f = y 2 i - y j + xz k and for the surface S
which is upper half of the sphere x 2 + y 2 + z 2 = a 2 and z ³ 0 (November 2004
Bharathiar)
8. Verify stoke’s theorem for f = xi + z 2 j + y 2 k over the surface x + y + z = 1 lying

in the first octant (November 2000: Bharathiyar)
9. Verify stoke’s theorem for f = (2 x - x 2 )i - ( x 2 - y 2 ) j and C is the boundary of

3
the region enclosed by the parabolas y2 = x and x2 = y (Ans: - )
5
10. Verify stoke’s theorem to evaluate ò f .dr where f = (sin x - y )i - cos x j and C is
(
the boundary of the triangle. Whose vertices are (0,0), p ,0 and p ,1
2 2
) ( )
Ans: p( 4
+2
p
)
8.7 References

83
Unit III
Lesson - 9
Fourier Series
Contents
9.1 Fourier Series
9.2. Examples
9.3 Let us sum up
9.5 Lesson-End activities
9.6 References
In this lesson we are going to study about the fourier series which is a
trigonometric series defined in various intervals (viz) (0,2B) (-B , B ) (0,2l), (- l,l) and
cosine and sine series of f(x) defined in (o,l).
9.1 Fourier Series:
an ¥ ¥
A trigonometric expression of the form f ( x) = + å an cos nx + å bn sin nx is
2 n =1 n =1
called a Fourier expansion of f(x) where f(x) is defined in a specified internal. a0 , an , bn

are called Fourier Coefficients:
Model :1
Let f(x) be defined in the interval (0,2p ) . Then its fourier expansion is given by
a o ¥ ¥
å
f ( x =) a c n+ on sx+ åsb
2 n =1 n =1
ni nn x --(1)
2p
1
Where a0 = ò f ( x)dx
p 0
84
2p
1
an = ò f ( x) cos nxdx
p 0
2p
1
bn = ò f ( x) sin nxdx
p 0
9.2. Examples
1. Determine a fourier series for f(x) = x2 in (0,2 p )
Step 1: To find a0
2p 2p
1 2 1 x3 ù 1 8p 3 8p 2
a0 = ò0 x dx = ú = . =
p p 3 û0 p 3 3
2p
1 2
an = òx cos nxdx
p 0
1
u = x2 dv = cos nxdx V1 = - cos nx
n2
1
u1 = 2x ò dv = ò cos nxdx V2 = - 2 sin nx
n
sin nx
u 11 = 2 V =
n
1
\ an =
p
[
uj - u 1j1 + u 11j 2 - - - - - ]
2p
1 é x 2 sin nx æ -1 ö æ -1 öù
= ê - 2 xç 2 cos nx ÷ + 2.ç 2 sin nx ÷ú
p ë n èn ø èn øû 0
2p
1 é x 2 sin nx 2 2 ù
= ê + 2 n cos nx - 2 sin nx ú
pë n n n û0
1 ìé 2 ù ü
= íê 2 2p ´ 1ú - [0]ý
p îë n û þ
4p 4
= 2 = 2
pn n
4
an = 2
n
To find bn
85
2p
1 2
bn = òx sin nxdx
p 0
1
u = x2 dv = sin nxdx V1 = - sin nx
n2
1
u1 = 2x ò dv = ò sin nxdx V2 = - 3 sin nx
n
cos nx
u 11 = 2 V =-
n
\ bn =
1
p
[
uv - ò udv ]
2p
1 é 2æ 1 ö æ -1 ö æ 1 öù
= ê x ç - n cos nx ÷ - 2 xç n 2 sin nx ÷ + 2.ç n 3 cos nx ÷ú
p ë è ø è ø è øû 0
2p
1 é - x 2 cos nx 2 2 ù
= ê + 2 x sin nx + 2 cos nx ú
pë n n n û0
1 ìé 1 2 ù é 2 ùü
= íê- (4p 2 ) + 3 ú - ê 3 ú ý
p îë n n û ë n ûþ
1 ì 4p 2 2 2ü
=í- + 3 - 3ý
pî n n n þ
- 4p
bn =
n
\ using in (1)
8p 2 ¥ ¥
3 + 4 4p
f ( x) =
2
å
n =1 n
2
cos nx - ån =1 n
sin nx
¥ ¥
2 cos nx sin nx
= 4p + 4å - 4p å
3 n =1 n 2
n =1 n
Model 2:
Let f(x) be a function of x defined in ( -p , p ) . Then its Fourier expansion is given

by
¥ ¥
a0
f ( x) = + å an cos nx + å bn sin nx
2 n =1 n =1
86
p
1
Where a0 = ò f ( x)dx
p -p
p
1
p -p
p
1
bn = f ( x) sin nxdx
p -òp
Example: f ( x) = x + x 2 ; (-p , p )
p
1
a0 = ò f ( x)dx
p -p
p
1
= ò ( x + x 2 )dx
p -p
p p
1é ù
= êò xdx + ò x 2 dx ú
p ë -p -p û
1 é x2 ö x3 ö ù
= ê ÷÷ + ÷÷ ú
p êë 2 ø -p 3 ø -p ú
û
1 é 1 3 3 ù
=
p
(
êë0 + 3 p - (-p ) úû )
1 1 2p 2
= . .2p 3 =
p 3 3
p
1
p -p
p
1
= ò ( x + x 2 ) cos nxdx
p -p
87
1
u = x + x2 dv = cos nxdx V1 = - cos nx
n2
1
u1 = 1 + 2x ò dv = ò sin nxdx V2 = - 3 sin nx
n
sin nx
u 11 = 2 V =
n
p p
1é ù
= êò xdx + ò x 2 dx ú
p ë -p -p û
1 é x2 ö x3 ö ù
= ê ÷÷ + ÷÷ ú
p êë 2 ø -p 3 ø -p ú
û
1 é 1 3 3 ù
=
p
(
êë0 + 3 p - (-p ) úû )
1 1 2p 2
= . .2p 3 =
p 3 3
p
1
p -p
p
1 2
an = ò (x + x ) cos nxdx
p -p
1
u = x + x2 dv = cos nxdx V1 = - cos nx
n2
1
u1 = 1 + 2x V2 = sin nx
n3
sin nx
u 11 = 2 V =
n
1
\ an =
p
[
uj - u 1j1 + u 11j 2 - - - - - ]
p
1é 2 sin nx æ -1 ö æ -1 öù
= ê( x + x ) n - (1 + 2 x)ç n 2 cos nx ÷ + 2.ç n 3 sin nx ÷ú
pë è ø è ø û -p
1ì sin nx 1 2
+ 2 (1 + 2 x) cos nx - 3 sin nx ]-p
p
= í( x + x 2 )
pî n n n
1 ìé 1 ù é 1 ù
= íê0 + 2 (1 + 2p )(-1) n - 0ú - ê0 + 2 (1 - 2p )(-1) n ú
p îë n û ë n û
88
1 ìé (-1) n ù
= íê 2 (1 + 2p - 1 + 2p )ú
p îë n û
(-1) n 4p (-1) n 4
= =
pn 2 n2
(-1) n 4
\ an =
n2
p
1
bn = ò f ( x) sin nxdx
p -p
p
1 2
= ò (x + x ) sin nxdx
p -p
1
u = x + x2 dv = sin nxdx V1 = - sin nx
n2
1
u1 = 1 + 2x ò dv = ò sin nxdx V2 = 3 cos nx
n
- cos nx
u 11 = 2 V =
n
bn =
1
p
[
uv - ò udv ]
p
1ì æ cos nx ö æ -1 ö æ 1 öü
= í( x + x 2 )ç - ÷ - (1 + 2 x)ç 2 sin nx ÷ + 2.ç 3 cos nx ÷ý
pî è n ø èn ø èn ø þ -p
p
1ì 2 cos nx 1 2 ü
= í- ( x + x ) + 2 (1 + 2 x) sin nx + 3 cos nx ý
pî n n n þ -p
1 ìé (p + p 2 )(-1) n 2 nù é (-p + p 2 )(-1) n 2 ùü

= íê - + 0 + 3
( - 1) ú ê-
- + 0 + 3 (-1) n ú ý
p îë n n û ë n n ûþ
1 ì (-1) n 2 2 ü
= í (-p - p 2 + p 2 - p ) + 3 (-1) n - 3 (-1) n ý
pî n n n þ
1 é (-1) n ù
= ê (-2p )ú
pë n û
- 2p - 2(-1) n
= (-1) n =
pn n
89
p2 ¥ é4 2 ù
\ f ( x) = + å (-1) n ê 2 cos nx - sin nx ú
3 n =1 ën n û
Model 3: Change of Interval:
Let f(x) be a function of x defined in the interval (0, 2l). Then its fouriers
expansion is
a0 ¥ æ npx ö ¥ æ npx ö
f ( x) = + å an cosç ÷ + å bn sin ç ÷
2 n =1 è l ø n =1 è l ø
2l
1
Where a 0 = ò f ( x)dx
l0
2l
1 æ npx ö
an = ò f ( x) cosç ÷dx
l0 è l ø
2l
1 æ npx ö
bn = ò f ( x) sinçè ÷dx
p 0
l ø
Problem:
Determine a fourier series for
x
f ( x) = ;0 < x < l
l
2l - x
= ; l < x < 2l
l
2 n =1 è l ø n =1 è l ø
2l
1
a 0 = ò f ( x)dx
l0
æL 2l
ö
= ç ò f ( x)dx + ò f ( x)dx ÷÷
ç
è0 0 ø
L 2l
1 éæ x) æ 2l - x ö ö÷ù
= êçç ò dx + ò ç ÷dx ÷ú
l êëè 0 l 0 è l ø øúû
90
l 2l
1 é x2 ö æ x2 ö ù
= 2 ê ÷÷ + çç 2lx - ÷÷ ú
l êë 2 ø0 è 2 øl ú
û
1 é l 2 æ 2 4l 2 ö 2 l2 ù
= 2 ê + çç 4l - ÷ - (2l - ú
l ë2 è 2 ÷ø 2û
1 é l 2 4l 2 3l 2 ù
= ê + - ú
l2 ë 2 2 2 û
1 é l 2 l 2 ù 1 2l 2
= ê + ú= . =1
l2 ë 2 2 û l2 2
a0 = 1
To find an
2l
1 æ npx ö
l0 è l ø
L 2l
1 éæ x æ npx ö 2l - x æ npx ö ö÷ù
ç
= êç ò cosç ÷dx + ò cosç ÷dx ú
l ëêè 0 l è l ø 0
l è l ø ÷øûú
1
= [I1 + I 2 ] --------(1)
l2
L 2l
x npx æ npx ö
I 1 = ò cos dx; I 2 = ò (2l - x) cosç ÷dx
0
l l 0 è l ø
To find I1
Integrate by Parts
npx
u=x dv = cosdx
l
npx
u1 = 1 ò dv = ò cos l dx
1 æ npx ö
v= sin ç ÷
np è l ø
\ I 1 = uv - ò vdx
91
l l
l npx ù l æ npx ö
= x. sin ú -ò sin ç ÷dx
np l û 0 0 np è l ø
l
é æ npx ö ù
ê - cosç ÷
l
ê è l ø úú
=-
np ê np ú
ê l ú
ë û0
l2
= - 2 2 [(cos(np ) - 1]
np
l2
=- 2 2
np
[
(-1) n - 1 ]
To find I2
2l
æ npx ö
I 2 = ò (2l - x) cosç ÷dx
0 è l ø
u = 2l - x dv = -dx
npx
u 1 = -1 ò dv = ò cos dx
l
1 æ npx ö
v= sin ç ÷
np è l ø
\ I 1 = uv - ò vdx
2l 2l
l npx ù l æ npx ö
= (2l - x) sin ú -ò sin ç ÷(-dx)
np l ûl l
np è l ø
2l
l é æ npx öù
= êsin ç l ÷ú
np ë è øû l
2l
l2 é æ npx öù
= 2 2 êcosç l ÷ú
n p ë è øû l
l2
= - 2 2 {cos 2np - cos np }
np
l2
=- 2 2
np
[
1 - (-1) n ]
92
\ using in (1)
1 é l2 n l2 ù
an = 2 ê 2 2
(( -1) - 1) - 2 2
(1 - (-1) n )ú
l ën p np û
2
1 l
[
= 2 . 2 2 (-1) n - 1 - 1 + (-1) n
l np
]
1
= 2 2 2(-1) n - 2
n p
[ ]
2
a n = 2 2 (-1) n - 1
n p
[ ]
When n is even, an = 0
2 -4
When n is odd, a n = 2 2
( -2) = 2 2
n p n p
Step 3: To find bn
2l
1 æ npx ö
bn = ò f ( x) sin ç ÷dx
l0 è l ø
l 2l
1 éæ æ npx ö æ npx ö ö÷ù
= êçç ò f ( x) sin ç ÷ dx + ò0 f ( x ) sin ç ÷dx ú
l êëè 0 è l ø è l ø ÷øúû
1 éæ l æ npx ö
2l
æ npx ö ö÷ù
= ç
êç ò x sin ç ÷ dx + ò f ( x ) sin ç ÷dx ÷ú
l2 ëêè 0 è l ø 0 è l ø øûú
1
= [I 3 + I 4 ] --------(2)
l2
Where
l 2l
æ npx ö æ npx ö
I 3 = ò x sin ç ÷dx; I 4 = ò (2l - x) sin ç ÷dx
0 è l ø l è l ø
l
æ npx ö
I 3 = ò x sin ç ÷dx
0 è l ø
93
æ npx ö
u=x dv = sin ç ÷dx
è l ø
æ npx ö
u1 = 1 ò dv = ò sinçè l ÷ødx
l æ npx ö -l2 npx
du = dx v=- cosç ÷; v1 = - 2 2
sin
np è l ø np l
I 3 = uv - u 1v1 + .......
l
é æ l ö æ npx ö æ -l2 ö æ npx öù
= ê xç - cos
÷ ç ÷ - 1 .çç 2 2 ÷÷ sin ç ÷ú
ë è np ø è l ø èn p ø è l øû 0
l
é l æ npx ö l2 æ npx öù
= ê- x x cosç ÷ + 2 2 sin ç ÷ú
ë np è l ø np è l øû 0
ì l ü
[ ]
= í- 2 2 (l (-1) n + 0) - [0]ý
î n p þ
-l2
= (-1) n
np
2l
æ npx ö
I 4 = ò (2l - x) sin ç ÷dx
l è l ø
æ npx ö
u = 21 - x dv = sin ç ÷dx
è l ø
æ npx ö
u 1 = -1 ò dv = ò sinçè l ÷ødx
l æ npx ö -l2 npx
v=- cosç ÷; v1 = 2 2
sin
np è l ø np l
I 3 = uv - u 1v1 + .......
2l
é 1 æ npx ö æ -l2 ö æ npx öù
= ê- (2l - x) cosç ÷ - 1.çç 2 2 ÷÷ sin ç ÷ú
ë np è l ø èn p ø è l øû l
2l
é l æ npx ö l2 æ npx öù
= ê- (2l - x) cosç ÷ - 2 2 sin ç ÷ú
ë np è l ø np è l øû l
94
ì é l2 ùü
= í[- (0 - 0)] - ê- 2 2 cos(np ) - 0ú ý
î ë np ûþ
l2 l2
= cos np = (-1) n
np np
Using in (2)
1 é- l2 n l2 ù
bn = 2 ê (-1) + (-1) n ú = 0
l ë np np û
bn = 0
¥
1 4 æ npx ö
f ( x) = - å 2 2 cosç ÷
2 n =1,3,5 n p è l ø
1 4 ¥ æ npx ö
= . 2 å cosç ÷
2 p 1,3,5 è l ø
Model 4:
Let f(x) be a function of x defined the interval (-l , l ) . Then its fourier series is
2 n =1 è l ø n =1 è l ø
l
1
a0 = f ( x)dx
l -òl
Where
l
1 æ npx ö
a n = ò f ( x) cosç ÷dx
l -l è l ø
l
1 æ npx ö
bn = ò f ( x) sinçè ÷dx
p -l
l ø
Problem: Determine a fourier series for f(x) = x2 in (-1,1)
Here l = 1
¥ ¥
a0
f ( x) = + å an cos(npx ) + å bn sin (npx )
2 n =1 n =1
1 -1
2 x3 ù 13 (-1) 3 2
a 0 = ò x dx = ú = - =
-1
3 û1 3 3 3
95
1
a n = ò x 2 cos npxdx
-1
u = x2 dv = cos npxdx
sin npx
u1 = 2x v=
np
1 1
u 11 = 2 v1 = - 2 2
cos npx; v 2 = - 3 3 sin npx
np np
a n = uv - u 1v1 + u 11v 2 .......
1
x 2 sin npx æ 1 ö æ 1 öù
= - 2 xç - 2 2 cos npx ÷ + 2ç - 3 3 sin npx ÷ú
np è n p ø è np øû -1
1
é x 2 sin npx 2 2 ù
=ê + 2 2 x cos npx - 3 3 sin npx ú
ë np np np û -1
ìé 2 ù é 2 ùü
= íê0 + 2 2 cos np - 0ú - ê0 - 2 2 cos np - 0ú ý
îë np û ë np ûþ
2 2
= 2
2
cos np + 2 2 cos np
n p n p
4
= 2 2 cos np
n p
4 n (-1) n 4
= ( -1) =
n 2p 2 n 2p 2
To find bn :
1
bn = ò f ( x) sin npxdx
-1
1
= ò x 2 sin npxdx
-1
u = x2 dv = sin npxdx
1
u1 = 2x v=- cos npx
np
1 1
u 11 = 2 v1 = - 2 2 sin npx; v 2 = 3 3 cos npx
np np
96
\ bn = uj - u 1j1 + u 11j 2 + u 111j 3 + ......
1
æ 1
2 ö æ 1 ö æ 1 öù
= x ç- cos npx ÷ - 2 xç - 2 2 sin npx ÷ + 2ç 3 3 cos npx ÷ú
è np ø è n p ø èn p øû -1
1
1 2 2 2 ù
=- x cos npx + 2 2 x sin npx + 3 3 cos npx ú
np np np û -1
ìé 1 2 ù é 1 2(-1) n ù ü
= íê- (-1) n + 0 + 3 3 (-1) n ú - ê- cos np + 0 + 3 3 ú ý
î ë np np û ë np n p ûþ
1 2 1 2
=- (-1) + (-1) n 3 3 + cos np - 3 3 (-1) n
np np np np
bn = 0
1 ¥ 4
\ f ( x) = + å (-1) n 2 2 cos(npx )
3 n =1 n p
Model 5: Cosine series and sine series
a). Let f(x) be a function defined in the interval (0,l) then its cosine series is defined
¥
a æ npx ö
f ( x) = 0 + å an cosç ÷
2 n =1 è l ø
l
2
a 0 = ò f ( x)dx
l 0
l
2 æ npx ö
l 0 è l ø
¥
æ npx ö
b. Its sine series is f ( x) = å bn sin ç ÷
n =1 è l ø
l
2 æ npx ö
Where a 0 = ò f ( x) sin ç ÷dx
l 0 è l ø
Examples
1. Determine a cosine series for f ( x) = ( x - 1) 2 ; in 0 < x < 1.
Here l = 1
97
¥
a0
f ( x) = + å an cos npx
2 1
1 1
a 0 = 2 ò f ( x)dx = 2 ò ( x - 1) 2 dx
0 0
1
( x - 1) 3 ù
=2 ú
3 û0
2 2
= (0 + 1) =
3 3
l
a n = 2ò f ( x) cos npxdx
0
l
= 2 ò ( x - 1) 2 cos npxdx
0
u = ( x - 1) 2 dv = cos npxdx
u 1 = 2( x - 1) ò dv = ò cos npxdx
sin npx
u 11 = 2 v=
np
1
v1 = - 2 2 cos npx
np
1
v 2 = - 3 sin npx
n p3
[
a n = 2 uj - u 1j1 + u 11j 2 .......... ]
1
é sin npx æ 1 ö æ 1 öù
= 2 ê( x - 1) 2 - (2)(n - 1)ç - 2 2 cos npx ÷ + 2ç - 3 3 sin npx ÷ú
ë np è np ø è np øû 0
1
é sin npx 2 2 ù
= 2 ê( x - 1) 2 + 2 2 (n - 1) cos npx - 3 3 sin npx ú
ë np np np û0
ì é 2 ùü
= 2í[0 + 0 + -0] - ê- 2 2 ´ 1 - 0ú ý
î ë np ûþ
4
an = 2 2
n p
1 ¥ 4
\ f ( x) = +å cos npx
3 n =1 n 2p 2
98
b) Sine series
f ( x) = ( x - 1) 2 ; (0,1).
¥
æ npx ö
f ( x) = å bn sin ç ÷;
1 è l ø
Here l = 1
1
\ bn = 2 ò ( x - 1) 2 sin (npx )dx
0
u = ( x - 1) 2 dv = sin npxdx
1
u = 2( x - 1) ò dv = ò sin npxdx
- cos npx
u 11 = 2 v=
np
1
v1 = - 2 2 sin npx
np
1
v 2 = 3 cos npx
n p3
[
bn = 2 uj - u 1j1 + u 11j 2 .......... ]
é æ 1 ö æ 1 ö æ 1 öù
= 2 ê( x - 1) 2 ç - cos npx ÷ - 2(n - 1)ç - 2 2 sin npx ÷ + 2ç 3 3 cos npx ÷ú
ë è np ø è np ø èn p øû
1
ì 1 2(n - 1) 2 ü
= 2 í- ( x - 1) 2 cos npx + 2 2 sin npx + 3 3 cos npx ý
î np np np þ0
ìé 2 ù é 1 2 ùü
= 2íê0 + 0 + 3 3 cos np ú - ê + 0 + 3 3 úý
îë np û ë np n p ûþ
é 2 2 ù
= 2 ê 3 3 ( -1) n - 3 3 ú
ën p np û
4
[
bn = 3 3 (-1) n - 1
np
]
When n is even, bn = 0
-8
When n is odd, bn =
n 3p 3
¥
8 1
\ f ( x) = -
p 3
+ å
n =1, 3, 5,.... n
3
sin npx
99
9.3. Let us sum up

