1. The document defines basic concepts related to solid geometry including direction ratios of lines, equations of straight lines and planes, different forms of the equation of a sphere, properties of spheres like touching spheres and tangent planes, right circular cones and cylinders. 2. It provides examples to find the equations of spheres, cones and cylinders given certain conditions like vertices, radii, axes etc. 3. Several exercises are given to find equations of spheres, cones and cylinders based on different scenarios like passing through points, intersecting planes or circles orthogonally.

1. Unit III : Solid Geometry
Basic Concepts
1.Direction ratios(drs) of line joining two points A and B are
.
e.g. A B
drs of line AB are 4 – 2, 1 – 3, 2 – (-1)
i. e. drs of line AB are 2, – 2, 3
2. Equation of straight line passing through point A and having drs a, b, c is
c
zz
b
yy
a
xx 111





e.g. (1)
2
2
1
1
2
1





 zyx
.This is equation of straight line passing through point
A and having drs 2, 1, 2.
(2) Equation of straight line passing through point A and having drs 2, 4, 3 is
3
2
4
1
2
2 




 zyx
3. Direction ratios(drs) of Two Parallel lines are same.
e.g. If line L is parallel to line
321
zyx
 then drs of line L are 1, 2, 3.
4.If line with drs is perpendicular to line with drs then
.
5. Direction ratios(drs) of line equally inclined to co-ordinate axis are 1, 1, 1.
6. i. Direction ratios(drs) of X-axis are 1, 0, 0.
ii. Direction ratios(drs) of Y-axis are 0, 1, 0.
iii. Direction ratios(drs) of Z-axis are 0, 0, 1.
7. If line L is perpendicular to plane then drs of line L are a, b, c.
e.g. L
2x – 3y + z – 5 =0 Plane drs of line L are 2, –3, 1.

2. 7. Angle between Two Lines with drs and with drs is given by
√ √
e.g. Angle between Two Lines L1 with drs 2, 1, and L2 with drs 3, 4, 0 is given by
√ √
8. Perpendicular distance from point A( , , ) to plane ax + by + cz + d = 0 is……..
|
√
|
1. SPHERE
(a) Centre and Radius Form:
Centre
(b) Standard Form:
Centre
(c) General Form:
Centre & Radius = dwvur 
222
(d) Intercept Form :The equation of the sphere which cuts off intercepts a, b, c from x,
y, z axis respectively is
(e) Diameter Form : Equation of sphere having ( , , ) and ( , , ) as end points of
its diameter is given by
TOUCHING SPHERES:
(a) Two spheres touch externally if the distance between their centre’s is equal to the sum of
their radii.
(b) Two spheres touch internally if the distance between their centre’s is equal to the
difference of their radii.
0
222
 czbyaxzyx
      2222
rczbyax 
2222
rzyx 
0222
222
 dwzvyuxzyx

3. Exercise 1
1) Find equation of the sphere with centre & passing through the point
.Ans:
2) Find equation of the sphere passing through origin & makes equal intercepts of unit
length with co-ordinate axes. Ans:
3) Find equation of the sphere where endpoints of diameter are and
. Ans:
4) Find the equation of the sphere passing through
& . Ans:
5) Find the equation of the sphere passing through &
Ans:
6) Find the equation of the sphere passing through &
Ans:
7) Find the equation of the sphere which touches the co-ordinate axe, whose centre is
in the positive octant and radius 4. Ans: √
8) Find the equation of the sphere passing through
9) Find the equation of the sphere whose centre is & which touches the
line Ans:
10) Prove that two spheres and
touch each other. Also find point of contact.
Ans: ( )
TANGENT PLANE to given SPHERE:
1.Equation of tangent plane to the sphere 0222
222
 dwzvyuxzyx
at point ( , , ) is
2. Given Sphere touches the plane if radius of the sphere is equal to perpendicular distance of
the plane from centre of the sphere.
Exercise 2
1) Find equation of tangent plane & normal to the sphere
Ans: Tangent Plane:
Normal:
2) Prove that the sphere touches the plane
Find point of contact.Ans:
3) Prove that plane touches the sphere
Find point of contact.Ans:( )
4) Prove that plane touches the sphere
Find point of contact.Ans:
5) Prove that plane touches the sphere
Find point of contact.Ans:
     11111
yyvxxuzzyyxx   01
 dzzw

