Z

                                                                                                      P(x,y,z)

                                                                      O                                             Y

                                                                                                     M

                                     X

P(x,y,z) 

r

z    N

                                                                      O             Y

                           M  Q

X

Cos = =

                                               Z =r cos 

From right angled triangle OQP,    OQ = OP sin 

       OQ = r sin 

In   ONQ cos  =  ON = OQ cos 

                 x= r sin  cos 

In  OMQ, sin  =  MQ = OQ sin 

                         y= r sin sin

                   P(x,y,z)

                                                                                                            M        

                                                                      O                                             X

                                                N      

                                                                                                  Q

                                        Y              

Cos  = 

  O M  = x =  cos 

  Sin  = 

  y =  sin 

  z = z 

Preliminaries

Distance Formula:

Distance between two points, P(x, y, z,) and Q(x2, y2, z2)

  PQ = 

Direction Cosines :

If a line OP makes an angle ,  and  with X, Y and Z  axes respectively then the quantities cos , cos , cos  are called direction cosines (d. c. s) of line OP, where O is the origin.Direction cosines of line are denoted by l, m, n

Direction ratios :

Numbers  a, b, c which are proportional to direction cosines  l, m, n respectively are called direction ratios of given line.

We can prove that  

  l = , m =  , n =

(x1 – x2, y1 – y2, z1 – z2) are d. r.’s of line joining the points and (x2, y2, z2).

Angle between two lines :

If l1, m, n and l2, m2, n2 are d. c. s of two line and if  is angle between those two lines then
cos  = l1l2 + m1 m2 + n1 n2

cos  =

Equation of Plane :

(1) General form :Ifa,b,c are direction ratios of normal to plane then equation of plane is                            

     ax + by + cz + d = 0

(2) Intercept form :If plane intersects with co-ordinate axes at a,b,c thenequation of plane is                            

(3) Normal (perpendicular) form :

If l, m, n are d. c.s  of normal to plane and p is perpendicular distance from origin to the plane, then Normal form of plane is,

 l x + my + nz = p

(4) Angle between two planes :

Angle between two planes :a1 x + b1 y + c1 z + d1 = 0 and a2x + b2 y + c2 z + d2 = 0

 is given by cos  = 

Length of Perpendicular :

Perpendicular distance from the point P(x1, y1, z1) to the plane ax + by + cz + d = 0 is

Given  

 We know that

Let the equation of sphere be

     x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0……………(1)

 Since, sphere pass through the point ( 1,3,1),therefore from equation ( 1 ), we get

1 + 9 + 1 + 2u+6v+2w+d=0 2u+6v+2w+d= - 11.........(2)

 Since, sphere pass through the point ( 2,-1,1),therefore from equation ( 1 ), we get

 4 + 1 + 1 + 4u-2v+2w+d=0 4u-2v+2w+d= - 6............(3)

 Since, sphere pass through the point ( 1,2,0),therefore from equation ( 1 ), we get

 1 + 4 + 0 + 2u+4v+0+d=0 2u+4v+d= - 5............(4)

 Since, sphere pass through the point ( 1,-1,1),therefore from equation ( 1 ), we get

 1 + 1 + 1 + 2u-2v+2w+d=0 2u-2v+2w+d= - 3............(5)

Eq(2) - Eq(3) gives   - 2u + 8v = - 5 ........(6)

Eq(2) – Eq(4) gives  2v + 2w = - 6.......(7)

Eq.(2)-Eq(5) gives 8v = - 8  v = -1

Using value of v in equation (6) we get  u = - 3/2

Using value of v in equation (7) we get w = - 2

Using values of u,v,w in equation(2) we get d = 2

From equation(1), we get   x2 + y2 + z2 + 2(-3/2)x + 2(-1)y + 2(-2)z + 2 = 0

   x2 + y2 + z2 - 3x - 2y - 4z + 2 = 0

Let equation of spherebe  x2 + y2 + z2 + 2ux + 2vy + 2wz + d=0……………(1)

  d = 0  ( sphere pass through origin)

