Vectors and Geometry Part-17

VECTORS AND GEOMETRY PART-17

PREPARED RASHMI PRESENTED AND CONTENT EDITED BY NANDAKUMAR

TANGENT PLANE Objectives 1.understands the concept of perpendicularity 2.introduces various cones such as right circular cone and enveloping cone 3.Familiarise the tangency of a plane and a cone Introduction In this chapter we introduce the condition for perpendicularity ,condition of tangency of a plane and a cone . introduce the concept of reciprocal cone ,right circular cone and enveloping cone.

Perpendicularity To find the condition, so that lines in which plane ux+ vy + wz = 0 cuts a cone

f( x , y , z )� ax2+ by 2 + cz 2 + 2 fyz + 2 gzx = 2 hxy 0 may be at right angles. The angle q between the two lines is given by

2P ( u2+ v 2 + w 2 ) 1/ 2 tanq = (a+ b + c )( u2 + v 2 + w 2 ) - f ( u , v , w )

1 If q = 900 , tanq = i.e., (a+ b + c )( u2 + v 2 + w 2 ) - f ( u , v , w ) = 0 This is the required condition. Example 1 Prove that the plane ax+ by + cz = 0 cuts the cone yz+ zx + xy = 0 in perpendicular lines if 1 1 1 + + = 0 a b c Solution The given cone is yz+ zx + xy = 0 … (1) and plane is ax+ by + cz = 0 … (2) Let one of the lines of intersection be x y z = = . l m n

This line lies in (2) and (1). Hence mn+ nl + lm = 0 … (3) And al+ bm + cn = 0 … (4) al+ bm From (4), n = - c Putting this value in (3), we get

骣 al+ bm 琪 (m+ l )琪 - + lm = 0 桫 c

i.e., al2+( a + b - c ) lm + bm 2 = 0 2 骣l l 琪 or a琪 +( a + b - c ) + b = 0 桫m m

2 l This is quadratic in . Let m

l1 l 2 , be its two roots, then m1 m 2

l1 l 2 b � m1 m 2 a

l1 l 2 m 1 m 2 n 1 n 2 i.e., = = 1/a 1/ b 1/ c (by symmetry) The angle between the lines will be a right angle if

l1 l 2+ m 1 m 2 + n 1 n 2 = 0 , 1 1 1 i.e., + + = 0 a b c Example 2 Find the equation to the lines in which the plane 2x+ y - z = 0 cuts the cone 4x2- y 2 + 3 z 2 = 0 . Solution The given cone is

4x2- y 2 + 3 z 2 = 0 … (5) and the plane is 2x+ y - z = 0 … (6) Let one of the lines of intersection of (5) and (6) be x y z = = l m n This line lies on (5) and (6), then

4l2- m 2 + 3 n 2 = 0 … (7) and 2l+ m - n = 0 … (8)

Eliminating n, we get

3 4l2- m 2 + 3(2 l + m ) 2 = 0 or 16l2+ 12 lm + 2 m 2 = 0 or 8l2+ 6 lm + m 2 = 0 or (2l+ m )(4 l + m ) = 0 When 2l+ m = 0 , from (8), we have n = 0 l m n i.e., = = 1- 2 0 Hence one line is x y z = = 1- 2 0 … (9) Again, when 4l+ m = 0,2 l + n = 0 [from (8)] l m n i.e., = = -1 4 2 Hence the other line is x y z = = -1 4 2 Example 3 Prove that the plane lx+ my + nz = 0 cuts cone (b- c ) x2 + ( c - a ) y 2 + ( a - b ) z 2 + 2 fyz + 2 gzx + 2 hxy = 0 in perpendicular lines if (b- c ) l2 + ( c - a ) m 2 + ( a - b ) n 2 + 2 fmn + 2 gnl + 2 hlm = 0 Solution Here sum of the coefficient of x2, y 2 , z 2 i.e (b- c ) + ( c - a ) + ( a - b ) = 0 This shows that the cone has three mutually perpendicular generators. Now if plane lx+ my + nz = 0 cuts the cone in two mutually perpendicular lines then normal to the plane through vertex (0,0,0) . x y z i.e., = = l m n should also be a generator of the cone i.e., (b- c ) l2 + ( c - a ) m 2 + ( a - b ) n 2 + 2 fmn + 2 gnl + 2 hlm = 0 which is the required condition.