4. 6) Find the equation of the sphere tangential to the plane
& passing through the point
Ans:
7) Find the equation of the sphere which passes through the point and touches
the plane at the point
Ans:
8) Find the equations of the spheres passing through the points
and touching the plane .
Ans:
Sphere passing through a Circle
1) Section of a Sphere by a Plane is a circle.
2) Intersection of two Spheres is a circle.
3) If S = 0 is the equation of a sphere and U = 0 is the equation of a plane, then
equations S = 0,U = 0 together represents a circle.
4) If S1 = 0 is the equation of a sphere and S2 = 0 is another equation of sphere, then
equations S1 = 0, S2 = 0 together represents a circle.
5) Equation of a Sphere passing through circle S = 0,U = 0:
6) Equation of a Sphere passing through circle S1 = 0,S2 = 0 :
ORTHOGONAL SPHERES
Two spheres 0222 1111
222
 dzwyvxuzyx and
0222 2222
222
 dzwyvxuzyx are said to be orthogonal if the tangent planes to
the two spheres at the points of intersection are at right angles.
Condition for orthogonality of two spheres:
Exercise 3
1) Find centre & radius of the circle ; .
Ans: Centre √
2) Find centre & radius of the circle ;
Ans: Centre ( )
3) Find centre & radius of the circle ;
Ans: Centre √
0 US 
021
 SS 
21212121
222 ddwwvvuu 

5. Exercise 4
Q. Find equation of the sphere passing through the circle
1)
Ans:
2)
the
Ans:
3) and meeting the plane
Ans:
4) touch the plane
Ans: ,
5) touches the
plane Ans: ,
.
6) touches the plane
Ans: ,
7)
orthogonally.
Ans:
8)
orthogonally. Ans:
9) the
sphere orthogonally.
Ans:
10) Find equation of the sphere which touches the sphere
& passes through the point
Ans: .
11) Find the equation of the sphere which passes through the point
Ans:
12) Find the equation of the sphere passing through the points &
cutting orthogonally.
Ans:
13) A sphere has
is a great circle.
Ans:

6. 2. CONE
Definition : A cone is a surface generated by a straight line which passes through a fixed
point and satisfies one or more condition e.g. it intersects a given curve or touches a given
surface.
The fixed point is called the vertex and the given curve (or surface) the guiding
curve of the cone. Any straight line lying on the surface of the cone is called its generator.
RIGHT CIRCULAR CONE:
Vertex Axis
A(x,y,z)
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 straight line through the vertex.
The fixed point is called vertex, the fixed line the Axis of the cone and the angle is
known as the semi-vertical angle of the cone.
Note : The section of right circular cone by any plane perpendicular to its axis is circle.
Necessary data for an equation of Right Circular Cone
1.Vertex of RCC
2.d.r.s of an axis
3.Semi-vertical angle.
V(a,b,c) Axis
A(x,y,z)
Let A(x,y,z) be any point on Right circular cone.
Drs of VA are x – a , y – b , z – c.
Drs of an axis are
Angle between VA & axis is given by
√ √
Simplifying above equation, we get required equation of right circular cone.