 Since sphere intersect at 1 on x-axis i.e. sphere intersects-axis at the point (1 , 0 ,0 )

  1 + 2u = 0  (  makes inter apt 1 on x-axis)

    u = –

 Similarly,  v = –   (sphere intersect y-axis at the point ( 0 ,1 ,0 )

 ,  w = –  (sphere intersect z-axis at the point ( 0 ,0 ,1 )

Using values of u,v,w  and d in equation(1) ,we get

   x2 + y2 + z2 – x – y – z  = 0                                         

Ex.3: Prove that the spheres

 x2 + y2 + z2 – 2x + 4y – 4z = 0 and x2 + y2 + z2 + 10 x + 2z + 10 = 0

 touch each other. Find the co-ordinates of point of contact.

Soln. :

 Let  S1 = x2 + y2 + z2 – 2x + 4y – 4z = 0 

 Centre  c1 = (1, – 2, 2)

 Radium,  r1 = = 3   

  S2 = x2 + y2 + z2 + 10x + 2z + 10 = 0

 Centre,  c2 = (– 5, 0, – 1)

 Radius,  r2 = = = 4

 Now  d (C1 C2) ==

               = = = 7

  r1 + r2 = 3 + 4 = 7

ie distance between  centers of two sphere is equal to sum of radii therefore the spheres touches each other externally.

Let P be the point of contact. We observe that P divides join of C1C2 in the ratio 4:3

Equation of line is given in symmetrical form.

The co-ordinates of any point on line are given by,

  = k

  x = – 5k – 1, y = 3k + 8, z = 4k + 4

The co-ordinates of any point on this line are of the form x = – 5k – 1, y = 3k + 8, z = 4k + 4

Let these are co-ordinates of point P   P (– 5 k – 1, 3k + 8, 4k + 4)………(1)

 Direction ratios of OP are (– 5k –3, 3k + 5, 4k + 5)…( direction of join of P(x1, y1, z1) and Q(x2, y2, z2) are (x1 – x2, y1 –y2, z1 – z2))

Also d. r.s of given line are (– 5, 3, 4)

If P is point of contact, OP is perpendicular to given line

–5 (– 5k – 3) + 3 (3k + 5) + 4 (4k + 5) = 0 

 25k + 9k + 16k + 15 + 50  = 0

 50k + 50  = 0    k = – 1

  co-ordinates of  P (4, 5, 0)  ( put k = -1 in eq.(1))

  OP = radius of sphere =

  OP =  = 3

 By using centre-radius form,

  (x – 2)2 + (y – 3)2 + (z + 1)2  = 9

  x2 + y2 + z2 – 4x – 6y + 2z + 5  = 0

 Let the equation of sphere be x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0……………(1)

Co-ordinates of center of sphere are ( -u , -v,-w)

The equations of co-ordinates planes x = 0, y = 0, z = 0.

The length of perpendicular from centre of sphere to these planes is equal to radius of sphere.

   =  = =    ( by formula of length of perpendicular                  

                                                                                               from point on given plane)

We have u = v = w  ( from first three ratios)

From last two ratios,

  u = 

  5u + 6 = 3u

  u = - 3

 (– u, – v, – w)  = (3, 3, 3)  ( u = v = w)

  radius = x-co-ordinate of center of sphere = - u = 3

  Equation of sphere

  (x – 3)2 + (y – 3)2 + (z – 3)2 = 9

  x2 + y2 + z2 – 6x – 6y – 6z + 18  = 0

Let sphere passes through origin and intercepts co-ordinate axes at say 

  A(x1,0,0),B(0,y1,0),C(0,0,z1)

Equation of sphere is given by

  x2 + y2 + z2 – x1x – y1y – z1z = 0

 Center of sphere is ( x1,y1,z1)

Radiaus = = r    (given)

  +  +  = 4r2   …(1)

Now equation  plane cutting axes at A, B, C is

   +  +  = 1 …(2)

Let (, , ) be foot of perpendicular from origin to the plane ABC (locus point).