We studied so far in finding fourier series in the intervals (0,2B),(-B,B), (0,2l),
(-l,l) etc.,
Due care should be taken in finding an and bn while for use the formula
“Bernonllis formula on Integration by parts.”

(1) Find a0 if f(x) =x; (0,2B)
(2) Find bn if f(x) = x2 (-B,B)
(3) Find an if f(x) = a, (0,2l) where a is a constant
1. If f ( x ) = 1, 0 < x < p ,
= -1, p < x < 2p , prove that
¥
4 sin nx
f ( x) = 3 å
p n =1,3,5,.... n
2. If f ( x ) = x ( 2p - n), show that
2p 2 ¥
1
f ( x) = - 4å 2 cos nx; (0,2p )
3 n =1 n
3. If f ( x ) = R, 0 < x < p ,
= - R, p < x < 2p ,
Prove that
4n ¥ 1
f ( x) = å sin nx
p n =1,3,5,.... n
4. If f ( x ) = -1 + x, - p < x < 0
= 1 + x, 0 < x < p
Prove that
2 ¥ 1
[ ]
f ( x) = å 1 - (-1) n (1 + p ) sin nx
p n =1 n
1. f ( x )= -p , - p < x < 0
= p, 0 < x < p
Show that
-p 2 æ cos 3 x cos 5 x ö
f ( x) = - ç cos x + 2
+ 2
+ ....... ÷
4 pè 3 5 ø
100
2. f ( x ) = 2, - p < x < 0
= 4, 0 < x < p
Prove that
4 ¥ sin nx
f ( x) = 3 + å
p n =1,3,5.... n
9.7 References
1. Fourier series by S. Narayanan T.K.M. Pillai
101
Lesson 10
Contents
10.0 Aims and Objective

10.1 Analytical Geometry
10.2 Examples
10.3 Let us sum up
10.7 References
In this lesson we are going to learn about the concept of the polar coordinates,
converting the castesian system to polar system, polar of a straight line, circle, a concern.
10.1 Analytical Geometry
10.1.1. Polar coordinates:
We shall study another type of coordinates called polar coordinates to represent

plane curves. By this system of coordinates, any point P is specified by its distance from
a fixed point O and the angle θ made by the join of P and O with a fixed line out.
θ
0 Fixed line A
The point P is represented by the polar coordinate P(r, θ); OP = r is called the
radius vector and θ is called the vectorial angle. The radius vector is considered positive
if measured from O along the line bounding the vectorial angle and negative in the
opposite direction.
102
10.1.2. Distance between the points (r1 , θ1 ) and (r2 , θ2 ).
θ2
θ1
0 A
Let the points be P (r1 , θ 1 ) and Q (r2 , θ2 ).
OP = r1 ; OQ=r2 ; AOP = q1 ; AOQ = q 2
POQ = J2 - J1
In D OPQ, we have
OP 2 + 0Q 2 - 2OP.OQ cos POQ = PQ 2 (Cosine formula)

2 2
r1 + r2 - 2r1 r2 cos(J 2 - J1 ) = PJ 2
2 2
\ PQ = r1 + r2 - 2r1 r2 cos(J2 - J1 )
10.1.3. Area of triangle in polar coordinates
R
Q
θ2 θ3
θ1
0 A
Let O be the pole:Let OA be the initial lline
Let PQR be the triangle whose vertices are P (r1 , θ 1 ) , Q (r2 , θ2 ) and R (r3 , θ3 )
Let OP = r1 ; OQ=r2 ; QR=r3
103
AOˆ P = q1 ; AOˆ Q = q 2 AOˆ Q = q 3
In D OPQ, POQ = J2 - J1
1
\ Area of DOPQ = .OP.OQ. sin(q 2 - q 1 )
2
1
= .r1 .r2 . sin(q 2 - q 1 )
2
In D QOR, QOR = J3 - J2
1
\ Area of DQOR = .OQ.OR. sin QOR
2
1
= .r2 .r3 . sin(q 3 - q 2 )
2
In D OPR, POR = J3 - J1
1
\ Area of DOPR = .OP.OR. sin POR
2
1
= .r1 .r3 . sin(q 3 - q 2 )
2
\ Area of DPQR =
= Area of D OPQ + Area of D QOP – Area of D OPR
1 1 1
= .r1 .r2 . sin(q 2 - q 1 ) + .r2 .r3 . sin(q 3 - q 2 ) - .r1 .r3 . sin(q 3 - q 1 )
2 2 2
1
=
2
[
. r1 .r2 . sin(q 2 - q 1 ) + .r2 .r3 . sin(q 3 - q 2 ) - .r1 .r3 . sin(q 3 - q 1 ) ]
10.1.4. Conversion of polar coordinates into Cartesian coordinates
Rule: To convert the Cartesian coordinate (x,y) into polar coordinate.
Put x = r cos θ; y = r sin θ
r= x2 + y2
æ yö
q = tan -1 ç ÷
èxø
Equation of a straight in polar coordinates
104
P
α A
x
0
Continue 110 pages:

Consider a straight line AB. Draw a perpendicular from the origin to the line B.
Let this perpendicular make an angle α with the x axis.
Then the equation of the straight line AB is
x cos α + y sin α = r ----(1)
Taking the origin as the pole and the x axis as the initial line
Put x = r cos θ, y = r sin θ in (1)
r cos q cos a + r sin q sin a = p
r [cos q cos a + sin q sin a ] = p
r cos(q - a ) = p
This is the polar equation of a straight line.
Equation of a straight line not passing through the pole.
Consider the equation of the straight line as ax+by+c = 0 ----(1)
Transforming into polar coordinates
Put x = r cos θ, y = r sin θ in (1)
ar cos q + br sin q + c = 0
c
a cos q + b sin q + = 0
r
105
c c
a cos q + b sin q = -= K where K = -
r r
Which is the equation of the line not passing through the pole.
The equation of the line passing through the pole is q = a constant.
Parallel lines:
If the equation of the line is

a cos q + b sin q = K , then the equation of the line parallel to this is
a cos q + b sin q = K '
Perpendicular lines:
Consider the perpendicular lines
Ax+By+C = 0 ----(1) and Bx-Ay-C' = 0 ---(2)
Put x = r cosθ, y = r sin θ in (1) & (2)
We get
A cosθ + B sinθ = K ----(1)

B cosθ – A sinθ = K' ----(2)
We observe that the line perpendicular to (1) is obtained by replacing θ by

p
+ q and K by K'.
2
Equation of a circle in polar coordinates
P
c
l
θ
α
A
0
106
Let 0 be the pole and OA be the initial line. Let C be the centre of the circle of radius a
units.
Let P be any point on the circle and OP = r; AOˆ P = J

Let AOˆ P = a ; let OC = c
\ \ COP = q - a
In DOCP
CP 2 = OC 2 + OP 2 - 2OC.OC. cos(COP )
a 2 = c 2 + r 2 - 2cr cos(q - a ) -----(1)
Cor1: If the pole lies on the circumference on the circle, then c = a
\ using in (1)
a 2 = a 2 + r 2 - 2ar cos(q - a )
\ r 2 = 2ar cos(q - a )
\ r = 2a cos(q - a ) ----(2)
10.1.5. Equation of chord joining the points P(q1 )andP(q 2 ) of the circle r = 2acos q
Proof:
Q P
O
r=2acosθ
Consider the equation of the circle where the pole lies on it and the initial line passes
through the centre of the circle.
The equation of the circle is r = 2acosθ --(1)
The points P(q1 )andP(q 2 ) lie in the circle (1)
107
\ the radii vectors of P and Q are 2acosθ1 , and 2acosθ2 respectively
\ P(2a cosq1 ,q1 ) and Q(2a cosq 2 ,q 2 )
Let the equation of the straight line PQ be
p = rc o-q s a( ) ----(2)
\ The points P,Q lie on (2)
\ p = 2ac o qs1 c- oq(1 s a ) ----(3)
o(2 s a
p = 2ac o qs2 c- q )
\ 2a cos q1 cos(q1 - a ) = 2a cos q 2 cos(q 2 - a )
2 cos q1 cos(q1 - a ) = 2 cos q 2 cos(q 2 - a )
But 2 cos A cos B = cos( A + B ) + cos( A - B )
\ cos(q1 + q1 - a ) + cos a = cos(q 2 + q 2 - a ) + cos a
cos(2q1 - a ) = cos(2q 2 - a )
\ 2q1 - a = ±2q 2 - a
\ q1 ¹ q 2 Taking negative sign 2q1 - a = -(2q 2 - a )
2q1 - a = -2q 2 + a
2q1 + 2q 2 = 2a
q1 + q 2 = a
Sub in (3)
p = 2a c qo-1 s q1 c -q
o1 s 2q ( )
= 2a cos q 1 cos(-q 2 )
p = 2ac o qs1 c oq2 s
108
Substitute the value of ** and α in (2) the equation of PQ is
2a cos q 1 cos q 2 = r cos(q - q 1 - q 2 ) ---------(A)
Cor 1: To find the equation of the tangent at ‘α’
Put q 1 = q 2 = a in (A)
2a cos a . cos a = r cos(q - 2a )
2a cos 2 a = r cos(q - 2a )
\ r cos(q - 2a ) = 2a cos 2 a
10.1.6.Polar equation of conic

L
P
M
r
θ
X
S N A α
L'
Let S be the focus and SX be the initial line

Let XM be the directrix. Let e be the eccentricity of the conic
Draw SX perpendicular to the directrix..
Let P be any point on the conic
Let P be (r,θ)
Let SP = r; XSP = θ
Draw PM perpendicular the directrix.
PN perpendicular the initial line.
109
Let LSL' be the latus rectum of the conic.

L is a point on the conic.
\ By the definition of the conic,
SL
=e
SX
SL = e.SX
But SL = l
\ l = e.SX
l
SX =
e
Also P is a point on the conic
SP
\ =e
PM
SP = e.PM
= e.NX
= e( SX - SN ) ---(1)
l
But SX =
e
SN
In DSPN , cos q =
SP
S N
c oq s=
r
\S N
= cr oq s
\ using in (1)
æl ö
r = eç - r c ÷oq s
èe ø
r = l -e cr oq s
110
r + e c o=rq s l
r ( +1 ce oq=s l )
l
= 1 +ce o qs
r
Note: Let SX makes an angle α with the initial line SA, then the radius vector SP makes
an angle q - a with the initial line
l
\ The equation of the conic in this case is = 1 ec+ o q-s a
( )
r
Note 2: Directrix corresponding to the poles.
Q
r
S X
Let Q (r, θ) be any point on the directrix.
SX
\ In DSQX ; cos q =
SQ
\ SX = SQ cos q
= r cos q
l
But SX =
e
l
\ = r cos q
e
111
l
\ = e cos q
r
l
Note 3: Consider the equation of the conic = e cos q + e cos(q - a )
r
l
\ Equating of the corresponding directrix is = e cos(q - a )
r
10.1.7.Equation of the chord joining the points A(a - b ) and B (a + b ) of the conic
l
= 1 + e cos q
r
l
Solution: The equation of the conic is = 1 + e cos q (1)
r
Let A and B be two points on the conic (1) whose vectorial angles are ‘α-β’ and ‘α+β’
respectively.
α+β A
S α-β
The equation of any chord not passing through the pole is

l
= A' cos q + B ' cos q -----(2)
r
The points A(SA, α-β) and B (SB, α+β) lie on (1) and (2)
112
l
\ = 1 + e cos(a - b ) ----(2)
SA
l
= 1 + e cos(a + b ) ----(3)
SB
Also
l
= A' cos(a - b ) + B ' sin(a - b ) ----(4)
SA
l
= A' cos(a + b ) + B ' sin(a + b ) ----(5)
SB
From (2) and (3)
1 + e cos(a - b ) = A' cos(a - b ) + B ' sin(a - b )
\ ( A'-e) cos(a - b ) + B ' sin(a - b ) = 1 ----(6)
From (2) and (5)
1 + e cos(a + b ) = A' cos(a + b ) + B ' sin(a + b )
\ ( A'-e) cos(a + b ) + B ' sin(a + b ) = 1 ----(7)
Solve (6) and (7)
(6) X sin (α+β) given
( A'-e) cos(a - b ) sin(a + b ) + B ' sin(a - b ) sin(a + b ) = sin(a + b )
(7) X sin (α-β) given
( A'-e) cos(a + b ) sin(a - b ) + B ' sin(a + b ) sin(a - b ) = sin(a - b )
Subtracting
( A'-e)[sin(a + b ) cos(a - b ) - cos(a + b ) sin(a - b )] = sin(a + b ) - sin(a - b )
( A'-e)[sin (a + b - a + b )] = 2 cos a sin b
2 cos a sin b
\ A'-e =
sin 2 b
113
2 cos a sin b
=
2 sin b cos b
A'-e = cos a sec b
To find B'
(6) X cos (α+β) given
( A'-e) cos(a - b ) cos(a + b ) + B ' sin(a - b ) cos(a + b ) = cos(a + b )
(7) X cos (α-β) given
( A'-e) cos(a + b ) cos(a - b ) + B ' sin(a + b ) cos(a - b ) = cos(a - b )
Subtracting
B' [sin(a - b ) cos(a + b ) - sin(a + b ) cos(a - b )] = cos(a + b ) - cos(a - b )
B' [sin (a - b - a - b )] = -2 sin a sin b
B ' sin( -2 b ) = -2 sin a sin b
- B ' sin 2 b = -2 sin a sin b
2 sin a sin b
B' =
2 sin b cos b
B ' = sin a sec b
\ A' = e + cos a sec b
Substitute A', B' in (2)
Equation of chord AB is
l
= (e + cos a sec b ) cos q + sin a sec b sin q
r
= e cos q + cos a sec b cos q + sin a sec b sin q
114
= e cos q + sec b [cos q cos a + sin q sin a ]
l
= e cos q + sec b cos(q - a ) -----(A)
r
Cor 1:
When b = 0 , the points A and B coincide. Chord AB becomes the tangent at ‘α’
\ Put b = 0 in (A)
Equation of the tangent at ‘α’ is
l
= e cos q + cos(q - a )
r
l
BW Equation of the normal at ‘α’ on the conic = 1 + e cos q
r
Let the equation of the conic be
l
= 1 + e cos q ---(1)
r
Equation of the tangent at ‘α’ is
l
= e cos q + cos(q - a ) ---(2)
r
\ The equation of the line perpendicular to the line (2) is
R æp ö æp ö
= e cosç + q ÷ + cosç + q - a ÷
r è2 ø è2 ø
R
= -e sin q - e sin(q - a ) ---(3)
r
If this is the normal at P(α), then P (SP, α) lies on (3)
R
= -e sin a ---(4)
SP
P (SP, α) lies on the conic.
115
l
\ = 1 + e cos a
SP
l
SP =
1 + e cos a
Using in (4)
- le sin a
R=
1 + e cos a
Using in (3)
Equation of the normal at ‘α’.
- le sin a 1
. = -e sin q - e sin(q - a )
1 + e cos a r
- le sin a
\ = e sin q + e sin(q - a )
r (1 + e cos a )
10.1.8. Properties
1. If the tangent at P to a Conic meets the directrix at K, prove that KSˆP = 90 0
Solution : P (α)
S x
directrix
116
Let the equation of the conic be
l
= 1 + e cos q ---(1)
r
Let P be any point on it whose vectorial angle is α
\ The equation of the tangent at P(α) is
l
= e cos q + cos(q - a ) ---(2)
r
The equation of the directrix is
l
= e cos q ---(3)
r
Solve (2) and (3)
e cos q = e cos q + cos(q - a )
\ cos(q - a ) = 0
p
q -a = ±
2
p
\q = a ±
2
\ KSP = ZSP = ZSK
æ pö
= a - ça ± ÷
è 2ø
p
KSˆP = ±
2
l
2. Prove that the tangents at the extremities of any focal chord of a conic = 1 + e cos q
r
intersect on the corresponding directrix.
l
Proof: Equation of the conic is = 1 + e cos q ---(1)
r
Let PQ be a focal chord of (1)
Let P be (SP, α). Then Q is (SQ, p + a )
117
\ The tangents at P,Q is