4 Example 4 Show that the cone whose vertex is at the origin and which passes through the curve of intersection of the sphere x2+ y 2 + z 2 = 3 a 2 , and any plane at a distance a from the origin has mutually perpendicular generators. Solution The given sphere is x2+ y 2 + z 2 = 3 a 2 … (10) Any plane at a distance a from the origin is lx+ my + nz = a … (11) Where l, m , n are actual direction cosines of normal to the plane. Making (10) homogeneous with the help of (11), equation of the cone whose vertex is origin and base the curve of intersection of (10) and (11) is

2 2 2 2 2 骣lx+ my + nz x+ y + z = 3 a 琪桫 a i.e., x2(1- 3 l 2 ) + y 2 (1 - 3 m 2 ) + z 2 (1 - 3 n 2= 0 ) - 6 lmxy - 6 mnzx - 6 nlzx Now, coeff. of x2 + coeff. of y2 + coeff. of z2

=(1 - 3l2 ) + (1 - 3 m 2 ) + (1 - 3 n 2 )

=3 - 3(l2 + m 2 + n 2 ) = 3 - 3 as l2+ m 2 + n 2 = 1 = 0 Hence plane (11) cuts the cone in three mutually perpendicular generators. Tangent Plane Let the equation of the cone be f( x , y , z )� ax2+ by 2 + cz 2 + 2 fyz + 2 gzx = 2 hxy 0 … (12) and that of a line through any point P(a , b , g ) be

5 x-a y - b z - g = = … (13) l m n Any point on (13) is (a+lr , b + mr , g + nr ) . The point in which the line (13) meets the cone (12) are determined by a(a+ lr )2 + b ( b + mr ) 2 + c ( g + nr ) 2 + 2 f ( b + mr )( g + nr )

+2g (g + nr )( a + lr ) + 2 h ( a + lr )( b + mr ) = 0 or r2[ al 2+ bm 2 + cn 2 + 2 fmn + 2 gnl + 2 hlm ] + 2 r [( aa + h b + g g ) l

+(ha + b b + f g ) m + ( g a + f b + c g ) n ] + f ( a , b , g ) = 0 … (14) This being quadratic in r, the line (13) in general meets the cone (12) in two points. If (a , b , g ) lies on (12), then f (a , b , g )= 0 and consequently from (13) one value of r is zero. Further if l( aa+ h b + g g ) + m ( h a + b b + f g ) + n ( g a + f b + c g ) = 0 … (15) Then the other root of (14) is also zero, i.e., the two points of intersection coincide at P. Hence in this case the line is a tangent to the cone at P. Here (15) is the condition for the line (13) to be a tangent line at P(a , b , g ) to the cone (12). The locus of all such tangent lines at P, i.e., the tangent plane to the cone at P is obtained by eliminating l, m , n between (15) and (13). Thus

(x-a )( a a + h b + g g ) + ( y - b )( h a + b b + f g )

+(z -g )( g a + f b + c g ) = 0 or x( aa+ h b + g g ) + y ( h a + b b + f g ) + z ( g a + f b + c g ) = 0 Since f (a , b , g )= 0

6 or aa x+ b b y + c g z + f( g yz + f ( g y + b z ) + g ( a z + g x ) + h ( b + a y ) = 0 This is the equation of the tangent plane to cone at (a , b , g ) . Condition of Tangency of a plane and a cone To find the condition that plane ux+ vy + wz = 0 may touch the cone ax2+ by 2 + cz 2 +2 fyz + 2 gzx + 2 hxy = 0 . The plane ux+ vy + wz = 0 will touch the cone if the angle between the lines of intersection of the plane and the cone is zero. if q be the angle between the lines of intersection of plane and the cone, then