 drs of  OP  , ,

 drs of  PQ  x – , y – , z –

 OP is normal to plane   OP  PQ

  (x – ) +  (y – ) +  (z – ) = 0

  x + y +  z =  +  +  = M (say)

    x + y+  z = M

   + +  = 1 …(3)

Comparing Equation (2) and  (3) we get,

x1 =  , y1 =  , z1 =

 Equation (1) 

   + +  = 4r2

  ( +  + )2= 4r2

i.e.  ( +  + )2 ( +  + ) = 4r2

 i.e.            (x2 + y2 + z2) (+ + ) = 4r2

  2x – y – 2z – 4 = 0

Let C be center of sphere and let sphere touch the plane 2x – y – 2z – 4 = 0 at the point

Say P(1, 2, – 2)

Line CP is normal to plane  2x – y – 2z – 4 = 0 therefore d,r,’s of line CP are 2,-1,2.

Line CP passing through point P(1, 2, – 2)

Equation of normal CP is, = = = k

x = 2k + 1,y = - k +2,z = - 2k-2

  The co-ordinates of any point on this line are of the form x = 2k + 1,y = - k +2,z = - 2k-2

 Let these are co-ordinates of point center C of sphere 

  C (2k+ 1, –k + 2, – 2k – 2)

 we have CP2 = CQ2

(2k + 1– 1)2 + (– k + 2 – 0)2 + (– 2k – 2 – 0)2

   = (2k + 1– 1)2 + (– k + 2 – 2)2 + (– 2k – 2+ 2)2

4k2 + (2 – k)2 + (2k + 2)2 = (2k)2 + (– k)2 + (– 2k)2

4 – 4k + k2 + 4k2 + 8k + 4 = k2 + 4k2

  4k = – 8

    k = –2

 co-ordinates of center of sphere is C( -3,4,2)

Radius of sphere =

Equation of sphere by using center-radius form is given by

Let     S   x2 + y2 + z2 + 2x – 4y – 2z – 3 = 0

  V  2x – 2y + z + 14 = 0

Co-ordinates of centre of sphere C  (– 1, 2, 1)

Radius of sphere = = 3

  l (CP) = Perpendicular from C(– 1, 2, 1) on the plane 2x – 2y + z + 14 = 0

   = 

   = 

   = 3

 Length of Perpendicular from centre of sphere on plane equal to radius of sphere therefore given sphere touch the plane

To find the Point of Contact :

Equation of line CP passing through the point C(– 1, 2, 1) and with dr’s 2, – 2, 1 is given by

   = =

   = = = t say

 x = 2t – 1, y = – 2t + 2, z = t + 1

These are co-ordinates of any general point on line CP.

Let these are co-ordinates of point P.

ie P P (2t – 1, – 2 t + 2, t + 1)

As P lies on given plane U = 0

ie on 2x – 2y + z + 14 = 0

 2 (2t – 1) – 2 (– 2 t + 2) + (t + 1) + 14 = 0

 4t – 2 + 4t – 4 + t + 1 + 14 = 0

 9t + 9 = 0  t = –1

 P = P (– 2 – 1, 2 + 2, –1 + 1)

= P (– 3, 4, 0)

Line CP is normal to the plane 2x – y – 2z = 4

dr’s of CP are 2, –1, 2

 Line CP pass through the point P (1, 2, – 2)

 Equation of line CP is

   = = = t say

   x = 2t + 1, y = – t + 2, z = 2t – 2

Let these are co-ordinates of point C

 C C (2t + 1, – t + 2, 2t – 2)

Now l (CP) = l (CA)=

   = 

  =   

  9t2 = 9t2 – 24 t + 26

  24 t = 26

  t = 

  C   

    C   

Radius of sphere  = l (CP)

 =   

   = 

   = 

  l (CP) = 

 Equation of sphere by centre-radius form is given by

 + + = 

Line CP is normal to plane

 x – 2y – 2z – 7 = 0

drs of CP are 1, – 2,  –2.