l
= e cos q + cos(q - a ) ---(2)
r
l
= e cos q + cos(q - a - p ) ---(3)
r
Let (r1 , q 1 ) be the point of intersection of the tangents at P,Q
l
\ = e cos q 1 + cos(q 1 - a )
r1
l
and = e cos q 1 + cos(q 1 - a - p )
r1
= e cos q 1 + cos(p + a - q 1 )
= e cos q 1 - cos(q 1 - a )
2l
\ Add . = 2e cos q 1
r1
l
\ = 2e cos q 1 ,
r1
l
\ The locus (r1 , q 1 ) is = e cos q which is the directrix
r
l
3. If the tangents at P and Q on the conic = 1 + e cos q meet at T, then prove the
r
following
^
a. ST bisects PSQ
b. If PQ intersects the directrix at K, than TSˆK = 90 0
c. ST2 = SP.SQ of the conic in a parabola
118
Solution :
Q T
P
S z
l
Let the equation of the conic be = 1 + e cos q ----(1)
r
Let P,Q be the points with vectorial angles α, β respectively. Then equations of the
tangent at P(α) and Q(β) are
l
= e cos q + cos(q - a ) ---(1)
r
l
= e cos q + cos(q - b ) ---(2)
r
Solve (1) and (2)
e cos q + cos(q - a ) = e cos q + cos(q - b )
\ cos(q - a ) = cos(q - b )
(q - a ) = ± (q - b )
Taking –ve, sign, q - a = -q + b
2q = a + b
119
a+b
q=
2
a+b
\ ZST =
2
a+b
Proof of (a): ZSP = a ; ZSQ = b , ZST =
2
a+b
\ PSˆT = ZSˆT - ZSˆP = -a
2
b -a
=
2
TSˆQ = ZSˆQ - ZSˆT
æa + b ö
=b -ç ÷
è 2 ø
2b - a - b b - a
= =
2 2
\ PSˆT = TSˆQ
Is ST bisects PSˆQ
b) The equation of PQ is
l æ a - b ö æç a+b ö
÷
= e cos q + secç ÷ cosçq - ÷ ---(1)
r è 2 ø è 2 ø
l
Equation of the directrix is = e cos q ---(2)
r
Let the chord PQ intersect the directrix at K.
\ solve (1) and (2)
æ a - b ö æç a+b ö
÷ = e cos q
e cos q + secç ÷ cosçq - ÷
è 2 ø è 2 ø
æa - b ö æç a+b ö
÷=0
secç ÷ cosçq - ÷
è 2 ø è 2 ø
120
æ a+b ö
\ cosççq - ÷=0
÷
è 2 ø
a+b p
\ q- =±
2 2
a+b p
\ q= ±
2 2
a+b p
LSK = ±
2 2
KSˆT = ZSˆT - ZSˆK
a + b æa + b p ö p
= -ç ± ÷=±
2 è 2 2ø 2
c) \ The conic is a parabola, e = 1
\ Equation of the tangent at ‘α’ is
l
= e cos q + cos(q - a ) ---(1)
r
æ a+bö
T is ç ST , ÷
è 2 ø
The point T lies on (1)
l æa + b ö æa + b ö
\ = cosç ÷ + cosç -a÷
ST è 2 ø è 2 ø
æa + b ö æa - b ö
= cosç ÷ + cosç ÷
è 2 ø è 2 ø
l æa ö æ b ö
= 2 cosç ÷ cosç ÷
ST è2ø è2ø
l
\ ST =
æa ö æ b ö
2 cosç ÷ cosç ÷
è2ø è2ø
l
But P, Q lis on = 1 + cos q
r
121
l 1
\ = 1 + cos a ; = 1 + cos b
SP SQ
l l
\ SP = ; SQ =
1 + cos a 1 + cos b
l2
SP.SQ =
(1 + cos a )(1 + cos b )
l2
=
a b
2 cos 2 2 cos 2
2 2
l2
=
a b
4 cos 2 cos 2
2 2
2
é ù
ê l2 ú
=ê ú
ê 2 cos a cos b ú
ëê 2 2 ûú
= ST 2
\ ST 2 = SP.SQ
10.2. Examples
10.3. Let us sum up

We have studied so far hints to find the equation straight lines, circle, conic in
polar coordinates.
(1) Find the equating the differ box to the conic l/r=1+e cosq
(2) Find the eccentricity of the cosq conic
l/r= 2+4 cosv
1) If PSQ is a focal chord of the conic l/r=1+e cosq, with the focus at
1 1 2
Prove that + =
s p s q l
122
2) If a chord of the conic l/r=1+ecos q subtends a right angle at the focus

2 2
æ 1 ö1 æ1 1ö e2
Prove that ç - ÷ + ç - ÷ = 2
è s pø e è s qø e e
(1) If the normal at L, one of the extremities of the latus sectum of the conic
l/r=1+e cosq mees the curve again in, Show that
l ( + 1e 2 +3e4 )
S Q=
1 + e 2 - e4
1. In any conic, prove that the sum of reciprocals of two perpendicular focal chord is
constant
Proof: Let the equation of the conic be
l
= 1 + e cos q ---(1)
r
Q(π/2+α)
P(α)
Q'
P'
Let PSP', QSQ' be the perpendicular focal chords of the conic (1)
Let the vectorial angle of P be α
p
Let the vectorial angle of Q be +a
2
Let the vectorial angle of P' be (p + a )
3p
Let the vectorial angle of Q' be +a
2
l
\ = 1 + e cos a
SP
123
l æp ö
= 1 + e cosç + a ÷ = 1 - e sin a
SP è2 ø
l
= 1 + e cos(p + a ) = 1 - e cos a
SP '
l æ 3p ö
= 1 + e cosç + a ÷ = 1 - e sin a
SQ' è 2 ø
l
\ SP =
1 + e cos a
l
SP ' =
1 - e cos a
l
SQ =
1 - e sin a
l
SQ ' =
1 - e sin a
l l
PP ' = SP + SP ' = +
1 + e cos a 1 - e cos a
l [1 - e cos a + 1 + e cos a ]
=
(1 + e cos a )(1 - e cos a )
2l
=
1 - e cos 2 a
2
l l
QQ ' = SQ + S ' Q = +
1 - e sin a 1 + e sin a
l [1 - e sin a + 1 + e sin a ]
=
(1 - e sin a )(1 + e sin a )
2l
=
1 - e sin 2 a
2
1 1 1 - e 2 cos 2 a 1 - e 2 cos 2 a
\ + = +
PP' QQ' 2l 2l
1
=
2l
[ (
2 - e 2 cos 2 a + sin 2 a )]
124
1
=
2l
( )
2 - e 2 = a constant
2. Prove that the perpendicular focal chords of a rectangular hyperbola are equal.
Proof:
Asian prob (1), pute = 2
1 1 1
+ = [2 - 2] = 0
PP ' QQ ' 2l
1 1
\ =-
PP ' QQ '
\ PP ' = -QQ '
\ PP ' = QQ ' ( in magnitude)
3. PSP', QSQ' are two focal chords of a conic cutting each other at right angles.
1 1
Prove that + = a constant
SP.SP ' SQ.SQ '
Solution : Let PSP' and QSQ' be two focal chords of the conic
l
= 1 + e cos q ---(1)
r
Let the vectorial angle of P be α
p æ 3p ö
\ Q is + a , P ' is p + a , Q ' is ç +a÷
2 è 2 ø
æ p ö
The points P ( SP, a ) Qç SQ, + a ÷,
è 2 ø
æ 3p ö
P ' ( SP ' , p + a ) Q ' ç SQ ' , +a ÷ ---(1) lie on
è 2 ø
l
\ = 1 + e cos a ;
SP
l æp ö
= 1 + e cosç + a ÷ = 1 - e sin a
SQ è2 ø
125
l
= 1 + e cos(p + a ) = 1 - e cos a
SP'
l æ 3p ö
= 1 + e cosç + a ÷ = 1 + e sin a
SQ' è 2 ø
l 1 + e cos a l 1 - e cos a
\ = ; =
SP l SP ' l
l l (1 + e cos a )(1 - e cos a )

\ . =
SP SP' l2
1
= 2
(1 - e 2 cos 2 a ) ---(2)
l
l l 1 - e sin a 1 + e sin a
. = ´
SQ SQ ' l l
1
= [(1 - e sin a )(1 + e sin a )]
l2
1
= 2
(1 - e 2 sin 2 a )
l
(2) + (3) gives
l l l l 1 - e 2 cos 2 a 1 - e 2 sin 2 a
. + . = +
SP SP' SQ SQ' l2 l2
1
= 2
[2 - e 2 (cos 2 a + sin 2 a ]
l
1
= 2
(2 - e 2 )
l
= a constant.
10.7. References
Analytical Geometry by T.K. Manickavasagam Pillai.
126
Unit IV
Lesson – 11
Analytical Geometry of Three Dimensions
Contents
11.1 Coplanar lines
11.2 Examples Model 1
11.3. Shortest distance between two lines.
11.4 Examples model I
11.5 Let us sum up
11.8 Points for Discussion
11.9 References
In this lesson we concentrate on the concept of the equations of a straight

live, sphere, including the fundamental of the planes and straight lives. Also we study the
shortest distance between two lines.
Results:
1) The distance between two points A(x1 ,y1 ,z1 ) and B(x2 ,y2 ,z2 ) is
AB = ( x 2 - x1 ) 2 + ( y 2 - y1 ) 2 + ( z 2 - z1 ) 2
2) If a straight line makes angles a , b , g with the coordinate axes, then the direction
cosines of the line are defined as cos a , cos b , cos g . These are denoted by l , m, n .
Note that l 2 + m 2 + n 2 = 1.
x y z
3) If OP = xi + y j + z k , then the direction cosines of OP are , ,
OP OP OP
4) A set of numbers a,b,c which are proportional to he direction cosines l,m,n of a line are
called the direction ratios of that line.
127
If a,b,c are direction ratios of a line then the direction cosines of that line are
a b c
l= ,m = ,n =
a2 + b2 + c2 a2 + b2 + c2 a2 + b2 + c2
5) The equation of a straight line passing through a point (x1 ,y1 ,z1 )having direction ratios
x - x1 y - y1 z - z1
l,m,n is = =
l m n
6) Equations of straight line passing through the points A(x1 ,y1 ,z1 ) and B(x2 ,y2 ,z2 ) is
x - x1 y - y1 z - z1
= =
x 2 - x1 y 2 - y1 z 2 - z 1
7) The direction ratios of the straight line joining the points A(x1 ,y1 ,z1 ) and B(x2 ,y2 ,z2 ) are
x 2 - x1 , y 2 - y1 , z 2 - z1
x - x1 y - y1 z - z1 x - x1 y - y1 z - z1
8) The condition for the lines = = and = = to be
l1 m1 n1 l2 m2 n2
l1 m1 n1
perpendicular is l1l 2 + m1 m 2 + n1 n 2 = 0 and to be parallel is = =
l 2 m2 n2
9) If a point P divides the line joining the points A(x1 ,y1 ,z1 ) and B(x2 ,y2 ,z2 ) internally in
æ mx + nx1 my 2 + yn1 mz 2 + nz1 ö
the ratio m:n, then P is given by Pç 2 , , ÷
è m+n m+n m+n ø
10) If the point P divides AB externally where A(x1 ,y1 ,z1 ) and B(x2 ,y2 ,z2 ), then
æ mx - nx1 my 2 - yn1 mz 2 - nz1 ö
P=ç 2 , , ÷
è m-n m-n m-n ø
11) Two planes intersect on a straight line. The general equation of a plane is of the form
ax+by+cz+d=0. a,b,c are called the direction rations of the normal to this plane.
12) Equation of any plane passing through the point (x1 ,y1 ,z1 ) is a(x- x1 ) + b(y-y1 ) + c(z-
z1 ) = 0.
13) Angle between two planes is defined as angle between their normals.
Let the equations of two planes be
a1 x+b1 y+c1 z+d1 =0 and a2 x+b2 y+c2 z+d2 =0
a). Then the angle between them is
128
a1 a 2 + b1b2 + c1 c 2
cos q =
2 2 2 2 2 2
a1 + b1 + c1 a 2 + b2 + c 2
a1 b1 c1
b) The condition for these planes to be parallel is = =
a 2 b2 c 2
c) The condition for these planes to be perpendicular is a1 a 2 + b1b2 + c1 c 2 = 0
x - x1 y - y1 z - z1
14) Equation of the plane containing the line = = = 0 is
l m n
a ( x - x1 ) + b( y - y1 ) + c( z - z1 ) = 0
Where al+bm+cn = 0.
15) Consider equations of the plane and the line as ax+by+cz+d=0 ---(1) and
x - x1 y - y1 z - z1
= = ---(2)
l m n
The condition for the plane and line to be parallel is al+bm+cn = 0 and to be
a b c
perpendicular is = =
l m n
16) The length of perpendicular from the point (x1 ,y1 ,z1 ) to the plane ax+by+cz+d=0 ---
(1) is
ax1 + by1 + cz1 + d

a2 + b2 + c2
d
a) The length of perpendicular from the origin to the plane (1) is
a2 + b2 + c2
b) The distance between the parallel planes ax+by+cz+d=0 and ax+by+cz+d1 =0
d - d1
is
a2 + b2 + c2
11.1 Coplanar lines
1. Symmetrical form of straight line
x - x1 y - y1 z - z1
The equation of the line = = is called they symmetrical
l m n
form of a straight line.
129
2. Since two planes intersect in a straight line, its equation can be put in the form
ax+by+cz+d=0 = a1 x+b1 y+c1 z+d1 =0
This is called non – symmetrical form of a straight line.
11.2 Examples Model 1:

1. Put in symmetrical form the lines
3x-2y+z-1 = 0 = 5x+4y-6z-2 ---(1)
Step 1: Let l,m,n be the direction rations of the given line
\ 3l - 2m + n = 0 ---(1)
5l + 4m - 6n = 0 ---(2)
By rule of cross multiplication
l m n
-2 1 3 -2
4 -6 5 4
l m n
= =
12 - 4 5 + 18 12 + 10
l m n
= =
8 23 22
Then the direction cosines of the line are proportional to 8,23,22 (which are nothing but
d.r’s)
Step 2: To find any point on the line (1).
Let the line (1) meet xy plane
\ put z = 0 in (1)
\ 3x-2y-1 = 0,
5x+4y-2 = 0
\ By rule of cross multiplication
x y 1
-2 -1 3 -2
4 -2 5 4
130
x y 1
= =
4 + 4 - 1 + 6 12 + 10
x y 1
= =
8 5 22
8 5
\ x= , y=
22 22
æ 8 5 ö
ç , , 0÷
ç 22 22 ÷
\ Any point on the lines ç ÷
çx, y1 , z1 ÷÷
ç 1
è ø
\ Equation of the line in symmetrical form is
8 5
x- y-
22 = 22 = z
8 23 - 22
2) Prove that the lines
3 x - 4 y + 2 z = 0 = -4 x + y + 3 z ----(1) and
x + 3 y - 5z + 9 = 0 = 7 x - 5 y - z + 7 ----(2) are parallel
Solution step 1:
Let l,m,n be the direction ratios of the line (1)
\ 3l - 4m + 2n = 0,
-4l + m + 3n = 0
\ By rule of cross multiplication
l m n
-4 2 3 -4
1 3 -4 1
l m n
= =
- 12 - 2 - 8 - 9 3 - 16
l m n l m n
= = or = =
- 14 - 17 - 13 14 17 13
\ The d.cs’ of line (1) are proportional to
131
+14 , +17, +13

a1 b1 c1
Step 2: Let l,m,n be the d.r’s of the line (2)
\ l - 3m - 5n = 0,
7l - 5m - n = 0
l m n
3 -5 1 3
-5 -1 7 -5
l m n
= =
- 3 - 25 - 35 + 1 - 5 - 21
l m n l m n
= = or = =
- 28 - 34 - 26 14 17 13
\ The d.cs’ of line (2) are proportional to
+14 , +17, +13
a2 b2 c2
a1 b1 c1
\ = =
a 2 b2 c 2
\ The lines (1) and (2) are parallel
3) Prove that the lines
2 x + y + 3z - 7 = 0 = x - 2 y + z - 5 and
4 x + 4 y - 8 z = 0 = 10 x - 8 y + 7 z are at right angles.
Solution: Proceed as before in example (2), check a1 a 2 + b1b2 + c1 c 2 = 0 for right angles
(ie. Perpendicular)
Condition for Two straight lines to be coplanar consider two straight lines
x - x1 y - y1 z - z1
= = ----(1) and
l1 m1 n1
x - x1 y - y1 z - z1
= = ----(2)
l2 m2 n2
Equation of the plane through the line (1) is
A( x - x1 ) + B( y - y1 ) + C ( z - z1 ) = 0 ----(3)
132
Where Al1 + Bm1 + Cn1 = 0 ----(2)

If the plane (3) contains the line (1), then (x2 ,y2 ,z2 ) lies on the plane (3)
\ Put x= x2 , y = y2 , z = z2 in (3)
A( x 2 - x1 ) + B( y 2 - y1 ) + C ( z 2 - z1 ) = 0 ----(4)
Also the line (2) is perpendicular to the normal to the plane (3)
Al 2 + Bm 2 + Cn 2 = 0
Where A( x 2 - x1 ) + B( y 2 - y1 ) + C ( z 2 - z1 ) = 0
Al + Bm + Cn = 0
Al1 + Bm1 + Cn1 = 0
Eliminate A,B,C between the above, we get
x 2 - x1 y 2 - y1 z 2 - z1
l1 m1 n1 = 0
l2 m2 n2
Equation of the plane containing the line (1) and (2) is

x - x1 y - y1 z - z1
l1 m1 n1 = 0
l2 m2 n2
Model 1: Two symmetrical form of straight lines are given.