2P ( u2+ v 2 + w 2 ) 1/ 2 tanq = (a+ b + c )( u2 + v 2 + w 2 ) - f ( u , v , w ) Now the plane will touch the cone if q = 0 , i.e., tanq = 0 or P = 0 or a h g u h b f v P2 = = 0 g f c w u v w 0 i.e., Au2+ Bv 2 + Cw 2 +2 Fvw + 2 Gwu + 2 Huv = 0 …….(16) where A, B , C , F , G , H are co- factors of a, b , c , f , g , h in a h g D = h b f g f c

=abc +2 fgh - af2 - bg 2 - ch 2 禗 i.e., A= = bc - f 2 a 禗 B= = ca - g 2 b 禗 C= = ab - h2 c

7 1 禗 F= = gh - af , 2 f

1 禗 G= = hf - bg , 2 g

1 禗 H= = fg - ch 2 h Thus (16) is the required condition. Reciprocal Cone The locus of the lines through the vertex of a cone of a cone normal to the tangent planes is a cone called the reciprocal cone. To find the equation of the cone reciprocal to ax2+ by 2 + cz 2 +2 fyz + 2 gzx + 2 hxy = 0 … (17) Let a plane be lx+ my + nz = 0 … (18) It will touch the cone (17) if

Al2+ Bm 2 + Cn 2 +2 Fmn + 2 Gnl + 2 Hlm = 0 … (19) Now a line normal to (18) through the vertex (0,0,0) is x y z = = l m n … (20) Now locus of (20) subject to the condition (19) is

Ax2+ By 2 + Cz 2 +2 Fyz + 2 Gzx + 2 Hxy = 0 …(21) This is the required cone reciprocal to (17). Note. We find that the cone reciprocal to (17) is (21). The name reciprocal cone is justified as the cone reciprocal to (21) comes out to be (17). Hence cones (17) and (21) are mutually reciprocal cones.

8 Since cone reciprocal to (21) is

Aⅱ x2+ B y 2 + C ⅱ z 2 +2 F yz + 2 G ⅱ zx + 2 H xy = 0 … (22) where A= BC - F2 =( ca - g 2 )( ab - h 2 ) - ( gh - af ) 2

=a( abc + 2 fgh - af2 - bg 2 - ch 2 ) =a D Similarly, Bⅱ= b D, C = c D , F ⅱ = f D , G = g D and H= h D . i.e., (22) becomes ax2+ by 2 + cz 2 +2 fyz + 2 gzx + 2 hxy = 0 . Which is same as (17). Thus cones (17) and (21) are reciprocal cones of each other and the name reciprocal cone is justified. Three mutually perpendicular tangent planes To show that the cone ax2+ by 2 + cz 2 +2 fyz + 2 gzx + 2 hxz = 0 … (23) will have three mutually perpendicular tangent planes if bc+ ca + ab = f2 + g 2 + h 2 The cone (23) will have three mutually perpendicular tangent planes if its reciprocal cone has three mutually perpendicular generators. The reciprocal cone to (23) is

Ax2+ By 2 + Cz 2 +2 Fyz + 2 Gzx + 2 Hxy = 0 where A= bc - f2, B = ac - g 2 , C = ab - h 2 etc. This will have three mutually perpendicular generators, if

9 A+ B + C = 0 or if (bc- f2 ) + ( ac - g 2 ) + ( ab - h 2 ) = 0 or if bc+ ca + ab = f2 + g 2 + h 2 This is the required condition. Example 5 Prove that the cones ax2+ by 2 + cz 2 = 0 and x2 y 2 z 2 + + = 0 are reciprocal. a b c Solution The reciprocal cone of ax2+ by 2 + cz 2 = 0 is