 line CP passing through the point (1, 1, – 3)

 Equation of line CP is given by

   =  =   = t say

 x = t + 1, y = – 2t + 1, z = – 2t – 3

 Let, C  C (t + 1, – 2t + 1, – 2t – 3)  

 Since, l(CP) = l(CA)   (Both are radius of sphere)

   = 

   =  

Squaring on both sides

  9t2 = (t2 – 4t + 4) + (4t2 – 8t + 4) + (4t2 + 8t + 4)

  9t2 = 9t2 – 4t + 12

    4t = 12    t = 3             

   C  (4, – 5, – 9)

Radius of sphere = l (CP)

  = 

  =  = 9

Equation of sphere by using center-radius form is given by

Here we find co-ordinates of centre of sphere then by using centre – radius form we find equation of sphere.

Given circle is section of required sphere taken by plane x + 2y – 2z + 11 = 0

Centre of this circle is P  (1, – 1, 5)

Radius of circle say l (PQ) = 4 

Line CP is normal to plane x + 2y – 2z + 11 = 0

dr’s of line CP are 1, 2, – 2

Line CP pass through point P  (1, –1, 5)

 Equation of line CP are

   = = = t say

   x = t + 1, y = 2t – 1, z = – 2t + 5

 Let  c   (t + 1, 2t – 1, – 2t + 5)    … (1)

Now from right angle triangle  CPQ,  

 l (CP)2 + l (PQ)2 = l (CQ)2

 [(t + 1) – 1]2 + [(2t – 1) + 1]2 + [(– 2t + 5) – 5]2 + 42 = l (CQ)2

  9t2 + 16 = l (CQ)2   … (2)

We have l (CQ)  = l (CA)   (∵ Radius of sphere)

 lCQ = lCA

 9t2 + 16 = [(t + 1) – 6]2 + [(2t – 1) – 4]2 + [(– 2t + 5) – 3]2

  = (t – 5)2 + (2t – 5)2 + (– 2t + 2)2

  = (t2 – 10t + 25) + (4t2 – 20t + 25) + (4t2 – 8t + 4) 

                              9t2 + 16 = 9t2 – 38 t + 54

   38 t = 38   t  =              1

  From equation (1)

  C  (2, 1, 3)   which is centre of sphere

  l (CQ)2 = 9 + 16 = 25   ( from equation (2))

  lCQ   Which is radius of sphere

 Equation of sphere by using centre-radius form is

  (x – 2)2 + (y – 1)2 + (z – 3)2 = 25             

  x2 + y2 + z2 – 4x – 2y – 6z – 11  = 0

Let    S    x2 + y2 + z2 + 4x – 6y + 2z – 10 = 0

Here we find equation of plane by which section of sphere S = 0 is taken co-ordinates of centre of sphere are C  (– 2, 3, –1)

Line CP is normal to required plane and line CP pass through the points P  (1, 2, 2) and C– 2, 3, – 1

dr’s of line CP are 1 – (– 2), 2 – 3, 2 – (– 1) ie 3, – 1, 3

 Equation of plane to which line CP is normal is of the form 3x – y + 3z + d = 0               … (1)

But point P  (1, 2, 3) lies on plane given by equation (1)

     3(1) – (2) + 3(2) + d  = 0

   d = 7

   From equation (1), 3x – y + 3z + 7  = 0

 Required equation of circle is

x2 + y2 + z2 + 4x – 6y + 2z – 10 = 0 ; 3x – y + 3z + 7 = 0

Equation of required sphere passing through given circle is

(x2 + y2 + z2) – 2x + 3y – 4z + 6 +  (3x – 4y + 5z – 15) = 0  … (1)

x2 + y2 + z2 + (– 2 + 3) x + (3 – 4) y + (– 4 + 5) z + (6 – 15) = 0

 Here  u1  = (– 2 + 3), v1 = (3 – 4), w1 = (– 4 + 5)

  d1 = 6 – 15

 For given sphere x2 + y2 + z2 + 2x + 4y – 6z + 11 = 0                               … (2)

u2 = 1, v2 = 2, w2 = – 3, d2 = 11

By given condition ie sphere given by equation (1) cuts the sphere given by (2) orthogonally.