Examples: 1. Show that the lines
x+3 y+5 z-7 x +1 y +1 z +1
= = ; = =
2 3 -3 4 5 -1
are coplanar and find the equation of the plane containing them.
x+3 y+5 z-7
Solution : Let L1 = = = = r1
2 3 -3
x+3 y+5 z-7
= r1 = r1 = r1
2 3 -3
x + 3 = 2r1 ; y + 5 = 3r1 ; z - 7 = -3r1 ;

x = -3 + 2r1 y = 3r1 - 5, z = -3r1 + 7
133
\ Any point on L1 is (2r1 - 3, 3r1 - 5, - 3r1 + 7) ----(A)
x +1 y +1 z +1
Let L2 : = = = r2
4 5 -1
x +1 y +1 z +1
= r2 = r2 = r2
4 5 -1
x + 1 = 4r2 ; y + 1 = 5r2 ; z + 1 = -r2
x = 4r2 - 1; y = 5r2 - 1; z = -r2 - 1
\ Any point on L2 is (4r2 - 1, 5r2 - 1, - r2 - 1) ----(B)
If the lines L1 and L2 are coplanar, then A and B represent the same point.
2r1 - 3 = 4r2 - 1; 3r1 - 5 = 5r2 - 1; -3r1 + 7 = -r2 - 1
4r2 - 2r1 = -2 - - -(1)

3r1 - 5r2 = 4 - - - ( 2)
- 3r1 + r2 = 8 - - -(3)
Solve (1) and (3)

r1 = 3; r2 = 1
Using in (3), the equation if statisfied.
\ The lines L1 and L2 are coplanar
Step 2: To find the point of intersection Put r2 =1 in (B), the point of intersection is (3,4,-
2)
Step 3: Equation of the plane containing them is
x+3 y+5 z-7
2 3 -3 =0
5 -1
( x + 3)(-3 + 15) - ( y + 5)(-2 + 12) + ( z - 7)(10 - 12) = 0

12( x + 3) - 10( y + 5) - 2( z - 7) = 0
12 x + 36 + 10 y - 50 - 2 z + 14 = 0
12 x - 10 y - 2 z = 0
134
¸ 2;
6x-5y- z=0
Model 2: Given a symmetrical form of a line and a non-symmetrical form of a line.
x +1 y +1 z +1
Prove that the lines = = and x + 2 y + 3 z - 8 = 0 = 2 x + 3 y + 4 z + 1
1 2 3
intersect and find their point of intersection. Find also the equation of the plane
containing them.
x +1 y +1 z +1
Solution: L1 : = = ---(1)
1 2 3
L2 : x + 2 y + 3z - 8 = 0 = 2 x + 3 y + 4 z - 11 ---(2)
Equation of any plane through the line (2) is
x + 2 y + 3 z - 8 + l ( 2 x + 3 y + 4 z - 11) = 0 ---(3)
L1 passes through (-1, -1, -1)
L1 lies on the plane (3) if (-1, -1, -1) lies on (3)
If ( -1-2-3-8) + l (-2-3-4-11) = 0
-7
Þl =
10
-7
Using l = in (3)
10
7
x + 2 y + 3z - 8 - (2 x + 3 y + 4 z - 11) = 0
10
Simplifying we get the equation of the plane (3) is
4x+y-2z-3=0 ---(4)
The d.r’s of the normal to this plane
4, 1, -2
a b c
The d.r’s of the line L1 : 1, 2, 3
l m n
\ al + bm + cn = 4 + 2 - 6 = 0
\ The line L1 lies in the plane (4)
135
\ L1 and L2 are coplanar
Step 2: To find the point of intersection.

Any point on L1 :
x +1 y +1 z +1
Let = = = r1 (say)
1 2 3
\ x = r1 - 1, y = 2r1 - 1; z = 3r1 - 1;
\ Any point on L1 is (r1 - 1, 2r1 - 1, 3r1 - 1) ----(A)
This point lies on the plane x+2y+3z-8=0
Then L1 is r1 - 1 + 2( 2r1 - 1) + 3( 3r1 - 1) - 8 = 0
r1 - 1 + 4r1 - 2 + 9r1 - 3 - 8 = 0
14r1 - 14 = 0
r1 = 1
Put r1 = 1 in (A), the point of intersection is (o,1,2).
Model 3: Two straight lines are given in non-symmetrical form prove that the straight
lines.
x + y + z - 3 = 0 = 2x + 3y + 4z - 5 = 0 ---(1) and
4x - y + 5z - 7 = 0 = 2x - 5 y - z - 3 ---(2) are coplanar. Find the point of
intersection and find the equation of the plane containing them.
Solution : Reduce the line (1) to symmetrical form and proved as in model 2.
Ans:
a) They are coplanar
b) Point of intersection is (4, -1, 0)
c) Equation of the plane containing them is x+2y+3z-2-0.
11.3. Shortest distance between two lines.

Definition 1: Two straight lines are called s kew if they are neither intersecting nor
parallel.
136
Definition 2: If two straight lines are skew, then there will be one and only are
perpendicular common to both the straight lines , then this common perpendicular is
known as the shortest distance between the two lines.
11.3.1. Find the short distance between the skew lines and find also find
its equation.
x - x1 y - y1 z - z1
L1 : = = and
l1 m1 n1
x - x1 y - y1 z - z1
L2 : = =
l2 m2 n2
Let AB and CD represent the two lines L1 and L2 respectively.
(x1 ,y1 ,z1 ) B

G
A
C (x2 ,y2 ,z2 )

H
D
Let GH be the shortest distance between the lines L1 and L2

L1 passes through A(x1 ,y1 , z1 ) and L2 passes through B (x2 ,y2 ,z2 )
Equation of any plane containing L1 is
a ( x - x1 ) + b( y - y1 ) + c( z - z1 ) = 0 ---(1)
Where al1 + bm1 + cn1 = 0 ---(2)
The plane (1) is parallel to L2
\ al 2 + bm 2 + cn 2 = 0 ---(3)
Solve (2) and (3)
al1 + bm1 + cn1 = 0
137
al 2 + bm 2 + cn 2 = 0
a b c
m1 n1 l1 m1
m2 n2 l2 m2
a b c
= =
m1 n 2 - m 2 n1 m1l 2 - l1 n 2 l1 m 2 - l 2 m1
\ Using in (1)
(m1 n 2 - m 2 n1 )( x - x1 ) + (m1l 2 - l1 n 2 )( y - y1 ) + (l1 m 2 - l 2 m1 )( z - z1 ) = 0
The SD = Perpendicular from the point B (x2 ,y2 ,z2 ) on the plane (4)
(m1 n 2 - m 2 n1 )( x 2 - x1 ) + (n1l 2 - n1l 2 )( y 2 - y1 ) + (l1 m 2 - l 2 m1 )( z 2 - z1 )

=
å (m n 1 2 - m 2 n1 ) 2
x 2 - x1 y 2 - y1 z 2 - z1
l1 m1 n1
l2 m2 n2
\ SD =
å (m n 1 2 - m 2 n1 ) 2
Step 2: Equation of the plane containing the line L1 and line GH is

x - x1 y - y1 z - z1
l1 m1 n1 = 0 ---(A)
l m n
Where l, m, n are the d.r’s of GH

Equation of the plane containing the line L2 and the line GH is
x - x2 y - y2 z - z2
l1 m1 n1 =0 ---(B)
l m n
Equation of A and B together representative equation of the line GH

11.4.Examples : Model 1:
Find the length of the SD between the lines
138
x -3 y -8 z -3 x+3 y+7 z-6

L1 : = = and = = and find its equation.
-3 1 -1 3 -2 -4
Solution: Step 1
x -3 y -8 z -3
L1 : = = ;
-3 1 -1
x+3 y+7 z-6
L2 : = =
3 -2 -4
Let GH be the SD between L1 and L2
Let l,m,n be the d.r’s of GH
GH perpendicular L1 \ -3l + m - n = 0
GH perpendicular L2 \ 3l + 2m - 4n = 0
By rules of cross multiplication

l m n
1 -1 -3 1
-2 -4 3 -2
l m n
= =
- 4 - 2 - 3 - 12 6 - 3
l m n
= =
- 6 - 15 - 3
l m n
or = =
-2 -5 1
\ The d.c’s of GH are proportional to -2, -5, 1

l , m, n
-2 -5 1
The d.c’s of GH are , , ,
30 30 30
L M N
139
x1 y1 z1
L1 passes through
(3, 8, 3)
(-3, - 7, 6)
L1 passes through
x2 y2 z2
\T h = e S(- 2D )l +1 x x(2 - m 1)y+ y2 ( n- 1z ) z(formula)
-2 5 1
= (-6) - (-7 - 8) + (6 - 3)
30 30 30
12 75 3
= + +
30 30 30
90
=
30
3 ´ 30
SD = = 3 30
30
Step 2: To find the equation of GH
The equation of SD is
x - x1 y - y1 z - z1 x - x2 y - y2 z - z2
l1 m1 n1 = 0 = l1 m1 n1 = 0
l m n l m n
x -3 y -8 z -3 x+3 y+7 z-6

-3 1 -1 = 0= 3 -2 4
-2 -5 1 -2 -5 1
4 x - 5 y + 17 z - 23 = 0 = 18 x - 11 y - 19 z + 91
Model 2: Find the SD between their lines
x-5 y-6 z -9
L1 : = =
3 -4 1
L2 : 2x - 2 y + z - 3 = 0 = 2x - y + 2z - 9
140
Solution: Any plane through L2 is

( 2 x - 2 y + z - 3) + l ( 2 x - y + 2 z - 9) = 0
2 x - 2 y + z - 3 + 2l x - l y + 2l z - 9l = 0
x( 2 + 2l ) - y ( 2 + l ) + z (1 + 2l ) - 3 - 9l = 0 ---(1)
The d.r’s of the normal to this plane are 2 + 2l ,-( 2 + l ),1 + 2l

Plane (1) is parallel to L1
\ 3( 2 + 2l ) + 4( 2 + l ) + 1 + 2l = 0
6 + 6l + 8 + 4l + 1 + 2l = 0
12l + 15 = 0
12l = -15
-5
l=
4
Using (1), the equation of the plane is 2x+3y+6z-33=0
L1 passes through (5, 6, 9)

\ SD = Perpendicular distance from (5, 6, 9) to the plane 2x+3y+6z-33=0
2(5) + 3(6) + 6(9) - 33

=
2 2 + 32 + 6 2
10 + 18 + 54 - 33
=
49
49
= = 7 unit.
7
Model 3:
Find the SD between the lines
L1 : 3x - 9 y + 5 z = 0 = x + y - z and
L2 : 6 x + 8 y + 3z - 10 = 0 = x + 2 y + z - 3
141
Solution: Reduce L1 to symmetric form and proceed as proceeded in model 2

13
Ans: SD =
342
11.5 Let us sum up
So far we have studied how to prove two lines are coplanar, how to find the
shortest distance between two skew lines.
(1) Find the condition of the lines
x - a x- b x- c -x a1 -y 1 b - z1 c
= 1 = a n =d = to be perpendicular
a1 b 1
c a b c
(2) Find the direction ratios of the normal to the plane x-y+2-1=0
1. Show that the lines
x -1 y - 2 z - 3 x-2 y-3 z -4
L1 : = = ; L2 : = = are coplanar. Find
2 3 4 3 4 5
their point of intersection and the equation of the plane containing them
Ans: x-2y+z = 0

x-a y-b z -c x - a ' y - b' z - c '
= = and = = are coplanar.
a' b' c' a b c

x -1 y - z z + b
= = and
2 -1 -3
x + 2 y + z + 2 = 0 = 4 x + 5 y + 3 z + 6 are coplanar. Prove also that their point
æ5 5 ö
of intersection is ç , ,-7 ÷ and the equation of the plane containing them is
è3 3 ø
2x+y+z+2=0/
142

1. Prove that the lines
x +1 y + z z - b
= = and
1 2 1
x - 2 y + 2 z = 3, x - 4 y + 5 z = 8 are coplanar.
2. Find the length and equation of the SD between the lines

x+3 y-6 z x+2 y z -7
= = ; = =
-4 6 2 -4 1 1
14
Ans: SD =
3
Equation : 16x+11y-z-18=0=2x+7y+z-3
11.9. References
Analytical Geometry of Three
Dimensions by N.P Bali
143
Lesson-12
Contents
12.1 Sphere
12.2 Examples
12.3 Tangent plane
12.4 Examples
12.5 Let us sum up
12.8 Points for Discussion
12.9 References

In this lesson we are going to learn in detail about sphere.
12.1 Sphere
The locus of a variable point in space whose distance from a fixed point is a
constant is called a sphere. The fixed point is called its centre and fixed distance is called
its radius.
12.2 Examples
1. Find the equation of a circle whose centre is at the point (a,b,c) and radius r units.
Solution :
P(x,y,z)
C (a,b,c)
Let C (a,b,c) be the centre. Let P(x,y,z) be any point on the sphere
Given CP = r
\ ( x - a ) 2 + ( y - b) 2 + ( z - c ) 2 = r
Squaring both sides
144
( x - a ) 2 + ( y - b) 2 + ( z - c ) 2 = r 2
Cor: If the centre is at the origin (0,0,0) then equation of the sphere is x2 +y2 +z2 =r2
2. Prove that the equation x2 +y2 +z2 +2ux+2vy+2wz+d=0 represents a sphere. Hence find
its centre and radius.
Solution: The given equation is
x2 +y2 +z2 +2ux+2vy+2wz+d=0
adding u2 +v2 +w2 to both sides.
(x2 +2ux+u2 )+(y2 +2vy+v2 )+(z2 +2wz+w2 )+d = u2 +v2 +w2
( x + u ) 2 + ( y + v) 2 + ( z + w) 2 = u 2 + v 2 + w 2 - d
[x - (-u )]2 + [ y - (-v)]2 + [z - (- w)]2 = ( u 2 + v 2 + w2 - d )

2
This is of the form ( x - a ) 2 + ( y - b) 2 + ( z - c) 2 = r 2

Which is a sphere.
\ The given equation represents sphere whose centre is at (-u, -v, -w) and
r = u 2 + v 2 + w2 - d .
Problems 1: Find the radius and the centre of the sphere
x 2 + y 2 + z 2 - 6 x - 2 y - 4 z - 11 = 0
Solution: The given equation of the form x 2 + y 2 + z 2 + 2ux + 2vy + 2 wz + d = 0
2u = -6 2v = -2 2 w = -4
\ d = -11
u = -3 v = -1 w = -2
\ Centre = (-u, - v, -w) = (3,1,2)
r = u 2 + v 2 + w2 - d
= 9 + 1 + 4 + 11
= 25
r = 5 unit.
145
2. Find the centre of and radius of the sphere 2 x 2 + 2 y 2 + 2 z 2 - 2 x + 4 y - 6 z + 5 = 0

Solution: The given equation of the sphere is
2x 2 + 2 y 2 + 2z 2 - 2x + 4 y - 6z + 5 = 0
¸ 2;
5
x 2 + y 2 + z 2 - x + 2 y - 3z + =0
2
5
2u = -1; 2v = 2; 2 w = -3; d=
2
-1 -3
u= ; v = 1; w=
2 2
\ Centre = (-u, - v, -w)
æ1 3ö
= ç ,-1, ÷
è2 2ø
r = u 2 + v 2 + w2 - d
1 9 5
= +1+ -
4 4 2
5 5
= +1-
2 2
r = 1 unit.
Model 2: Equation of the sphere passing through four given points.
1. Find the equation of the sphere passing through the points (2,3,1), (5,-1,2) (4,3,-1) and
(2,5,3)
Solution: Let the equation of the sphere be
x 2 + y 2 + z 2 + 2ux + 2vy + 2 wz + d = 0 ---(1)
(1) Passes through the point (2,3,1)
4+9+1+4u+6v+2w+d=0
4u+6v+2w+d=-14 ---(2)
(1) Passes through the point (5,-1,2)
146
25+1+4+10u-2v+4w+d=0
10u-2v+4w+d=-30 ---(3)
(1) Passes through the point (4,3,-1)
16+9+1+8u+6v-2w+d=0
8u+6v-2w+d=-26 ---(4)