Ax2+ By 2 + Cz 2 =2 Fyz + 2 Gzx + 2 Hxy = 0 … (24) where a 0 0 D =0b 0 = abc 0 0 c and 禗禗 A= = bc, B = = ac 抖a b

禗禗 C= = ab, F = = 0 抖c f

1禗 1 禗 G= =0, H = = 0 2抖g 2 h By putting these values, (24) becomes bcx2+ cay 2 + abz 2 = 0 x2 y 2 z 2 or + + = 0 a b c Example 6 Show that the general equation of a cone which touches the coordinate planes is f2 x 2+ g 2 y 2 + h 2 z 2 -2 ghyz - 2 hfzx - 2 fgxy = 0 or (fx )北 ( gy ) ( hz )= 0 Solution

10 The required cone is the reciprocal to the cone which passes through the three coordinate axes. The general equation of a cone through coordinate axes is fyz+ gzx + hxy = 0 Now replacing l, m , n by f, g , h in example 2 above, we get the required cone as f2 x 2+ g 2 y 2 + h 2 z 2 -2 ghyz - 2 hfzx - 2 fgxy = 0 or (fx+ gy + hz )2 = 4 fgxy or fx+ gy - hz = 2 ( fgxy ) or fx+ gy�2 ( fgxy ) hz or (fx )� ( gy ) ( hz ) or (fx )北 ( gy ) ( hz )= 0 . Example 7 Find the condition that the lines of intersection of the plane lx+ my + nz = 0 and cones fyz+ gzx + hxy =0, ax2 + by 2 + cz 2 = 0 should be coincident. Solution Note that the common generators are nothing but the intersection of the two given cones. Any cone through the intersection of the two cones is ax2+ by 2 + cz 2 +l( fyz + gzx + hxy ) = 0

Since the lines of section of the given cone with lx+ my + nz = 0 are coincident, for some value of l , the above equation must represent a pair of planes of which one plane is lx+ my + nz = 0 . Let the other plane be lⅱ x+ m y + n z = 0 . Then

11 ax2+ by 2 + cz 2 +l( fyz + gzx + hxy ) = ( lx + my + nz )( lⅱ x + m y + n z ) Equating the coefficients of like terms, we get a= llⅱ, b = mm , c = nn , or a b c lⅱ=, m = , n = l m n and cm bn cm2+ bn 2 l f= mnⅱ + m n = + = n m mn an2+ cl 2 and lg = nl am2+ bl 2 lh = lm Now, eliminating l , we get the required conditions as cm2+ bn 2 an 2 + cl 2 am 2 + bl 2 = = fmn gnl hlm . 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 through that fixed point. The fixed point is called the vertex, the fixed line the axis, and the constant angle the semi-vertical angle of the cone. Enveloping cone The locus of the tangent lines drawn from a given point to a given surface is called the enveloping cone of that surface with the given point as vertex. This cone is also known as the tangent cone of the surface. Equation of the enveloping cone of a sphere To find the equation of the enveloping cone of the sphere

12 x2+ y 2 + z 2 +2 ux + 2 vy + 2 wz + d = 0 with its vertex at (a , b , g ) . Let the equations to any line through the point A(a , b , g ) be x-a y - b z - g = = … (25) l m n The coordinates of the point on it whose distance from A is r are (a+lr , b + mr , g + nr ) . If the line (25) meets the given sphere in the point, then (a+lr )2 + ( b + mr ) 2 + ( g + nr ) 2 + 2 u ( a + lr ) + 2 v ( b + mr ) +2w (g + nr ) + d = 0 or r2( l 2+ m 2 + n 2 ) + 2 r (a l + b m + g n + lu + mv + nw )

+(a2 + b 2 + g 2 + 2u a + 2 v b + 2 w g + d ) = 0 … (26) This is a quadratic equation in r. It gives two values of r corresponding to two points on the sphere where the line (25) meets it. In case the line (25) touches the sphere, the two points must coincide and for this, the roots of the equation (26) should be equal. Hence