(– 2 + 3) + 2(3 – 4) – 3 (– 4 + 5) = (6 – 15) + 11

   = –

 Put value of  in equation (1) we get 5(x2 + y2 + z2) – 13x + 19y – 25z + 45 = 0

Sphere: Orthogonal sphere, Great circle

Equation of sphere passing through given circle x2 + y2 + z2 = 4 ; z = 4 is given by

 (x2 + y2 + z2 – 4) +  (z – 4) = 0     … (1)

  x2 + y2 + z2 + z + (– 4 – 4) = 0     … (2)

Here u1 = 0, v1 = 0, w1 = , d1 = – 4 – 4

Given sphere is S1 = x2 + y2 + z2 + 10y – 4z – 8 = 0  … (2)

Here U2 = 0, v2 = 5, w2 = – 2, d2 = – 8

As spheres given by equation (1) cuts the sphere (2) orthogonally we have

2u1u2 + 2v1v2 + w1w2 = d1+ d2

 2 (0) (0) + 2 (0) (5) + 2 (– 2) = (– 4 – 4) – 8

  – 2 = – 4 – 12

   2 = – 12

   = – 6

Using value of  in equation (1) we get

(x2 + y2 + z2) – 6 (z – 4) = 0

 x2 + y2 + z2 – 6z + 8 = 0

Equation of sphere passing through given circle is

 (x2 + y2 + z2 + 7y – 2z + 2) +  (2x + 3y + 4z – 8) = 0                                … (1)

 x2 + y2 + z2 + 2x + (7 + 3) y + (– 2 + 4) 2 + (2 – 8) = 0

Centre of this sphere is given by

  C  

Since given circle is a great circle of the

Sphere x2 + y2 + z2 + 7y – 2z + 2 = 0

 co-ordinates of centre of sphere satisfies equation of plane 2x + 3y + 4z = 8

 2 [– ] + 3  + 4 = 8

   – 2 – – + 4 – 8 

  = –1

Using value of  in equation (1) we get

(x2 + y2 + z2 + 7y – 2z + 2) – 1 (2x + 3y + 4z – 8) = 0

 x2 + y2 + z2 – 2x + 4y – 6z + 10 = 0

Equation of sphere by diameter form is given by,

(x – 1) (x – 4) (y + 2) (y – 0) + (z – 3) (z – 6) = 0

 ie   x2 + y2 + z2 – 5x + 2y – 9z + 2z = 0      … (1)

The intersection of the sphere (1) by a plane x + y – 2z + 6 = 0 is given by,

(x2 + y2 + z2 – 5x + 2y – 9z + 22) +  (x + y – 2z + 6) = 0

x2 + y2 + z2 + (– 5 + ) x + (2 + ) y + (– 9 – 2) z + 22 + 6 = 0

Centre of this sphere is

Since given circle is great circle, this centre must lie on given plane x + 2y – 2z + 6 = 0

 + 2 – 2 + 6 = 0

 =

 Equation of required sphere is,

2 (x2 + y2 + z2) – 11x + 3y – 16 z + 38 = 0

In above, equation  +  = 1

L.H.S. is of degree 2 but r.h.s. is of degree zero.

 By z = c we have   = 1

   =  1

   +  = 

    + –  = 0 

is required equation of cone

Given plane  +  +   = 1 ents

Co-ordinate planes at a, b, c respectively. Equation of sphere OABL is given by plane and a sphere.

x2 + y2 + z2 – ax – by – cz = 0

we make this equation homogeneous,

as   x2 + y2 + z2  – (ax + by + cz)  = 0

simplifying we get,

Type II: Equation of cone with vertex other than origin

The equation of any generator of cone with vertex at (3,1,2) is,

=  =   (i)