4+25+9+4u+10v+6w+d=0
4u+10v+6w+d=-38 ---(5)
Step (2) – (3) gives
4u + 6v + 2w + d = -14
10u – 2v + 4w + d = -30
Sub -6u + 8v – 2w = 16
¸ 2;
-3u + 4v – w = 8 ---(6)
(3) – (4) gives
10u – 2v + 4w + d = -30
8u + 6v – 2w + d = -26
Sub 2u – 8v + 6w = -4
¸ 2;
u – 4v + 3w = -2 ---(7)
(4) – (5) gives
8u + 6v – 2w + d = -26
4u + 10v + 6w + d = -38
Sub 4u – 4v – 8w = 8
¸ 4;
or u – v – 2w = 2 ---(8)
147
Step 3: Solve (6). (7), (8) by cramers’ solution (or) by solving
-3u + 4v – w = 8 ---(6)
u – 4v + 3w =-2 ---(7)
u – v – 2w = 2 ---(8)
-3u + 4v – w = 8
u – 4v + 3w = -2
Add - 2u + 2w = 6
-u + w = 3 ---(9)
(7) is u – 4v + 3w = -2
(8) x 4 ; 4u - 4v - 8w = 8
Sub -3u + 11w = -10 ---(10)

Solve (9) and (10)
3 x (9) -3u + 3w = -9
-3u + 11w = -10
Sub -8w = 19
- 19
w=
8
Using in (9)
19
-u - =3
8
19 5
-u = -3=-
8 8
5
u=
8
148
5 - 19 27
Put u = , w= in (8) , v =
8 8 8
Substitute the values of u, v, w in (4)
d = - 56
using in (1)
10 54 38
x2 + y2 + z2 + x+ y+ z - 56 = 0
8 8 8
8( x 2 + y 2 + z 2 ) + 10 x + 54 y + 38 z - 448 = 0
Which is the required sphere.
2. Find the equation of the sphere passing through the points (1,3,4), (1,-5,2), (1, -3, 0)
and having its centre on the plane x + y + z = 0.
Solution: Let the equation of the sphere be

x 2 + y 2 + z 2 + 2ux + 2vy + 2 wz + d = 0 ---(1)
2u+6v+8w+d=-20 ---(2)
2u-10v+4w+d=-30 ---(3)
2u+6v+d=-10 ---(4)
The centre (- u, -v, -w ) lies on the plane
x + y + z = 0.
\ -u - v - w = 0 \ u + v + w = 0 ---(5)
Solve (2), (3), (4), (5)
u = -1, v = 3, w = -2, d = 10
we get the equation of the sphere is
x 2 + y 2 + z 2 + 2 x + 6 y + 4 z + 10 = 0 .
3. Find the equation of the sphere passing through the points (1,1,-2), (-1,1,2) and having
its centre lies on the line
149
x + y – z – 1 = 0 = 2x – y + z – 2
Proof: Let the equation of the sphere be
x 2 + y 2 + z 2 + 2ux + 2vy + 2 wz + d = 0 ---(1)

(1) Passes through the point (1,1,-2)
2u+2v-4w+d=-6 ---(2)
(1) Passes through the point (-1,1,2)
-2u+2v+4w+d=-6 ---(3)
The centre (- u, -v, -w ) lies on the plane
x + y - z -1= 0 = 2x – y + z - 2
\ -u - v + w - 1 = 0 -2u + v - w - 2 = 0
\ u + v - w =1 or 2u - v + w + 2 = 0 ---(5)
Solve (2), (3), (4), (5)

1 1
u = -1; v = - ; w = - ; d = -5
2 2
Using in (1) Equation of the sphere is
x 2 + y 2 + z 2 - 2x - y - z - 5 = 0 .
Equation of a sphere on the line joining the points (x1 ,y1 ,z1 ) as diameter.
Let the points be A (x1 ,y1 ,z1 ) and B(x2 ,y2 ,z2 )
P(x,y,z)
A B
Let P(x,y,z) be any point on the sphere join AP,PB, APB = 900
The d.r’s of AP are x - x1 , y - y1 , z - z1
The d.r’s of BP x - x 2 , y - y 2 , z - z 2
150
\ AP perpendicular PB
( x - x1 ), ( x - x 2 ) + ( y - y 2 )( y - y 2 ) + ( z - z1 )( z - z 2 ) = 0
Which is the required equation of the sphere.
Problems (1) Find the equation of the sphere described on the line joining the points
A(4,6,8) and B(-1,3,7) as diameter
Proof: Equation of the sphere is

( x - x1 )( x - x 2 ) + ( y - y1 )( y - y 2 ) + ( z - z1 )( z - z 2 ) = 0
( x1 , y1 , z1 ) = (4,6,8)
( x 2 , y 2 , z 2 ) = (-1,3,7)
\ ( x - 4)( x + 1) + ( y - 6)( y - 3) + ( z - 8)( z - 7) = 0
x 2 + y 2 + z 2 - 3 x - 9 y - 15 z + 70 = 0
12.3 Tangent plane
The locus of all tangent lines drawn to a sphere at a point is called the tangent plane.
12.4.Examples: Find the condition for the plane lx+my+nz = p to be tangent plane to the
sphere x 2 + y 2 + z 2 + 2ux + 2vy + 2 wz + d = 0
Proof:
lx+my+nz=p
c (-u,-v,-w)
The radius of the sphere u 2 + v 2 + w2 - d = r

P = The perpendicular from the centre of the sphere to the plane lx+my+nz = p
(-lu - mv - nw - p )
=±
l 2 + m2 + n2
151
(lu + mv + nw + p )
=±
l 2 + m2 + n2
If the plane lx+my+nz = p is a tangent plane to the sphere,
Then r = p
(lu + mv + nw + p )
u 2 + v 2 + w2 - d = ±
l 2 + m2 + n2
u 2 + v 2 + w 2 - d l 2 + m 2 + n 2 = ±(lu + mv + nw + p )
Squaring both sides

(u2 +v2 +w2 -d) (l2 +m2 +n2 ) = (lu+mv+nw+p)2
Which is required condition.
Find the equation of the tangent plane to the sphere

x 2 + y 2 + z 2 + 2ux + 2vy + 2 wz + d = 0 at the point (x1 ,y1 ,z1 )
Solution: Equation of the sphere is
x 2 + y 2 + z 2 + 2ux + 2vy + 2 wz + d = 0 ---(1)
Equation of any line through (x1 ,y1 ,z1 ) is
x - x1 y - y1 z - z1
= = = r (say) ---(2)
l m n
Any point on the line (2) is ( x1 + lr , y1 + mr , z1 + nr )
ì x - x1 = lr ; y - y1 = mr; z - z1 = nr - 1;ü
í ý
î x = x1 + lr y = y1 + mr; z = z1 + nr þ
\ If this point lies on the sphere (1) then

( x1 + lr ) 2 + ( y1 + mr ) 2 + ( z1 + nr ) 2 2 + 2u ( x1 + lr ) + 2v( y1 + mr ) + 2 w( z1 + nr ) + d = 0
r 2 (l 2 + m 2 + n 2 ) + 2r [lx1 + my1 + lu + mv + nw] + x1 + y1 + z1 + 2ux1 + 2vy1 + 2 wz 1 + d = 0

2 2 2
\ ( x1 , y1 , z 1 ) lies on the sphere

2 2 2
x1 + y1 + z1 + 2ux1 + 2vy1 + 2 wz1 + d = 0
\ r 2 (l 2 + m 2 + n 2 ) + 2r [l ( x1 + n) + m( y1 + r ) + n( z1 l + w)] = 0
152
This is a quadratic in ‘r’, since the line is a tangent line to the sphere then
( x - x1 )( x1 + u ) + ( y - y1 )( y1 + v) + ( z - z1 )( z1 + w) = 0
x( x1 + n) - x1 ( x1 + n) + y ( y1 + v) - y1 ( y1 + v) + z ( z1 + w) - z1 ( z1 + w) = 0
2 2 2
xx1 + ux - x1 - ux1 + y1 y + vy - y1 - vy1 + zz1 + wz - z1 - z1 w = 0
2 2 2
xx1 + yy1 + zz1 + ux + vy + wz = x1 + y1 + z1 + ux1 + vy1 + wz 1 = 0
Add ux1 + vy1 + wz1 + d to both sides
xx1 + yy1 + zz1 + u ( x + x1 ) + v( y + y1 ) + w( z + z1 ) + d = 0

Cor: Equation of the tangent plane at (x1 , y1 , z1 ) to the sphere x2 +y2 +z2 = a2 is
xx1 + yy1 + zz1 = a 2
Plane section of a sphere
Prove that the plane section of a sphere is a circle.
π
H
P
C
Let the centre of the sphere be C. Whose radius is r units

Let π be the plane.
Draw CN perpendicular to the plane π
Let P be any point on the section of the sphere by a plane
Then OP = radius of the sphere
In D CNP CNP = 90 0
\ CP 2 = CN 2 - NP 2
NP 2 = CP 2 - CN 2
= r2 - p2 where CN = P
153
\ NP = r 2 - p 2 = a constant
\ The focus of P is a circle and N is the centre of the circle.

\ The plane section of a sphere is a circle. Such a circle is called a small circle.
Note:
1) If S=0, P=0 represent the equation of a sphere and the plane respectively, then the
equation of the circle is S=0, P=0
2) If the plane π passes through the centre of the sphere, then the circle is called a
great circle.
3) Equation of any sphere passing through the circle S=0, P=0 is
S + lP = 0 where l is a constant to be determined.
1) Find the centre and radius of the circle

x 2 + y 2 + z 2 - 2 y - 4 z - 11 = 0, x + 2 y + 2 z = 15
Proof: The equation of the sphere is x 2 + y 2 + z 2 - 2 y - 4 z - 11 = 0

This is of the form
x 2 + y 2 + z 2 + 2ux + 2vy + 2 wz + d = 0
2u = 0, 2v = -2; 2 w = -4; d = -11
u = 0, v = -1, w = -2, d = -11
\ Centre = C(0,1,2)
r = u 2 + v 2 + w2 - d
= 0 + 1 + 4 + 11 = 16 = 4
r=4
p = The length of perpendicular distance from the centre (0,1,2) of the sphere to the plane
x+2y+2z-15=0,
(0 + 2 + 4 - 15) (-9)
=± =±
1+ 4 + 4 9
(-9)
=±
3
p=3
154
\ Radius of the circle
= r2 - p2
= 4 2 - 32
= 16 - 9
= 7
Step 2: To find the centre of the circle.
The equation of the plane is
x+2y+2z=15
The d.r’s of the normal are 1,2,2
l , m, n
\ The d.r’s of CN are 1, 2, 2
The line CN passes through ( 0, 1, 2 )
x1, y1, z1
x - x1 y - y1 z - z1
\ Equation of CN is = =
l m n
x - 0 y -1 z - 2
\ = = = R (say)
1 2 2
Any point on this line is (R, 2R+1, 2R+2) ---(A)
Let this be the point N.
If this point lies on the plane
x+2y+2z=15, then
R+2(2R+1)+2(2R+2)-15=0
R+4R+2+4R+4-15=0
9R-9=0
9R=9
R=1
Put R = 1 in (A)
The centre if the circle is (1, 3, 4)
2. Find the equation of the sphere for which the circle
x 2 + y 2 + z 2 + 7 y - 2 z + 2 = 0, 2 x + 3 y + 4 z = 8
155
Solution : S = x 2 + y 2 + z 2 + 7 y - 2 z + 2 = 0
P = 2x + 3y + 4z - 8 = 0
Equation of any sphere through the circle
S=0, P = 0 is S + lP = 0
x 2 + y 2 + z 2 + 7 y - 2 z + 2 + l ( 2 x + 3 y + 4 z - 8) = 0 ---(1)
x 2 + y 2 + z 2 + 2lx + y (7 + 3l ) + z (4l - 2) + 2 - 8l = 0
æ - (7 + 3l ) - (4l - 2) ö
Centre = ç - l , , ÷
è 2 2 ø
This lies on the plane 2x+3y+4z-8=0
1 1
\ Put n = -l ; y = - (7 + 3l ) : z = - (4l - 2)
2 2
3 1
\ -2l - (7 + 3l ) - 4. ( 4l - 2) - 8 = 0
2 2
- 4l - 3(7 + 3l ) - 4.( 4l - 2) - 16 = 0
- 4l - 21 - 9l - 16l + 8 - 16 = 0
- 29l - 29 = 0
\ l = -1
Put l = -1 in (1)
x 2 + y 2 + z 2 + 7 y - 2 z + 2 - 1( 2 x + 3 y + 4 z - 8) = 0
x 2 + y 2 + z 2 + 7 y - 2z + 2 - 2x - 3y - 4z + 8 = 0
x 2 + y 2 + z 2 - 2 x + 4 y - 6 z + 10 = 0
3. Prove that the circles
x 2 + y 2 + z 2 - 2 x + 3 y + 4 z - 5 = 0, 5 y + 6 z + 1 = 0
x 2 + y 2 + z 2 - 3 x - 4 y + 5 z - 6 = 0, x + 2 y - 7 z = 0
lie on the same sphere. Find its equation?
Step 1: Equation of the sphere through the first circle
x 2 + y 2 + z 2 - 2 x + 3 y + 4 z - 5 + l ( 5 y + 6 z + 1) = 0
x 2 + y 2 + z 2 - 2 x + 3 y + 4 z - 5 + 5ly + 6lz + l = 0
156
x 2 + y 2 + z 2 - 2 x + y (3 + 5l ) + z (4 + 6l ) - 5 + l = 0 ---(1)
Equation of the sphere through the second circle is
x 2 + y 2 + z 2 - 3x - 4 y + 5 z - 6 + m ( x + 2 y - 7 z ) = 0
x 2 + y 2 + z 2 - 3 x - 4 y + 5 z - 6 + mx + 2 my - 7 mz = 0
x 2 + y 2 + z 2 + x( m - 3) + y (2 m - 4) + z (5 - 7 m ) - 6 = 0 ---(2)
The two circles lie on the same sphere if (1) and (2) represent the same sphere
-2 3 + 5l 4 + 6l -5+l
\ = = =
m -3 2m - 4 5 - 7m -6
(i ) (ii ) (iii ) (iv)
From (i) and (ii),
- 4 m + 8 = ( m - 3)(3 + 5l )
- 4 m + 8 = 3m + 5lm - 9 - 15l
\ - 15l + 7 m + 5lm - 17 = 0 ---(3)
From (i) and (ii)
-2 -5+l
=
m -3 -6
12 = ( -5 + l )( m - 3)
12 = -5m + 15 + lm - 3l
\ 3l + 5m - lm - 3 = 0 ---(4)
- 15l + 7 m + 5lm - 17 = 0
(4) x 5: 15l + 25 m - 5lm - 15 = 0
Add 32 m - 32 = 0
32 m = 32
m =1
Put m = 1 in (i) and (ii)
- 2 3 + 5l
=
-2 2-4
- 2 = 3 + 5l
157
5l = -5
l = -1
Using (iii) and (iv)
4 + 6l 4-6 -2
= = =1
5 - 7m 5-7 -2
\ The two circles lie on the same sphere
To find its equation
Step 2: Put m = 1 in (2)
x 2 + y 2 + z 2 - 2x - 2 y - 2z - 6 = 0
4. Find the equation of the tangent plane to the sphere
x 2 + y 2 + z 2 + 6 x - 2 y - 4 z = 35 at the point (3,4,4)
Solution:
2x = 6 2v = -2 2 w = -4
u =3 v = -1 w = -2
Equation of the tangent at (x1 ,y1 ,z1 ) is

xx1 + yy1 + zz1 + 3 x + 3 x1 - y - y1 - 2 z - 2 z1 - 35 = 0
\ Equation of the tangent plane at (3,4,4) is
3 x + 4 y + 4 z + 3 x + 9 - y - 4 - 2 z - 8 - 35 = 0
6 x + 3 y + 2 z - 28 = 0
5. Find the tangent planes to the sphere x 2 + y 2 + z 2 - 4 x - 2 y - 6 z + 5 = 0 which are

parallel to the plane x+4y+8z=0 and find the point of contact.
Solution: Equation of any plane parallel to the plane x+4y+8z=0 is x+4y+8z+R=0 ---(1)
The centre of the sphere is (2,1,3)
r = 4 +1+ 9 - 5 = 9 = 3
\ P = the length perpendicular from (2,1,3) on the plane x+4y+8z+R=0
2 + 4 + 24 + R
=
1 + 16 + 64
158
(30 + R )
=± = 3 (radius)
9
= ±(30 + R) = 27 -(30 + R) = 27
30 + R = 27 30 + R = -27
R = -3 R = -57
Using in (1) the equations of tangent planes are x + 4 y + 8 z - 3 = 0 and
x + 4 y + 8 z - 57 = 0
Step 2: To find the point of contact
Let P be the point of contact of the plane and the sphere
The d.r’s of CP are 1,4,8