(al+ b m + g n + lu + mv + nw )2 = ( l 2 + m 2 + n 2 )( a 2 + b 2 + g 2 + 2 u a

+2vb + 2 w g + d ) or [(a+u ) l + ( b + v ) m + ( g + w ) n ]2

=(l2 + m 2 + n 2 )(a 2 + b 2 + g 2 + 2 u a + 2 v b

+2wg + d ) … (27) Eliminating l, m , n between (25) and (27), the equation to the required enveloping cone is

[(a+u )( x - a ) + ( b + v )( y - b ) + ( g + w )( z - g )]2

13 =[(x -a )2 + ( y - b ) 2 + ( z - g ) 2 ][ a 2 + b 2 + g 2

+2ua + 2 v b + 2 w g + d ] … (28) If S� x2+ y 2 + z 2 +2 ux + 2 vy + 2 wz d ,

2 2 2 S1 �a+ b + g +2 u a + 2 v b + 2 w g d and T=a x + b y + g z + u( x + a ) + v ( y + b ) + w ( z + g ) + d . Then the equation (28) of the enveloping cone can be written as

2 (T- S1 ) = ( S - 2 T + S 1 ) S 1 which on simplification takes the form 2 SS1 = T Example 8 Find the equation to the right circular cone whose vertex is at origin the axis along x-axis and semi- vertical anglea . Solution Let P( x , y , z ) be any point on the surface of the cone, so that the direction ratios of the line OP are x, y , z ; O being the origin. The direction cosines of x-axis are1,0,0 . \ x��1 y 0 z 0 cosa = x2+ y 2 + z 2 or (x2+ y 2 + z 2 )cos 2a = x 2 or y2+ z 2 = x 2tan 2 a . Example 9 Find the enveloping cone of the sphere x2+ y 2 + z 2 -2 x + 4 z = 1 with its vertex at (1,1,1) . Solution Here

14 S� x2+ y 2 - z 2 +2 x - 4 z 1

2 2 2 S1 �1+ 1 - 1 � 2 � 1 = 4 1 1 4 and T鹤1 x+ 1 � y � 1 + z + ( x + 1) - 2( z 1) 1 Therefore the equation of the enveloping cone is

(x2+ y 2 + z 2 - 2 x + 4 z - 1)(4) = ( y + 3 z ) 2 or 4x2+ 3 y 2 - 5 z 2 - 6 yz - 8 x + 16 z - 4 = 0 . Example 10 The enveloping cone of the sphere x2+ y 2 + z 2 = 2 az is cut by the plane z = 0 in a plane z = 0 in a parabola. Prove that the locus of the vertex of the cone is the surface z2( x 2+ y 2 + z 2 - 4 az + 4 a 2 ) = 2 az ( x 2 + y 2 ) Solution Let the vertex of the enveloping cone be (a , b , g ) . Therefore the equation of the enveloping cone will be 2 (SS1 = T )

(x2+ y 2 + z 2 - 2 az )(a 2 + b 2 + g 2 - 2 a g ) = [ a x + b y + g z - a ( z + g )] 2 It is cut by the plane z = 0 , which gives

(x2+ y 2 )(a 2 + b 2 + g 2 - 2 a g ) = ( a x + b y - a g ) 2 or (b2+ g 2 - 2a g ) x 2 + ( a 2 + g 2 - 2 a g ) y 2 - 2 a b xy + 2 a a yx + 2 abg y - a 2 g 2 = 0 If this represents a parabola, the product of the coefficient of x2 and y2 should be equal to the square of the coefficient of 2xy ,( h2 = ab ) , i.e.,

(b2+ g 2 - 2a g )( a 2 + g 2 - 2 a g ) = ( ab ) 2 or g2( a 2+ b 2 + g 2 - 4a g + 4 a 2 ) - 2 a g ( a 2 + b 2 ) = 0 Hence the locus of (a , b , g ) is

15 z2( x 2+ y 2 + z 2 - 4 az + 4 a 2 ) - 2 az ( x 2 + y 2 ) = 0 Summary. Now let us summarise the session here we discussed some special types of cone such as reciprocal cone, right circular cone and enveloping cone,but in the beginning we come across the condition of perpendicularity That is, the condition so that the lines in which plane cuts a cone may be at right angles.Also condition of tangency of a plane and a cone are discussed with numerous examples.