From given equation z =0

 =  = 

x = 3 – , y = 1 –

If this point has to be on the guiding curve ther it should satisfy 2x2 + 3 y2 = 1

i.e. 2  + 3  = 1     (2)

we eliminate l, m, n from equation (2)

from equation (1)

x – 3 = (z – 2)  = 

y – 1 = (z – 2)  =

From equation (2),

2 + 3   = 1

 2 = (z – 2)2

 2+ 3  = (z – 2)2

 2 + 3  = z2  – 4z + 4

 8 x2 + 12y2 + 20z2 – 24xz – 12yz + 4z – 4 = 0

                                             Generator                                        Guiding curve(Base)

Semivertical angle

Vertex                                                                                                          Axis

P( x,y,z)                                     Guiding curve(Base)

Semivertical angle

       A

        Vertex                                                    

                                              generator                                                Axis  =   = 

P( x,y,z)                                     Guiding curve(Base)

Semivertical angle

       A =

        Vertex                                                    

                                           generator                                                             Axis

Semivertical angleP( x,y,z)                                                     Guiding curve(Base)

=

A

Vertex

generator                                                    Axis

Here vertex of cone is P  (1, – 1, 2) dr’s of axis are 2, 1, – 2

Let Q (x, y, z) be any point on cone. dr’s of PQ are x – 1, y – (– 1), z – 2

i.e. x – 1, y + 1, z – 2.

Since angle between generator PQ and axis is 45  (Given)

   cos  45 = 

   = 

Squaring on both sides and simplifying

9[(x – 1)2 + (y + 1)2 + (z – 2)2]  =  (2x + y – 2z + 3)2

9[(x2 – 2x + 2) + (y2 + 2y + 1) + (z2 – 4z + 4)]

                                                   = 2[4x2 + y2 + 2z2 + 9 + 4xy – 9xz + 12 x – 4yz + 6y – 12z]

9 x2 + 9 y2   + 9 z2  – 18x + 18y – 36 z + 63

                                           = 8x2 + 2 y2 + 8z2  + 8xy – 16xz – 8yz + 24x + 12y–24z + 18

x2 + 7y2 + z2 – 8xy + 16 xz + 8yz – 42x + 6y – 12z + 45 = 0.

P( x,y,z)                                     Guiding curve(Base)

Semivertical angle

       A =

        Vertex                                                    

                                              generator                                                    Axis d.r.’s 2,1, - 2

Since axis of cone is perpendicular to plane 2x + y – 2z + 1 = 0, therefore dr’s of axis are

2, 1, – 2 . Let Q (x, y, z) be any point on cone.

dr’s of PQ are x – 1, y + 1, z – 1  

Angle between any generator say PQ and axis is semi vertical angle which is given to be    

45

    cos 45 = 

   = 

Squaring on both sides and simplifying weget,

9[(x – 1)2 + (y + 1)2 + (z – 1)2] = 2 (2x + y – 2z + 3)2

9[x2 + y2 + z2 – 2x + 2y – 2z + 3] =

                                              2 [4 x2 + y2 + 4z2 + 9 + 4xy – 8xz + 12x – 4yz + 6y –   12z]

9 x2  + 9 y2  + 9 z2  – 18x + 18y – 18z + 27 =

                                             8 x2 + 2y2 + 8z2 + 8xy – 16xz – 8yz + 24x + 12y – 24z + 18

 x2 + 7y2 + z2 – 8xy + 16 xz + 8yz – 42z + 6y + 6z + 9 = 0

Given line is 2x + 3y = 6,    z = 0

These equations can be written as

2x = 6 – 3y = 3(2 – y),    z = 0

i.e.    =   =  

 i.e.  =  =   =  = 

Since given cone is generated by rotating line  =  =  therefore this line is generator and since this line is rotated about y-axis, therefore y-axis is the axis. Whose dr’s are 0,1,0.