The line CP passes through C(2,1,3)
x - 2 y -1 z - 3
Equation of CP is = = = R (say)
1 4 8
Any point on CP is ( R+2, 4R+1, 8R+3) ---(A)
If this lies on the plane x+4y+8z=57
\ R+2+4(4R+1)+8(8R+3)-57=0
R+2+16R+4+64R+24-57=0
8R-27=0
8R = 27
27 1
R= =
81 3
1 æ1 4 8 ö
Put R = in (A), the point of conduct is ç + 2, + 1, + 3 ÷
3 è3 3 3 ø
159
æ 7 7 17 ö
=ç , , ÷
è3 3 3 ø
Step 3: The other tangent plane is x+4y+8z-3=0
To find the point of contact.
If the point (A) lies on this plane

x+4y+8z-3=0, then
R+2+4(4R+1)+8(8R+3)-3=0
R+2+16R+4+64R+24-3=0
81R+27=0
81R = -27
+1
R ==
3
1 æ 5 -1 1ö
Put R = - in (A), the point of conduct is ç , , ÷
3 è3 3 3ø
6. Find the equation of the spheres which pass through the circle x2 +y2 +z2 =5, x+2y+3z=3
and touch the plane 4x+3y=15.
Solution: Equation of any sphere through the given circle is
x2 +y2 +z2 -5+ l (x+2y+3z-3)=0 ---(1)

x2 +y2 +z2 + l x+2 l y+3 l z-3 l -5=0
æ-l - 3l ö
Centre = ç , - l, ÷
è 2 2 ø
l2 9l 2
r= + l2 + + 3l + 5
4 4
14l 2
= + 3l + 5
4
14l 2 + 12l + 20
=
4
160
æ-l - 3l ö
P = Perpendicular from ç , - l, ÷ to the plane 4x+3y-15=0
è 2 2 ø
æ l ö
ç - 4 ´ - 3l - 15 ÷
è 2 ø
=±
16 + 9
=±
(- 2l - 3l - 15)
5
=±
(- 5l - 15)
5
r=p
14l 2 + 12l + 20
= ±(l + 3)
4
14l 2 + 12l + 20 = 4(l + 3) 2
2 4
On simplification l = -
1 5
Using in (1) the equations of the spheres are x 2 + y 2 + z 2 + 2 x + 4 y + 6 z - 11 = 0 and
5 x 2 + 5 y 2 + 5 z 2 - 4 x - 8 y - 12 z - 13 = 0
12.5 Let us sum up

So far we have studied and learnt about equations of a sphere in different forms,
the equation of the tangent plane to a sphere and in finding the radium centre of the circle
which is the intersection of the plane and the sphere.
(1) Find the centre of the sphere
x2 +y2 +i2 -2x+4y-6z-11=0
(2) Find the equation of the sphere having (1,2,3) and (3,2,1) as diameter.
161

1. Find the center and radius of the sphere 3x 2 + 3 y 2 + 3 z 2 + 12 x - 8 y - 10 z + 10 = 0
47 æ 4 5ö
r= ; C = ç - 2, , ÷
3 è 3 3ø
2. Find the equation of the sphere through the points (1,1,1) (1,2,1), (1,1,2) and (2,1,1)
Ans: x 2 + y 2 + z 2 - 3x - 3 y - 3z + 6 = 0
3. Find the equation of the sphere which passes through (0,7,7),(1,8,11), (-3,10,7) and
centre lies on the plane 2x+2y-z-7=0
Ans: x 2 + y 2 + z 2 + 2 x - 18 y - 18 z + 154 = 0
4. Find the equation of the sphere passing through the points (2,1,1) and (0,3,2) and
centre lies on the line 2 x + y + 3 z = 0 = x + 2 y + 2 z
Ans: 9( x 2 + y 2 + z 2 ) + 28 x + 7 y - 21z - 96 = 0
5. Find the equation of the sphere through the points (0,0,0) (a,0,0) (0,b,0) and (0,0,c)
Ans: x 2 + y 2 + z 2 - ax - by - cz = 0
6. Find the centre and radius of the circle

x 2 + y 2 + z 2 - 2 x - 4 y - 11 = 0; x + 2 y + 2 z - 14 = 0
Ans: r = 7 ; C = (2,4,2)
7. Find the equation of the sphere in which the circle
x 2 + y 2 + z 2 - 6 x + 3 y - z - 8 = 0, 2 x + 3 y - z + 6 = 0 is a great circle
Ans: x 2 + y 2 + z 2 - 4 x + 6 y + 2 z - 2 = 0
12.8. Points for discussion.
1. Show that the two circles x 2 + y 2 + z 2 - y + 2 z = 0, x - y + z - 2 = 0 and
x 2 + y 2 + z 2 + x - 3 y + z - 5 = 0, 2 x - y + 4 z - 1 = 0 lie on the same plane.

2. Find the equations of the spheres which pass through the circle
x 2 + y 2 + z 2 - 4 x - y + 3z + 12 = 0, 2 x + 3 y - 7 z = 0 and find the plane x-2y+2z=1
Ans: x 2 + y 2 + z 2 - 2 x + 2 y - 4 z = 2
x 2 + y 2 + z 2 - 6 x + 4 y - 10 z = 22
162
3. Find the condition for the plane lx+my+nz=P to be a tangent plane to

x2 + y2 + z2 = a2
4. Find the equation of the tangent plane at(x1 ,y1 ,z1 ) on the sphere x 2 + y 2 + z 2 = a 2
12.9. References
1. Analytical Geometry of Three Dimensions by N.P. Bali.
163
Unit V
Lesson - 13
Contents
13.1 Cone
13.2 Examples
13.3 Enveloping cone of a sphere.
13.4 Examples
13.5 Right circular cone
13.6 Examples
13.7 Cylinder
13.8 Examples
13.9 Right circular cylinder
13.10 Examples
13.11 Let us sum up.
13.15 References

Our Aim is to learn lesson, the concept of a cone, right, circular cone, cylinder and
conicoids.
13.1. Cone
Definition:
A cone is a surface generated by a line through a fixed point, (the fixed point is
called the vertex of the cone) which satisfies one more condition is intersecting a given
curve or founding a given surface.
The given curve is called the base curve or the guiding curve and the variable line
is called a generator of the cone.
13.2. Examples
164
Book work 1: Find the equation of a cone with a given vertex and given base
Solution : Let the vertex of the cone be (a , b , g ) .
Let the equation of the base be

ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0, z = 0 ---(1)
Any generator through (a , b , g ) is
x -a y - b z - 0
= = ---(2)
l m n
This line meets the plane z = 0
\ Put z = 0 in (2)
x -a y - b -g
= =
l m n
x -a -g y - b -g
= ; =
l n m n
-l -m
x -a = g; y-b = g
n n
-l -m
x =a g; y=b g
n n
Using in (1)
2 2
æ l ö æ l öæ m ö æ m ö æ l ö æ m ö
ç a - r ÷ + 2hç a - r ÷ç b - r ÷ + bç b - r ÷ + 2 g ç a - r ÷ + 2 f ç b - r ÷ + c = 0
è n ø è n øè n ø è n ø è n ø è n ø
Eliminate l,m,n between (2) and (3)
2 2
æ x -a ö æ x -a öæ y-b ö æ y-b ö
açç a - r ÷÷ + 2hçç a - r ÷÷çç b - r ÷÷ + bçç b - r ÷÷
è z -g ø è z -g øè z -g ø è z -g ø
æ x -a ö æ y-b ö
+ 2 g çç a - r ÷÷ + 2 f çç b - r ÷÷ + c = 0
è z -g ø è z -g ø
(az - xg ) 2 (az - xg )( b z - yg ) ( b z - yg ) 2 (az - xg ) ( b z - yg )

2
+ 2h 2
+b 2
+ 2g +2f +c = 0
(z - r) (z - r) (z - r) (z - r) (z - r)
165
Multiply both sides by ( z - r ) 2
(az - xg ) 2 + 2h(az - xg )( b z - yg ) + b( bz - yg ) 2 + 2 g (az - xg )( z - r ) +

2 f ( bz - yg )( z - r ) + c( z - r ) 2 = 0
Which is the required equation of the cone.
2) Prove that the equation of a cone with its vertex at the origin is a homogeneous
equation
Proof: Consider the equation
ax 2 + 2hxy + by 2 + 2 fyz + 2 gzx + 2hxy + 2ux + 2vy + 2 wz + d = 0 ---(1)
Let P (x1 ,y1 ,z1 ) be any point, on the cone
\ Equation of OP is
x y z
= = where (1) = (0,0,0)
x1 y1 z1
x y z
\ = = = r (say)
x1 y1 z1
x = x1 r ; y = y1 r ; z = z1 r
\ Any point on OP is (rx1 , ry1 , rz1 )
If this point lies on the cone (1)
Then
2 2
ar 2 x1 + 2hrx1 ry1 + 2br 2 y1 + 2 fr 2 y1 z1 + 2 gr 2 z1 x1 + 2hr 2 x1 y1 + 2urx1 + 2vry1 + 2 wrz 1 + d = 0
2 2
r 2 (ax1 + 2hx1 y1 + 2by1 + 2 fy1 z1 + 2 gz1 x1 + 2hx1 y1 ) + 2r (ux1 + vy1 + wz1 ) + d = 0
Treating the above as an identify.

2 2
ax1 + 2hx1 y1 + by1 + 2 fy1 z1 + 2 gz1 x1 + 2hx1 y1 = 0 ---(2)
ux1 + vy1 + wz1 = 0 ---(3)
d=0 ---(4)
If u,v,w be not all zero, then (3) represents a plane.

\ u = v = w = 0.
166
\ (2) becomes
2 2
ax1 + 2hx1 y1 + by1 + 2 fy1 z1 + 2 gz1 x1 + 2hx1 y1 = 0
\ The locus of (x1 , y1 , z1 ) is

ax 2 + 2hxy + by 2 + 2 fyz + 2 gzx + 2 fxy = 0
Which is a homogeneous equation second degree 2 in x, y, and z.

Bookwork 3: A homogenous equation of second degree in x, y and z always represents a
cone with the vertex at the origin.
Solution: Let the homogenous equation of second degree in x, y and z be
ax 2 + 2hxy + by 2 + 2 fyz + 2 gzx + 2hxy = 0 ---(1)
Let P (x1 , y1 , z1 ) be any point on the surface represented by (1)
2 2
\ ax1 + 2hx1 y1 + by1 + 2 fy1 z1 + 2 gz1 x1 + 2hx1 y1 = 0 ---(2)
Let O be the point (0,0,0)
x y z
\ The equation of OP is = = = r (say)
x1 y1 z1
\ Any point on OP =(x1 r, y1 r, z1 r)
This point lies on (1) , using (2)
\ (1) represents a cone whose vertex is at the origin.
13.3. Enveloping cone of a sphere.
Definition:
The locus of the tangent lines to a sphere drawn from a given point is a cone
called the enveloping cone of the sphere having the given point as its vertex.
13.4. Examples
Bookwork 4: Find the equation of the enveloping cone of the sphere x 2 + y 2 + z 2 = a 2

with its vertex at (x1 , y1 , z1 )
167
Solution: Let the equation of the sphere be x 2 + y 2 + z 2 = a 2 ---(1)

Let the given point be A (x1 , y1 , z1 )
Let P (x,y,z) be any point on a tangent drawn from A to the sphere
The coordinates of a point dividing AP in the ratio m=1 is
æ mx + x1 my + y1 mz + z1 ö
ç , , ÷
è m +1 m +1 m +1 ø
If this lies on (1)
(mx + x1 ) 2 (my + y1 ) 2 (mz + z1 ) 2
Then 2
+ 2
+ 2
= a2
(m + 1) (m + 1) (m + 1)
(mx + x1 ) 2 + (my + y1 ) 2 + (mz + z1 ) 2 = a 2 (m + 1) 2

2 2 2
m 2 ( x 2 + y 2 + z 2 - a 2 ) + 2m( xx1 + yy1 + zz1 - a 2 ) + x1 + y1 + z1 - a 2 = 0 ---(2)
The line AP is a tangent line to (1)
\ The roots are equal.
2 2 2
\ 4( xx1 + yy1 + zz1 - a 2 ) = 4( x 2 + y 2 + z 2 - a 2 ) ´ ( x1 + y1 + z1 - a 2 ) = 0
2 2 2
\ ( xx1 + yy1 + zz1 - a 2 ) = ( x 2 + y 2 + z 2 - a 2 ) ´ ( x1 + y1 + z1 - a 2 ) = 0
Which is the required equation.
Problems (1) Find equation of the line whose vertex is at (a , b , g ) and base
y 2 = 4ax, Z = 0
Solution: Any line through (a , b , g ) is
x -a y - b z -g
= = (---1)
l m n
Equation of the base is y 2 = 4ax, Z = 0 (---2)
The line (1) meets the plane Z =0
\ Put Z =0 in (1)
x -a y - b -g
= =
l m n
-l -m
x -a = g; y-b = g
n n
168
-l -m
x =a g; y=b g
n n
Using in (1)
2
æ m ö æ l ö
ç b - r ÷ + 4aç a - r ÷ (---3)
è n ø è n ø
2
æ y-b ö æ x -a ö
çç b - r ÷÷ + 4açç a - r ÷÷
è z -g ø è z -g ø
( bz - yg ) 2 (az - xg )
2
+ 4a
(z - r) (z - r)
Multiply both sides by ( z - r ) 2
\ ( bz - yg ) 2 = +4a (az - xg )( z - r )
2. Find the enveloping cone of the sphere x 2 + y 2 + z 2 + 2 x - 2 y - 2 = 0 with its vertex

(1,1,1)
Solution:
S = x 2 + y 2 + z 2 + 2x - 2 y - 2
T = xx1 + yy1 + zz1 + x + x1 - y - y1 - z - z1 - 2
Here (x1 , y1 , z1 ) = (1,1,1)

=x+y+z+x+1–y–1–z–1–2
= 2x + z – 2
S 1 = 12 + 12 + 12 + 2 - 2 - 2
=1
FORMULA T 2 = SS1
(2 x + z - 2) 2 = x 2 + y 2 + z 2 + 2 x - 2 y - 2
\ 4 x 2 + z 2 + 4 + 4 xz - 4 z - 8 x - x 2 - y 2 - z 2 - 2 x + 2 y + 2 = 0
\ 3 x 2 - y 2 + 4 xz - 10 x + 2 y - 4 z + 6 = 0
169
13.5. Right circular cone

Definition:
A right circular cone is a surface generated by a straight line which passes through
a fixed point and makes a constant angle with a fixed line. The fixed point is called the
vertex of the cone. The constant angle is called the semivertical angle of the cone. The
fixed line is called the axis of the cone.
13.6. Examples
1) Standard Equation of a right circular cylinder.
M P (l , m , g )
α
Let P (l , m , g ) be any point on the cone
Given that POˆ Z = a .
P
Draw PM perpendicular to OZ in DOPM
PM
tan a =
OM
α
\ PM 2 = OM 2 tan 2 a ---(1) 0
M
But PM 2 = OP 2 - OM 2
170
= l 2 + m 2 + g 2 - q projection of OP on OZ
= l2 + m 2 + g 2 - g 2
PM 2 = l 2 + m 2
Using in (1)
l 2 + m 2 = g 2 tan 2 a
\ The locus (l , m , g ) is x 2 + y 2 = z 2 tan 2 a

(2) Find the equation of a right circular cone with its vertex at (a , b , g ) , it’s axis the line
x -a y - b z -g
= = and its semivertical angle q , l , m, n being d.r’s of the axis.
l m n
Solution:
0
θ
A (a , b , g )
P (l , m , g )
Let A (l , m , g ) be the vertex of the curve

Let P (l , m , g ) be any point on the cone
Let AP wake an angle q with the axis AO of the cone.
The d.r’s of AP are l - a , m - b , g - g
The d.r’s of OA are l,m,n
l (l - a ) + m( m - b ) + n(g - g )
cos q =
l 2 + m2 + n2 (l - a )2 + (m - b )2 + (g -g)
2
171
\ l (l - a ) + m( m - b ) + n(g - g ) = cos q é l 2 + m 2 + n 2 (l - a )2 + (m - b )2 + (g -g) ù

2
êë úû
Squaring both sides

[l (l - a ) + m( m - b ) + n(g - g )]2 = cos 2 q [(l 2 + m 2 + n 2 ) ((l - a )2 + (m - b )2 + (g - g )2 )]
\ The focus (l , m , g ) is
[l (l - a ) + m( m - b ) + n(g - g )]2 = cos 2 q [(l 2 + m 2 + n 2 ) ((l - a )2 + (m - b )2 + (g - g )2 )]
Which is the required equation of the cylinder
(3) Find the equation of the right circular cone whose vertex is the point (1,1,1) the axis is
x -1 y -1 z -1
the line = = and semi vertical angle is 300
-1 2 3
Solution: Let P (l , m , g ) be any point on the cone the vertex of the cone is A(1,1,1)
\ The d.r’s of AP are l - 1, m - 1, g - 1
The d.r’s of the axis are -1,2,3
- 1(l - 1) + 2( m - 1) + 3(g - 1)
cos q =
1+ 4 + 9 (l - 1)2 + (m - 1)2 + (g - 1)
2
3 - l + 1 + 2 m - 2) + 3g - 3)
=
2 14 (l - 1) + (m - 1) + (g - 1)
2 2 2
- l + 2 m + 3g - 4
=
14 (l - 1)2 + (m - 1)2 + (g - 1)
2
\ 3 14 (l - 1)2 + (m - 1)2 + (g - 1) = (- l + 2 m + 3g - 4 )
2
Squaring both sides
[
42 (l - 1) + (m - 1) + (g - 1)
2 2 2
] = (- l + 2 m + 3g - 4 )
2
The focus of (l , m , g ) is
172
[
42 ( x - 1) + ( y - 1) + ( z - 1)
2 2 2
] = (- x + 2 y + 3 z - 4 )
2
On simplification, the equation of the right circular cone is

19 x 2 + 13 y 2 + 3z 2 + 8 xy + 12 xz - 24 yz - 58 x - 10 y + 6 z + 31 = 0
(4) Find the condition that the general equation of the second degree may represent a
cone
Solution:
ax 2 + by 2 + cz 2 + 2 fyz + 2 gzx + 2hxy + 2ux + 2vy + 2 wz + d = 0 ---(1)
Represent a cone with vertex at the point (x1 ,y1 ,z1 )

Shifting the origin to the point (x1 ,y1 ,z1 ) (1) becomes
a ( x + x1 ) 2 + b( y + y1 ) 2 + c( z + z1 ) 2 + 2 f ( y + y1 )( z + z1 ) + 2 g ( z + z1 )( x + x1 )
+2h( x + x1 )( y + y1 ) + 2a ( x + x1 ) + 2v( y + y1 ) + 2 w( z + z1 ) + d = 0
ax 2 + by 2 + cz 2 + 2 fyz + 2 gzx + 2hxy +