Assignment 1.Find the enveloping cone of the sphere x2+y2+z2- 2x+4z=1 with its vertex at (1,1,1)

2.Find the equation to the right circular cone whose vertex is at origin ,the axis along x-axis and semi vertical angle a

3. Prove that the tangent planes to the cone ayz+bzx+cxy=0 are at right angles to the generators of the cone a2 x 2+ b 2 y 2 + c 2 z 2 -2 bcyz - 2 cazx - 2 abxy = 0

4. Find the equation to the right circular cone whose vertex is p(2,-3,5) axis PQ

16 which makes equal angles with the axis semi vertical angle is 300

FAQ 1.Say whether the given cones are reciprocal

x2 y 2 z 2 2x2 + 3 y2 + 4 z 2 = 0 and + + = 0 2 3 4

Answer

The given cones are reciprocal ,because this is similar to the example (5) the only difference is that the values of a,b,c are respectively 2,3 and 4

17 Compairing with that example A=12 B=8 C=6 F=0

G=0 H=0

2.Is it true that every cone will have three mutually perpendicular tangent planes

Answer The cone ax2+ by 2 + cz 2 +2 fyz + 2 gzx + 2 hxz = 0 will have three mutually perpendicular tangent planes if bc+ ca + ab = f2 + g 2 + h 2 That is the reciprocal cone has three mutually perpendicular generators The reciprocal cone) is Ax2+ By 2 + Cz 2 +2 Fyz + 2 Gzx + 2 Hxy = 0 This will have three mutually perpendicular generators, if A+ B + C = 0 or bc+ ca + ab = f2 + g 2 + h 2 This is the condition.

Glossary Reciprocal Cone The locus of the lines through the vertex of a cone of a cone normal to the tangent planes

18 is a cone called the reciprocal cone

Right Circular Cone 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 through that fixed point.

Enveloping cone The locus of the tangent lines drawn from a given point to a given surface is called the enveloping cone of that surface with the given point as vertex.

Quiz 1.The condition that the plane ux+vy+wz=0 may touch the cone 2 2 2 ax +by +cz =0 骣u2 v 2 w 2 琪 + + = 0 (b) (a) 琪桫a b c u v w + + = 0 (c) u+ v + w = 0 a b c (d)a+b+c=0

2.Find the right circular cone having its vertex at the origin and passing through the circle y2+z2=25 , x=4 (a)x2=y2+z2 (b) 25x2=16(y2+z2) (c) x2=16(y2+z2)(d) 25x2= (y2+z2)

3.The enveloping cone of the of the sphere x2+y2+z2- 2x+4z=1 with its vertex at (,1,1) (a)4x2+3y2-5z2-6yz-8x+16z- 4=0 (b) x2+y2-z2-6yz-8x+16z-

19 4=0(c) 4x2+3y2-5z2-6yz- x+z=0 (d) 4x2+3y2-5z2-6yz- 8x+16z=0

Answer 骣u2 v 2 w 2 琪 + + = 0 1(a) 琪桫a b c

2(b) 25x2=16(y2+z2)

Reference 1 S.L.Loney The Elements of Coordinate Geometry ,Macmillian and company, London 2 Gorakh Prasad and H.C.Gupta Text book of coordinate geometry, Pothisala pvt ltd Allahabad 3 P.K.Mittal Analytic Geometry Vrinda Publication pvt Ltd,Delhi.