Semivertical angle is angle between generated and axis which is given by

  cos  = 

  cos  = 

                                                              P(x,y,z)                               Guiding curve                                                                            

Semivertical angle

A

        Vertex                                                    

                                              generator                                                    Axis d.r.’s   0,1,0

Let P(x,y,z) be any point on the cone

dr’s of AP are x – 0,  y – 2,   z – 0   i.e. x, y – 2,  z

    cos  = 

  –  = 

 4[x2 + (y – 2)2 + z2] = 13(y – 2)2

  4[x2 + y2 – 4y + 4 + z2] = 13(y2 – 4y + 4)

  4 x2 – 9 y2 + 4 z2  + 36y – 36  =  0

P( x,y,z)                                     Guiding curve(Base)

Semivertical angle

       A

        Vertex                                                    

                                              generator                                                Axis  =   = 

P( x,y,z)                                     Guiding curve(Base)

Semivertical angle

       A =

        Vertex                                                    

                                              generator                                                Axis

Take any point p(x,y,z) on given cone. The d.r.’s of AP are x, y, z.

The d.r.’s of axis are 1,2,3.

We have cos 30 = 

  = 

 i.e. = 2x + 4y + 6z

Squaring on both sides we get,

 42(x2 + y2+  z2) = (2x + 4y + 3z)2

  = 4 x2 + 16 y2 + 19 z2 – 48xy – 72yz + 36zx = 0

P( x,y,z)                                     Guiding curve(Base)

Semivertical angle

       A =

        Vertex                                                    

                                              generator                                                    Axis

Here vertex of cone is P  (1, – 1, 2) dr’s of axis are 2, 1, – 2

Let Q (x, y, z) be any point on cone. dr’s of PQ are x – 1, y – (– 1), z – 2

i.e. x – 1, y + 1, z – 2.

Since angle between generator PQ and axis is 45  (Given)

   cos  45 = 

   = 

Squaring on both sides and simplifying

9[(x – 1)2 + (y + 1)2 + (z – 2)2]  =  (2x + y – 2z + 3)2

9[(x2 – 2x + 2) + (y2 + 2y + 1) + (z2 – 4z + 4)]

 = 2[4x2 + y2 + 2z2 + 9 + 4xy – 9xz + 12 x – 4yz + 6y – 12z]

9 x2 + 9 y2   + 9 z2  – 18x + 18y – 36 z + 63 = 8x2 + 2 y2 + 8z2  + 8xy – 16xz – 8yz + 24x + 12y

– 24z + 18

x2 + 7y2 + z2 – 8xy + 16 xz + 8yz – 42x + 6y – 12z + 45 = 0.

P( x,y,z)                                     Guiding curve(Base)

Semivertical angle

       A =

        Vertex                                                    

                                              generator                                                    Axis d.r.’s 2,1, - 2

Since axis of cone is perpendicular to plane 2x + y – 2z + 1 = 0, therefore dr’s of axis are

2, 1, – 2 . Let Q (x, y, z) be any point on cone.

dr’s of PQ are x – 1, y + 1, z – 1  

Angle between any generator say PQ and axis is semi vertical angle which is given to be 45

    cos 45 = 

   = 

Squaring on both sides and simplifying weget,

9[(x – 1)2 + (y + 1)2 + (z – 1)2] = 2 (2x + y – 2z + 3)2

9[x2 + y2 + z2 – 2x + 2y – 2z + 3] = 2 [4 x2 + y2 + 4z2 + 9 + 4xy – 8xz + 12x – 4yz + 6y – 12z]

9 x2  + 9 y2  + 9 z2  – 18x + 18y – 18z + 27 = 8 x2 + 2y2 + 8z2 + 8xy – 16xz – 8yz + 24x + 12y –

24z + 18

 x2 + 7y2 + z2 – 8xy + 16 xz + 8yz – 42z + 6y + 6z + 9 = 0

Given line is 2x + 3y = 6,    z = 0

These equations can be written as

2x = 6 – 3y = 3(2 – y),    z = 0

i.e.    =   =  

 i.e.  =  =   =  = 

Since given cone is generated by rotating line  =  =  therefore this line is generator and since this line is rotated about y-axis, therefore y-axis is the axis. Whose dr’s are 0,1,0.