2[x(ax1 + hy1 + gz1 + u ) + y (hx1 + by1 + fz1 + v) + z ( gx1 + fy1 + cz1 + w)]
2 2 2
+ ax1 + by1 + cz1 + 2 fy1 z1 + 2 gz1 x1 + 2hx1 y1 + 2ux1 + 2vy1 + 2 wz1 + d = 0
This equation is referred to (x1 ,y1 ,z1 ) which is the new origin
\ ax1 + hy1 + gz1 + u = 0 ---(2)
hx1 + by1 + fz1 + v = 0 ---(3)

gx1 + fy1 + cz1 + w = 0 ---(4)
2 2 2
ax1 + by1 + cz1 + 2 fy1 z1 + 2 gz1 x1 + 2hx1 y1 + 2ux1 + 2vy1 + 2 wz1 + d = 0 ---(5)
This can be written as

x1 (ax1 + hy1 + gz1 + u ) + y1 (hx1 + by1 + fz1 + v) + z1 ( gx1 + fy1 + cz1 + w) + ux1 + vy1 + wz1 + d = 0
\ using (2), (3), (4); ux1 + vy1 + wz1 + d = 0 ---(6)
173
Eliminate x1 ,y1 ,z1 between (2), (3), (4),(6)
a h g u
h b f v
=0
g f c w
u v w d
Which is the required condition.
(5) Prove that the equation
7 x 2 + 2 y 2 + 2 z 2 + 10 zx + 10 xy + 26 x - 2 y + 2 z - 17 = 0 represents a cone/
Proof: The given equation is of the form
ax 2 + by 2 + cz 2 + 2 fyz + 2 gzx + 2hxy + 2ux + 2vy + 2 wz + d = 0
a = 7; b = 2; c = 2; 2f =0 2 g = -10 2h = 10
f =0 g = -5 h=5
2u = 26 2v = -2 2w = 2 d = -17
u = 13 v = -1 w = -1
a h g u 7 5 - 5 13
h b f v 5 2 0 -1
=
g f c w -5 0 2 1
u v w d 13 - 1 1 - 17
2 0 -1 5 0 -1 5 2 -1 5 2 0
=7 0 2 1 - 5- 5 2 1 - 5- 5 0 1 + 13 - 5 0 2
- 1 1 - 17 13 1 - 17 13 - 1 - 17 13 - 1 1
= 0 (on simplification)
\ The given expression represents cone.
174
Method 2: To find the vertex of a cone where f(x1 ,y1 ,z1 ) as a homogenous function of
¶f ¶f ¶f ¶f
degree in variable x,y,z,t find , , , and square to zero after substituting t = 1.
¶x ¶y ¶z ¶t
Solve fx = 0, fy = 0, fz = 0, ft = 0 and t = 1, if the equations are consistent, then the given
equation represents a cone and the value of (x,y,z) denote the vertex of the cone.
(1) Prove that the equation
2 x 2 + 2 y 2 + 7 z 2 - 10 yz + 10 zx + 2 x + 2 y + 2 y + 26 z - 17 = 0 represents a cone with
vertex at (2,2,1)
Solution:
Let f(x,y,z) = 2 x 2 + 2 y 2 + 7 z 2 - 10 yz + 10 zx + 2 x + 2 y + 2 y + 26 z - 17 = 0
f (x,y,z,t) = 2 x 2 + 2 y 2 + 2 z 2 - 10 yz - 10 zx + 2 xt + 2 yt + 26 zt - 17t 2
¶f ¶f
= 4 x - 10 z + 2t ; = 4 y - 10 z + 2t ;
¶x ¶y
¶f
= 14 z - 10 y + 26t - 10 x
¶z
¶f
= 2 x + 2 y + 26 z - 34t
¶t
Put t = 1, in the above partial derivation and equate these to Zero.
\ 4 x - 10 z + 2 = 0 ie 2x - 5z + 1 = 0 ---(1)
4 y - 10 z + 2 = 0 \ 2 y - 5z + 1 = 0 ---(2)
- 10 x - 10 y + 14 z + 26 = 0 ie - 5 x - 5 y + 7 z + 13 = 0 ---(3)
2 x + 2 y + 26 z - 34t = 0
\ x + y + 13 z - 17 = 0 ---(4)
Solve (1), (2) , (3)

2x – 5z + 1 = 0
2y – 5z + 1 = 0
175
Sub 2x – 2y = 0
\x= y ---(5)
Put x = y in (3) -5y – 5y + 7z + 13 = 0

\ -10 y + 7 z + 13 = 0 ---(6)
But (2) is 2y – 5z + 1 = 0 ---(2)
Solve (2) and (6); (2) x 5; 10 y – 25 z + 5 = 0
(6) is -10y + 7z + 13 = 0
Add -18z + 18 = 0
\ z =1
Put z = 1 in (2) 2y – 5 + 1 = 0
2y – 4 = 0
2y = 4
y=2
\x=2
Hence x = 2 , y = 2, z = 1
Using in (4)
LHS = x + y + 13z – 17
= 2 + 2 + 13 – 17
= 0 = RHS
\ Equations (1), (2), (3), (4) are consistent
\ The given equation represents a sphere with the vertex (2,2,1)
176
13.7. Cylinder
The Cylinder is the surface generated by a variable straight line which remains
parallel to a fixed straight line and satisfies one more condition, ie. It may intersect a
given curve or it may touch a given surface.
13.8. Examples
(1) To find the equation of a cylinder whose generators are parallel to the line
x y z
= = and base the conic ax 2 + by 2 + 2 fyz + 2hxy + 2 gx + 2 fy + c = 0, z = 0
l m n
Solution: The equation of the base is ax 2 + by 2 + 2 fyz + 2hxy + 2 gx + 2 fy + c = 0, z = 0
--(1)
x y z
Let (x1 ,y1 ,z1 ) be any point on the generator, which is parallel to the line = =
l m n
x - x1 y - y1 z - z1
\ Equation of the generator is = = ---(2)
l m n
This meets the plane z = 0
Put z = 1 in (2)
x - x1 y - y1 z1
\ = =
l m n
x - x1 z1 -m
\ = ; y - y1 = z1
l n n
-l m
x= z1 + x1 ; y = y1 - z1
n n
Using in (1)
2 2
æ l ö æ m ö æ l öæ m ö æ l ö æ m ö
aç x1 - z1 ÷ + bç y1 - z1 ÷ 2hç x1 - z1 ÷ç y1 - z1 ÷ + 2 g ç x1 - z1 ÷ + 2 f ç y1 - z1 ÷ + c = 0
è n ø è n ø è n øè n ø è n ø è n ø
a(nx1 - lz1 ) + b(ny1 - mz1 ) + 2h(nx1 - lz1 )(ny1 - mz1 ) + 2 gn(nx1 - lz1 ) + 2 fn(ny1 - mz1 ) + cn 2 = 0
2 2
Which is the equation of the cylinder.
177
y z
(1) Find the equation of a cylinder whose generators are parallel to the line x = - =
2 3
and whose guiding curve is the ellipse x 2 + 2 y 2 = 1, z = 3
Solution: The equation of the guiding curve is x 2 + 2 y 2 = 1, z = 3 ---(1)

Let (x1 ,y1 ,z1 ) be any point on the generator. Since the generator is parallel to the
x y z
line = = ,
1 -2 3
x - x1 y - y1 z - z1
Equation of the generator is = =
1 -2 3
This meets the plane z =3
x - x1 y - y1 3 - z1
\ = =
1 -2 3
3 - z1 2
\ x = x1 + , y = y1 - (3 - z1 )
3 3
3x1 + 3 - z1 3 y1 - 6 + 2 z 1
; y=
3 3
Using in (1) x 2 + 2 y 2 = 1, we get
2 2
æ 3x1 + 3 - z1 ö æ 3 y - 6 + 2 z1 ö
ç ÷ + 2ç 1 ÷ =1
è 3 ø è 3 ø
(3x1 + 3 - z1 )2 + 2(3 y1 - 6 + 2 z1 )2 = 9
On simplification
The focus of (x1 ,y1 ,z1 ) is 3x 2 + 6 y 2 + 3z 2 + 8 yz - 2 zx + 6 x - 24 y - 18 z + 24 = 0
13.9 Right circular cylinder
The right circular cylinder is the surface generated by a straight line which is
parallel to a fixed line and is at a constant distance from it.
The variable line is called the generator of the cylinder. The fixed line is called
the axis of the cylinder. The constant distance is called the radius of the cylinder.
178
13.10. Examples
(1) Find the equation of a right circular cylinder whose axis is the line
x -a y - b z -g
= = and where radius is R units.
l m n
P (l , m , g )
A M
(a , b , g )
x -a y - b z -g
The equation of the axis are = = ----(1)
l m n
Let P (l , m , g ) be any point on the cylinder
Draw PM perpendicular to the axis.

PM = R
Join PA where A = (a , b , g )
AM = The projection of AP on the axis of the cylinder
l (l - a ) + m( m - b ) + n(g - g )
=
l 2 + m2 + n2
From DAPM , AP 2 + AM 2 + PM 2
(l - a ) 2 + ( m - b ) 2 + (g - g ) 2 =
[l (l - a ) + m( m - b ) + n(g - g )]2
l 2 + m2 + n2
( )[ ]
\ l 2 + m 2 + n 2 (l - a ) 2 + ( m - b ) 2 + (g - g ) 2 = [l (l - a ) + m( m - b ) + n(g - g )] + R 2
2
(
= [l (l - a ) + m( m - b ) + n(g - g )] + R 2 l 2 + m 2 + n 2
2
)
[ 2 2
\ The focus of (l , m , g ) is (l - a ) + ( m - b ) + (g - g ) 2
](l 2 2
+m +n 2
)=
(
= [l (l - a ) + m( m - b ) + n(g - g )] + R 2 l 2 + m 2 + n 2
2
)
which is the required equation.
179
(1) The radius of a normal section of a right circular cylinder is 2 units, the axis lie along
x -1 y + 3 z - 2
the line = = . Find its equation
1 -1 5
Solution: The equation of the axis is
x -1 y + 3 z - 2
= =
1 -1 5
The d.r’s are 1, -1, 5
1 -1 5
\ Its d.r’s are , ,
27 27 27
P (x1 ,y1 ,z1 )
M
A(1,-3,2)
Let P(x1 ,y1 ,z1 ) be any point on the cylinder
AP = ( x1 - 2) 2 + ( y1 + 3) 2 + ( z1 - 2) 2
R= 2 = Radius of the cylinder

AM = The projection of AP on the axis
1
= [( x1 - 1) + ( y1 + 3) + ( z1 - 2)]
27
1
= [x1 + y1 + z1 ]
27
1
AM 2 = [x1 + y1 + z1 ]2
27
In DAPM , AP 2 + AM 2 + PM 2
180
1
( x1 - 1) 2 + ( y1 + 3) 2 + ( z1 - 2) 2 = (x1 + y1 + z1 )2 + 2 2
27
[ ] 2
27 ( x1 - 1) 2 + ( y1 + 3) 2 + ( z1 - 2) = ( x1 + y1 + z1 ) + 108
2
Simplifying, the focus of (x1 ,y1 ,z1 ) is

29 x 2 + 29 y 2 + 29 z 2 + 2 xy - 2 xz + 2 yz - 30 x + 150 y - 93z + 75 = 0
2. Find the equation of the right circular cylinder whose guiding circle is
x 2 + y 2 + z 2 = 29; x- y+ z =3
Solution:
The circle is nothing but the plane section of the sphere x 2 + y 2 + z 2 = 9 , by the plane
x- y+z =3
NP = Radius of the circle.
= ( radius of the sphere ) 2 - ( perpendicu lar dis tan ce from the centre of the sphere to the plane ) 2
= r2 - p2 ---(1)
Radius of the sphere = 3
Centre of the sphere = (0,0,0)
p = perpendicular distance from the centre (0,0,0) to the plane x – y + z -3 =0
(-3) 3 3 3
=± = = = 3
12 + 12 + 12 3 3
\ using in (1)
R = 3 2 - ( 3) 2 = 6
M is the centre of the circle.

The d.r’s of the normal to the plane are (1,-1,1).
x-0 y+0 z-0
Equation of PMB = =
1 -1 1
x y z
= = = l (say)
1 -1 1
181
x = l , y = -l , z = l
Any point on PM = (l , - l , l ) = M (say)

This lies on the plane x – y + z = 3.
\ l +l +l = 3
3l = 3
l =1
\ Centre of the circle is (1,1,1)
Step 2: Let P (x,y,z) be any point on the cylinder
P (x1 ,y1 ,z1 )
N
A
A(0,0,0) P (x1 ,y1 ,z1 )
2 2 2
AP 2 = x1 + y1 + z1
NP = 6 \ NP 2 = 6
AN = The projection of AP on AN
1
= ( x1 + y1 + z1 )
3
In DAPN ,
AP 2 = AN 2 + NP 2
2
2 2 2 é 1 ù
x1 + y1 + z1 = 6 + ê ( x1 + y1 + z1 )ú
ë 3 û
1
=6+ ( x1 + y1 + z1 ) 2
3
182
2 2 2 2 2 2
3 x1 + 3 y1 + 3 z1 = 18 + x1 + y1 + z1 + 2 x1 y1 + 2 y1 z1 + 2 z1 x1
2 2 2
2 x1 + 2 y1 + 2 z1 - 2 x1 y1 - 2 y1 z1 - 2 z1 x1 - 18 = 0
¸ 2;
2 2 2
x1 + y1 + z1 - x1 y1 - y1 z1 - z1 x1 - 9 = 0
The focus of (x1 ,y1 ,z1 ) is
x 2 + y 2 + z 2 - xy - yz - zx - 9 = 0
13.11. Let us sum up.
We have leant the different types of cone, cylinder and the dimension their
equations in different – forms.
(1) Find the equation of a right circular cylinder of radius 2 and which axis is x
(2) What is the general equating a give passing through the axis
Cone:
1. Find the equation to the cone whose vertex is the origin and which passes through
the curve of intersection of
(i) ax 2 + by 2 + cz 2 = 1, lx + my + nz = p
Ans: p (ax 2 + by 2 + cz 2 ) = ( lx + my + nz ) 2
(ii) ax 2 + by 2 = 2 z; lx + my + nz = p
Ans: p (ax 2 + by 2 ) = 2 z ( lx + my + nz )
x y z
2. If = = is a generator of the cone, represented by f(x,y,z) = 0, then prove
l m n
that f(l,m,n) = 0
3. Find the equation of the cone whose vertex is the point (a , b , g ) and whose
x2 y2
generating line pass through the conic + = 1, z = 0
a2 b2
1
Ans:
2
(az - xr )2 + 12 (bz - yv )2 = ( z - v) 2
a b
4. Prove that the equation of the cone whose vertex is the point (1,1,0) and whose
generating curve y = 0, x 2 + z 2 + = 4 is x 2 + 3 y 2 + z 2 - 2 xy + 8 y - 4 = 0
183
5. Find the equation of a right circular cone whose vertex is at the origin, whose axis
x y z
is the line = = and which has a semivertical angle of 600
1 2 3
Ans : 19 x 2 + 13 y 2 + 3 z 2 - 24 yz - 12 zx - 8 xy = 0
6. Prove that the
equation 7 x 2 + 2 y 2 + 2 z 2 - 10 zx + 10 xy + 26 x - 2 y + 2 z - 17 = 0 represents a cone
whose vertex is at (1, -2, 2)
7. Find the equation of a right circular cone whose axis is the line x = y = z and the
generator is 2x = y = -3z ( APRIL 2004; Bharathiyar)
(Hint: To find semi vertical angle: The d.r’s of the axis are 1 , 1, 1
a1 , b1 , c1
1 -1
The d.r’s of the generator are ,1,
2 3
a2 , b2 , c2
a1 a 2 + b1b2 + c1c 2
The semivertical angle is given by cos q =
2 2 2 2 2 2
a1 + b1 + c1 a 2 + b2 + c 2
8. Find the general equation of the cone which touches the coordinate planes (NOV
2000, Bharathiyar)
1. Find the equation of the right circular cone with vertex at the origin semi
x y z
vertical angle 300 and the line = = being the axis of the conic (April
1 2 3
2004, Bharathiar)
2. Find the equation of the cone with the vertex at (1,1,1) and base curve
x 2 + 2 y 2 = 1, z = 0 (April 2004; Bharathiar)
x y z
3. The plane + + = 1 meets the coordinates axes in A,B,C. Prove that the
a b c
equation to the cone generated by lines drawn from 0 to meet the circle ABC
æb cö
is å yzçè c + b ÷ø = 0 .
184
4. Find the equation of the cylinder whose generators are parallel to the line x =
y = z and whose guiding curve is the circle.
x 2 + y 2 + z 2 - 2 x - 3 = 0, 2 x + y + 2 z = 0
Ans: 17 x 2 + 18 y 2 + 17 z 2 - 18 zy - 16 zx - 18 yz - 40 x + 10 y + 20 z - 75 = 0
5. Find the equation of the right circular cylinder of radius 2 whose axis passes
through (1,2,3) and has direction cosines proportional to 2, -3, 6
Ans: 45 x 2 + 40 y 2 + 13 z 2 + 12 xy + 36 yz - 24 zx - 42 x - 280 y - 126 z + 294 = 0
6. Find the equation of the cylinder whose generators are parallel to the line
x y z
= = and whose guiding curve is x 2 + 2 y 2 = 1, z = 3
1 -2 3
Ans: 3 x 2 + 6 y 2 + 3 z 2 + 8 yz - 2 zx + 6 x - 24 y - 18 z + 24 = 0
7. Find the equation of the right circular cylinder of radius 3 whose axis passes
through (2,3,4) and has direction cosines proportional to (2,1,-2)
Ans: 5 x 2 + 8 y 2 + 5 z 2 - 4 xy + 4 yz + 8 xz - 40 x - 56 y - 68 z + 179 = 0
8. Find the equation of a right circular cylinder of radius 3 with axis
x + 2 y - 4 z -1
= = (April 2005, Bharathiar)
3 6 2
13.15 References
Analytical Geometry of Three Dimension by N.P. Bali
185
Lesson – 14
Contents
14.1 Concicoids
14.2 Nature of a conicoid
14.3 Enveloping cone
14.4 Examples
14.5 Director Sphere
14.6 Examples
14.7 Let us sum up
14.11 References

In this lesson we are going to study about use definition coin cord which is a new
concept, the standard equation of coin cord, tangent plane, enveloping cone, director
sphere.
14.1 Concicoids
The general equation of second degree in x,y,z ie.
f ( x, y, z ) = ax 2 + by 2 + cz 2 + 2 fyz + 2 gzx + 2hxy + 2ux + 2vy + 2 wz + d = 0
represents a locus called a coincoid or a quadric.
Note: (1) The equation of a conicoid contains only line disposable constants
(2) We can reduce this equation to some standard form by the suitable change of the
axes.
In other words, we know a conic is intersected by a straight line at two points.
\ The curve of intersection of a plane and a quadric surface is a conic.
\ Quadric surfaces are called coincoids
186
Centre of the coincoid: If every chord of the conicoid passes through the origin , it is
called the centre of the coincoid central conicoid. The conic having a centre is called a
central conicoid.
14.2. Nature of a conicoid:

The standard equation of a central conicoid is ax 2 + by 2 + cz 2 = 1
This represents i) an ellipsoid if a,b,c are all positive ii) a hyperboloid of one sheet if two
are positive and the one is negative. (iii)a hyperboloid two sheets if two are negative and
one positive. (iv) a virtual ellipsoid if all are negative.
Note (1). The conicoid ax 2 + by 2 + cz 2 = 1 has origin as centre
(2). The equation ax 2 + by 2 + cz 2 = 1 is called the standard equation of the conicoid.