7. Exercise 5
Q. Find the equation of the right circular cone
1) with vertex at origin, whose axis is the line
321
zyx
 and semi vertical
angle 0
30 .
2) with vertex at )3,2,1(  , semi vertical angle 





3
1
cos
1
and the line
1
1
2
2
2
1





 zyx
as axis.
3) vertex is )3,2,1(  , axis is the line zyx  2 and semi vertical angle 0
45 .
4) vertex is )3,2,1( , axis has the drs 4,1,2  and semi vertical angle .60
0
5) vertex is )1,1,1(  , axis is parallel to the line z
yx

21
and one of its generator
has direction cosines proportional to .1,2,2
6) with vertex at )2,1,1(  ,axis the line
2
2
1
1
2
1





 zyx
and semi vertical angle .45
0
7) which passes through the point )4,3,1( with vertex at )1,2,2( and axis parallel to the
line
3
2
2
1
2
1 




 zyx
.
8) which passes through the point )1,2,2(  with vertex at the origin and axis parallel to
the line
1
2
1
1
5
2 



 zyx
.
9) which passes through the point )2,1,1( has its axis the line zyx 436  and vertex at
origin.
10) whose vertex is origin and making equal angles with the coordinate axes, and having
generator with direction cosines proportional to .2,2,1 
11) having vertex at )3,0,0( and passing through the circle 0,16
22
 zyx .
12) Find the equation of the right circular cone which has its vertex at the point )10,0,0(
and whose intersection with the xy - plane is the circle of diameter 10.
13) vertex is at the point )1,0,1( which pass through the point )1,1,1( and axis of the cone is
equally inclined to the co-ordinate axes.
14) whose vertex is )1,1,1( and base of the circle 2,4
22
 zzx .
15) generated when the straight line 0,632  xzy revolves about z - axis.
16) with vertex at )1,1,1(  ,axis is perpendicular to the plane 0122  zyx and semi
vertical angle 0
45 .
17) generated by rotating the line
321
zyx
 about the line
21
z
y
x


.

8. 3.CYLINDER
Definition :A cylinder is a surface generated by a straight line which is always parallel to a
fixed straight line and satisfies one more condition e.g. it intersects a given curve or touches a
given surface.
Right Circular Cylinder
Axis
Definition :A right circular cylinder is a surface generated by a straight lines(called as
generators) which are parallel to a fixed line and having a fixed distance from fixed line.
Fixed line is called the axis of the cylinder and fixed distance is called radius of the cylinder.
Section of a right circular cylinder by plane perpendicular to the axis is called normal
section; which is a circle.
Note :The length of the perpendicular from any point on right circular cylinder to its axis is
equal to the radius of the cylinder.
Necessary data for an equation of Right Circular Cylinder
1.Radius of RCC
2.d.c.s of an axis
3.Point on axis.
A(x,y,z)
M Axis
P(a,b,c)
Let r be the radius of RCC. Here AM = r.
Let A (x,y,z) be any point on RCC.
Let be an equation of an axis of RCC.
Drs of PA are x – a , y – b , z – c.
Drs of an axis are
Dcs of an axis are
√
,
√ √
.
PM = Projection of PA on axis of RCC.
√
+
√ √
.
From ---------------------------(1)
Putting values of PA, AM & PM in equation (1) & then simplifying it, we get required
equation of right circular cylinder.

9. Exercise 6
Q.Find the equation of right circular cylinder whose
1) radius is 2 and axis is the line
6
3
3
2
2
1 




 zyx
.
2) radius is 3 and axis is the line
1
5
2
3
2
1





 zyx
.
3) radius is 5 and axis is the line
1
1
1
3
3
2 



 zyx
.
4) radius is 2 and axis is the line
5
2
1
3
2
1 




 zyx
.
5) radius is 4 and axis is the line zyx  2 .
6) radius is 3 and axis is the line zyx  2)1(2 .
7) radius is 2 , axis passes through )3,2,1( and has direction cosines proportional to
.
8) radius is 2 , axis passes through )3,2,1( and has direction cosines proportional to
.
9) radius 4 whose axis passes through origin and makes equal angles with the coordinate
axes.
10) axis is
31
1
2
2 zyx




and which passes through the point )3,0,0( .
11) Find the equation of the right circular cylinder described on the circle through
)3,0,0(),0,3,0(),0,0,3( .
12) Find the equation of the right circular cylinder which passes through the section of the
sphere 25
222
 zyx made by the plane 022  zyx .
13) Find the equation of the right circular cylinder which passes through the section of the
sphere made by the plane