Semivertical angle is angle between generated and axis which is given by

  cos  = 

  cos  = 

P( x,y,z)                                     Guiding curve(Base)

Semivertical angle

       A

        Vertex                                                    

                                              generator                                                    Axis d.r.’s   0,1,0

Let P(x,y,z) be any point on the cone

dr’s of AP are x – 0,  y – 2,   z – 0   i.e. x, y – 2,  z

    cos  = 

  –  = 

 4[x2 + (y – 2)2 + z2] = 13(y – 2)2

  4[x2 + y2 – 4y + 4 + z2] = 13(y2 – 4y + 4)

  4 x2 – 9 y2 + 4 z2  + 36y – 36  =  0

Let P(x1,y1,z1) be any point on the cylinder.

Equation of generator which pas through point P is given by

   =  =    (1)

Since guiding curve is

  x2 + y2 = 9;   z = 1

Put     z = 1   in equation (1) we get

   =   =

   =    x  =  – x1  +

     x = 

 Also  =  y = y1  + 

      = 

Using values of x and y in another equation of guiding curve we get,

   + = 9

   +

Changing x1, y1, z1 to x, y, z respectively

9x2 + 9y2 + 5z2 + 6xz – 12yz – 6z + 12y – 10z – 76 = 0

i.e.

Equation of Right Circular Cylinder:

                                                         P(x,y,z)

                                                                 Radius

                                 A    M                                          Axis d.r.’s

                                                       P(x,y,z)

                                                                 Radius = 2

                                 A    M                                          Axis d.r.’s

Let P(x, y, z) be any point on given cylinder.

The d.r.s of axis are 2,1,2. It’s d.c.’s are , ,

i.e.  ,   , .

Since axis of cylinder pass through the point A(1,2,3)

d.r’s of AP ax x – 1, y – 2, z – 3

  AM = Project of AP on Axis

  AM =  (x – 2) +   (y – z) +  (z – 3)

From Fig,

  AP2 = AM2+ PM2

  (x – 1)2 + (y – 2)2 + (z – 3)2 = + 4

 5x2 + 8y2 + 5z2 - 4yz – 8xz – 4xy + 22x – 16y – 14z – 10 = 0

Axis of given right circular cylinder is 2(x – 1) = y + 2 = z which can be

written as   =   = 

This line pass through the point say

A  (1, – 2, 0)

                                                       P(x,y,z)

                                                                 Radius = 3

                                 A    M                                          Axis  d.r.’s

Axis,    =   = 

dr’s 1,2,2

Let P(x,y,z) ne any point on the cylinder

l(AP) =   (1)

(∵ by distance formula)

L (PM) = 3  (2) (Radius of cylinder)

dr’s of axis are 1,2,2

dc’s of axis are   , ,

i.e. ,  , 

 Now l(AM)  =length of projection of line AP in axis.

   =  (x – 1) +  (y + 2) +  (z – 0)

  l(AM) =  (3)

From  PMA,

 l(AM)2 + l(PM)2  = l(AP)2

 + (3)2 = (x – 1)2 + (y + 2)2 + z2

 (x2 + 4y2 + 4z2 + 9 + 4xy + 4xz + 6x + 8yz + 12y + 12z) + 81

           = 9 [(x2 – 2x + 1) + (y2 + 4y + 4) + z2]

 8x2 + 5y2 + 5z2 – 4xy + 4xz +8yz – 24x + 24y – 122 – 90 = 0

Since direction cosines of axis of dr’s of cylinder is proportional to 2, – 3,6, there for

dc’s of axis are ,  ,

i. e. ,  ,

                                                       P(x,y,z)

                                                                 Radius = 2

                                 A    M                                          Axis d.r.’s

  l(AP) =  (by distance  (1) formula)

  l(PM) = Projection of AP on axis

   =  (x – 1) +  (y – 2) +  (2 – 3)

  l(AM) =  (3)

Now From figure

l(AM)2 + l(PM)2 = l(AP)2