14.3. Enveloping cone
The locus of the tangent lines drawn from a given point to a given conicoid is a
cone called the enveloping cone of the conicoid or tangent cone of the concoid having the
given point as its vertex.
14.4. Examples
(1) Find the equation of the enveloping cone of the conic coid ax 2 + by 2 + cz 2 = 1
with its vertex at A(x1 ,y1 ,z1 )
Solution: The equation of the conicoid is
ax 2 + by 2 + cz 2 = 1 ---(1)
Let A be the point (x1 ,y1 ,z1 )
Equation of any line through A(x1 ,y1 ,z1 ) is
x - x1 y - y1 z - z1
= = = R (say) ---(2)
l m n
Any point on (2) is ( x1 + Rl ), y1 + Rm, z1 + Rn)
If this point lies on (1), then
a ( x1 + Rl ) 2 + b( y1 + Rm) 2 + c( z1 + Rn) 2 = 1
2 2 2
R 2 ( al 2 + bm 2 + cn 2 ) + 2 R ( alx1 + bmy1 + cnz1 ) + ( ax1 + by1 + cz1 ) = 0
The line (2) is a tangent line to (1)
\ The above quadratic equation has equal roots.
187
2 2 2
\ 4( alx1 + bmy1 + cnz1 ) 2 = 4( al 2 + bm 2 + cn 2 )l (ax1 + by1 + cz1 - 1) ---(3)
[a( x - x1 ) x1 + b( y - y1 ) y1 + c( z - z1 ) z1 ]2 = [a( x - x1 ) 2 + b( y - y1 ) 2 + c( z - z1 ) 2 ](ax1 2 + by1 2 + cz1 2 - 1)

[(axx1 (
+ byy1 + czz1 - 1) - ax1 + by1 + cz1 - 1
2 2 2
)] 2
=
[(ax 2 2
+ by + cz 2
) - 2(axx1 (
+ byy1 + czz1 - 1) + ax1 + by1 + cz1 - 1
2 2 2
)] --(4)
´(ax - 1)
2 2 2
1 + by1 + cz1
Let T = axx1 + byy1 + czz1 - 1
S = ax 2 + by 2 + cz 2 -1
S1 = ax12 + by12 + cz12 - 1

Using in (4)
(T-S1 )2 =[S-2T+S1 ]S1
T2 -2TS1 +S12 =SS1 -2TS1 +S12
T2 =SS1
[axx1 + byy1 + czz1 - 1]2 = (ax 2 + by 2 + cz 2 - 1)(ax1 2 + by1 2 + cz1 2 - 1)
Which is the required equation.
Tangent plane
(2) Find the equation of the tangent plane at (x1 ,y1 ,z1 ) of the conicoid ax 2 + by 2 + cz 2 = 1
Solution:
The equation of the central conicoid is ax 2 + by 2 + cz 2 = 1 ---(1)
Equations of any line through (x1 ,y1 ,z1 ) is
x - x1 y - y1 z - z1
= = = R (say)
l m n
Any point on the line is
x = x1 + Rl , y = y1 + Rm; z = z1 + Rn
i.e (x1 + Rl , y1 + Rm; z1 + Rn )
If this lies on (1) then
a ( x1 + Rl ) + b( y1 + Rm ) + c ( z1 + Rn )2 = 1
2 2
[ 2
] [ 2
] [ 2
a x1 + 2 Rl n1 + R 2 l 2 + b y1 + 2 y1 mR + m 2 R 2 + c z1 + 2 y1 nR + n 2 R 2 = 1 ]
188
R 2 ( al 2 + bm 2 + cn 2 ) + 2 R (lx1 + my1 + nz1 ) + ax1 + by1 + cz1 - 1 = 0

2 2 2
But (x1 ,y1 ,z1 ) lies in (1)

2 2 2
\ ax1 + by1 + cz1 - 1 = 0
\ R 2 (al 2 + bm 2 + cn 2 ) + 2 R(lx1 + my1 + nz1 ) = 0

This is a quadratic equation in R.
Whose roots are R1 and R2
Since the line (2) is a tangent line in quadratic equation
ax 2 + by + c = 0,
b 2 - 4ac = 0
But c = 0
\b2 = 0
b=0
The roots are equal
\ coefficient of R = 0
\ lx1 + my1 + nz1 = 0 ---(3)

(x - x1 )x1 + ( y - y1 ) y1 + (z - z1 )z1 = 0
2 2 2
xx1 + yy1 + zz1 = ax1 + by1 + cz1
xx1 + yy1 + zz1 = 1
Which is the required equation of the tangent plane at (x1 ,y1 ,z1 )
Condition for tangency
(3) Find the condition for the plane lx+my+nz = P to be a tangent plane to the conicoid
2 2 2
ax1 + by1 + cz1 = 1
Solution: Let the line lx+my+nz = P ---(1) touch the conicoid of (x1 ,y1 ,z1 )
\ Equation of the tangent plane at (x1 ,y1 ,z1 ) is axx1 + byy1 + czz1 = 1 ---(2)
But equations (1) and (2) represent the same tangent plane
\ their corresponding coefficients are proportional
189
ax1 by1 cz1 1

\ = = =
l m n p
(i ) (ii ) (iii ) (iv )
From (i) and (iv)
ax1 1 m n 1
\ = , y1 = ; z1 = ; x1 = ;
l p bp p ap
But (x1 ,y1 ,z1 ) lies on ax12 + by12 + cz12 = 1
l2 m2 n2
\ a. + b. + c. =1
a2 p2 b2 p 2 c2 p2
l2 m2 n2
+ + =1
ap 2 bp 2 cp 2
1 æ l 2 m2 n2 ö
çç + + ÷÷ = 1
p2 èa b c ø
l 2 m2 n2
\ + + = p2
a b c
Which is the required condition.
æ l m nö
Cor 1: The point of contact is çç , , ÷÷
è ap bp cp ø
Cor 2: Find the equation of two tangent planes to the central conicoid
ax 2 + by 2 + cz 2 = 1 which are parallel to the plane lx+my+nz = 0
Solution: Equation of the plane parallel to the plane lx+my+nz = 0 is
lx+my+nz = P --- (1)
(1) touches the conicoid ax 2 + by 2 + cz 2 = 1
l 2 m2 n2
\ p2 = + +
a b c
l 2 m2 n2
\p=± + +
a b c
Using in (1), Equations two tangent planes are
l 2 m2 n2
lx+my+nz = ± + +
a b c
190
14.5. Director Sphere

The locus of the point of intersection of three mutually perpendicular tangent
planes to a conicoid is called a director sphere.
Find the equation of the director sphere of the coincoid ax 2 + by 2 + cz 2 = 1 (April
2006, Bharathiyar)
Proof: The equation of the central conicoid is ax 2 + by 2 + cz 2 = 1
Let the three mutually perpendicular planes be
2 2 2
l m n
1l x+m1 y+n1 z = ± 1 + 1 + 1 ---(2)
a b c
2 2 2
l m n
12 x+m2 y+n2 z = ± 2 + 2 + 2 ---(3)
a b c
2 2 2
l m n
13 x+m3 y+n3 z = ± 3 + 3 + 3 ---(4)
a b c
Where li,mi,ni, i = 1,2,3 are d.c’s of the normals to these planes
2 2 2
\ å l1 = 1; å m = 1; å n
1 1 = 1;
åll 12 = 0; å1 m = 0
1 2
2 2 2
l1 + m1 + n1 = 1 etc.
\ By eliminating 1l ,m1,n1; 12,m2,n2,; 13 ,m3,n3
Between (2), (3), (4)
1 1 1
We get x 2 + y 2 + z 2 = + +
a b c
Which is the equation of the director sphere of the conicoid ax 2 + by 2 + cz 2 = 1
14.6. Examples
1. Prove that equation of two tangent planes to the conicoid ax 2 + by 2 + cz 2 = 1 --(1)
which pass through the line u = lx + my + nz - p = 0 ,
u ' = l ' x + m' y + n' z - p ' = 0 is
2æ l '2 m'2 n'2 2ö æ ll ' mm' nn' ö 2 æl

2
m2 n2 ö
ç
u ç + + ÷
- p ' ÷ - 2uu ' ç + + ç
- pp ' ÷ + u ' ç + + - p 2 ÷÷ = 0
èa b c ø èa b c ø èa b c ø
Solution: Equation of any polar through the given line is u + lu ' = 0 ---(A)
191
(lx + my + nz - p ) + l (l ' x + m' y + n' z - p') = 0

x (l + l l ' ) + y ( m + l m ' ) + z ( n + l n ' ) - p - l p ' = 0
\ x (l + l l ' ) + y ( m + l m ' ) + z ( n + l n ' ) = p + l p ' ---(2)
This is of the for Lx+My+Nz =P
\ L = l + ll ' ; M = m + lm' ; N = n + ln' ; P = p + lp '
The plane (2) touches the coni coid (1)
L2 M 2 N 2
\ + + = P2
a b c
1
(l + ll ')2 + 1 (m + lm')2 + 1 (n + ln')2 = p + lp' ---(B)
a b c
\ These are two roots
Hence there are two tangent planes to the conicoid (1)
Eliminate l between (A) and (B)
(u ' l - ul ')2 + (u ' m - um')2 + (u ' n - un')2 = (u ' p - up ' ) 2
a b c
æ l '2 m'2 n'2 ö æ ll ' mm' nn' ö æ l 2 m2 n2 ö
ie u 2 çç + + - p '2 ÷÷ - 2uu ' ç + + - pp ' ÷ + u '2 çç + + - p 2 ÷÷ = 0
èa b c ø èa b c ø èa b c ø
which is the required equation.
2. Find the equations of two tangent plane to the coincoid 2 x 2 - 6 y 2 + 3z 2 = 5 which pass
through the line x + 9y – 3z = 0, 3x – 3y + 6z – 5 =0 (April 2006, Bharathiyar, November
2005, April 2004)
Proof: Equation of the conicoid is 2 x 2 - 6 y 2 + 3z 2 = 5
¸ 5;
2 2 6 2 3 2
x - y + z =1
5 5 5
2 6 3
\a= ; b=- ; c=
5 5 5
Equation of any plane through the given line is
x + 9 y - 3 z + l (3 x - 3 y + 6 z - 5) = 0
x (1 + 3l ) + y (9 - 3l ) + z (6l - 3) - 5l = 0
192
x(1 + 3l ) + y (9 - 3l ) + z (6l - 3) - 5l ---(1)

This is of the form lx + my + nz = p
\ l = 1 + 3l ; m = 9 - 3l ; n = 6l - 3; P = 5l
Formula
l 2 m2 n2
\ P2 = + +
a b c
5 5 5
25l = (1 + 3l ) 2 - (9 - 3l ) 2 + (6l - 3) 2
2 6 3
On simplification, l = ±1
Using in (1), the two tangent planes are
4x + 6y + 3z – 5 = 0; 2x – 12y + 9z – 5 = 0
14.7. Let us sum up

So far we have studied the new concept of the conicord, tangent plane, enveloping
cone and director sphere.
1) What is the director sphere of the conicord 2x2 -by2 +3y2 =5
2) Write down the equation of the tangent plant at (1,1,1) to the conicord
x2 +y2 -z2 =1
1. Find the equations of the tangent planes to the conicoid x 2 + y 2 + 4 z 2 = 1 which
intersect the line 12x-3y- z=0, z = 1 (April 2006 Bharathiyar)
2. Find the equations of two tangent planes to the conicoid 2 x 2 + 2 y 2 + y 2 = 2 which
pass through the line z = 0, x + y =0. (April 2005, Bharathiyar)
3. Find the equations of the tangent plane to the coincoid 7 x 2 + 5 y 2 + 3z 2 = 60 which
pass through the line 7x + 10y – 30 = 0, 5y – 3z = 0 (April 2005, November
2004,Bharathiyar)
4. Find the point of contact of the plane x + 2y + z – 2 = 0 which touches the
conicoid x 2 + 2 y 2 - z 2 - 2 = 0 (April 2004, Bharathiyar)
193

a. Find the equations of two planes that can be drawn through the line x = 4, 3y + 4z
= 0 to touch the conicoid x 2 + 3 y 2 - 6 z 2 = 4 .
b. (Ans: x + 9y + 12z = 4, x – 9y – 12z = 4)
c. Show that the plane 9x + 8y – 5z = 38 touches the conicoid
3 x 2 + 4 y 2 - 5 z 2 = 38 and find the point of contact (November 2002 Bharathiar)
d. Find the equations of the tangent planes to x 2 + y 2 + 4 z 2 = 1 which intersect in the

line whose equations are 12x – 3y – 5 = 0, z = 1. (November 2005 Bharathiar).
14.11 References
Analytical Geometry of Three Dimension by N.P. Bali
194

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TEXT BOOK OF TRIGONOMETRY, VECTOR CALCULUS

1.0 Aims and Objectives

Equate real and imaginary parts

(1) Expand cos 6 q inpowers of cos q

1.3 Let us sum up

2.0 Aims and Objectives

= sin α cos hβ + i cos α sin hβ

Real part = cos α cos hβ

= cos (iα) cos β + sin (iα) sin β

= 2 cos A sin B; A = β +iα

= cos (A+B) + cos (A-B)

= e7 q - e-7 q -7 e5 q + 7.6 e 3q - 7.6.5 eq + 7.6.5.4 e -q - 7.6 e -3q + 7e -5q

=( e7 q - e-7 q )- 7 (e5 q - e-5 q ) + 21 (e3 q - e-3 q ) +– 35 (e q - e- q )

6 e 7q - e -7q æ e 5q - e -5q ö æ e 3q - e -3q ö æ eq - e -q ö

= sin q cosh q + cos q (i sinh q )

é1 + cos 2q ù é cosh 2f - 1ù é1 - cos 2q ù é cosh 2f + 1ù

sin x cos (iy) + cos x sin (iy) = cos q + i sin q

sin x coshy + cos x (i sinhy) = cos q + i sin q

sin (x+iy) = sin x coshy + i cosx sinhy

tanx sinh2b = sin2a tanhy =

2.3. Let us Sum up

2.4 Check your progress

3.0 Aim and Objectives

3.1 To find log e (x+iy)

Let log e (x+iy) = α+iβ

(1)2 + (2)2 gives

L o exg +(i ) y =l xo +gi ( y+) n2p i

amplitude = tan -1 æç y ö÷ = tan -1 (1) =i p

but log (1 + Z) = Z - Z 3 + Z 3 ...........

8. If i x+iy = x+iy, prove that

3.4 Check you progress

3. Prove that Log (1- i) =

7. Show that log

Multiply both sides by 2 sin æç b ö÷

2 sin æç b ö÷ s n = 2 sin a sin b + 2 sin b sin(a + b ) +

\ 2 sin a sin b = cosæç a - b ö÷ - cosæç a + b ö÷

2 sin(a + b ) sin b = cosæç a + b ö÷ - cosæç a + 3b ö÷

2 sin(a + 2 b ) sin b = cosæç a + 3b ö÷ - cosæç a + 5b ö÷

But we know that

Multiply both sides by 2 sin b

\ 2 sin b S n = 2 cos a sin b + 2 cos(a + b ) sin b + 2 cos(a + 2 b ) sin b

2 cos a sin b = sin(a + b ) - sin(a - b )

2 cos(a + b ) sin b = sin(a + 3b ) - sin(a + b )

2 cos(a + 2 b ) sin b = sin(a + 5 b ) - sin(a + 3b )

2 cos(a + n - 1b ) sin b = sin(a + (n - 1 / 2) b ) - sin(a + (n - 3 / 2) b )

But we know that

2. Find the sum to n terms of the series

But 1 + cos 2 q =2cos2 q

1é cos( n + 1)a sin( na ) ù

3. Find the sum of the series

Sin3 a +sin3 2 a +sin3 3 a +……………..+sin3 n a

4. Find the sum to n terms of

Type 1: Problems based on ex and e-x

C + is = e x cos b cos(a + x sin b ) + ie x cos b sin(a + x sin b )

Equate imaginary part

2. Find the sum of the series

C + is = e cos a cos(sin a ) + ie cos a sin(sin a )

Equate imaginary parts