Chapter 3

Analytical Solid Geometry

We have learned points and lines in two-dimensional rectangular coordinate system. In this chapter, we will extend the system to three dimensions.

3.1 Coordinates of a Point in Space

In the plane, each point is associated with an order paired of real number. In space, each point is associated with an ordered triple of real numbers. Through a fixed point, called the origin O, draw three mutually perpendicular lines: the x-axis, they-axis and the z-axis. A point P in space is determined by an ordered triple (x, y, z) of real numbers as shown in the diagram. These numbers x, y, z are called the coordinate of P.

fig-3.1

The xy-plane consists of the x-axis and the y-axis, the z-axis is perpendicular to the xy-plane. The coordinates of the points on the xy-plane are of the form (x, y, 0).

The yz-plane consists of the y-axis and z-axis, and the x-axis is perpendicular to the yz-plane. The coordinates of the points on the yz-plane are of the form (0, y, z).

fig-3.2

The zx-plane consists of the x-axis and the z-axis, and the y-axis is perpendicular to the zx-plane. The coordinates of the points on the zx-plane are of the form (x, 0, z).

The equation of xy-plane, yz-plane and zx-plane are as shown in the following table.

Plane	Equation	Coordinate
xy-plane	z = 0	(x, y, 0)
yz-plane	x = 0	(0, y, z)
zx-plane	y = 0	(x, 0, z)

Planes parallel to xy-plane: Equation of any plane parallel to the xy-plane is z = c, and the coordinates of the points on that plane are of the form (x, y, c).
fig-3.3

Planes parallel to yz-plane: Equation of any plane parallel to the xy-plane is x = a, and the coordinates of the points on that plane are of the form (a, y, z).

Planes parallel to zx-plane: Equation of any plane parallel to the xy-plane is y = b, and the coordinates of the points on that plane are of the form (x, b, z).
table-3.2

Lines perpendicular to xy-plane: Equation of any line perpendicular to the xy-plane and passing through the point (a, b, c) is x = a, y = b, and the coordinates of the points on that line are of the form (a, b, z).
fig 3.7

Lines perpendicular to yz-plane: Equation of any line perpendicular to the yz-plane and passing through the point (a, b, c) is y = b, z = c, and the coordinates of the points on that line are of the form (x, b, c).

Lines perpendicular to zx-plane: Equation of any line perpendicular to the zx-plane and passing through the point (a, b, c) is z = c, x = a, and the coordinates of the points on that line are of the form (a, y, c).

Example 1.
Find the equation of the line through the point (-3, 5, 7) and perpendicular to
(a) xy-plane (b) yz-plane (c) zx-plane

Find the point of intersection of the line and plane.

Solution
(a) The equation of the line through the point (-3, 7, 5) and perpendicular to
x = -3, y = 5 or (-3, 5, z)

The point of intersection of the line and xy-plane is (-3, 5, 0).

(b) The equation of the line through the point (-1, 5, 7) and perpendicular to yz-plane is
y = 5, z = 7 or (x, 5, 7)
The point of intersection of the line and yz-plane is (0, 5, 7).

(c) The equation of the line through the point (-3, 5, 7) and perpendicular to zx-plane is
z = 7, x = -3 or (-3, y, 7).

The point of intersection of the line and zx-plane is (-3, 0, 7).

Note that

• equation of x-axis: y = 0, z = 0
• equation of y-axis: x = 0, z = 0
• equation of z-axis: x = 0, y = 0

Distance between two points: From the right triangle PAB, we have PB² = AB² + PA².
fig 3.7

Then from the right triangle PBQ, we have

PQ² = PB² + BQ²
= AB² + PA² + BQ²
= (x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²

So, the distance between P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is

PQ = √ (x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²

Exercise 3.1

Find the equation of the plane containing the point (1, -2, 3) and parallel to

Find the equation of the line through the point (2, 3, -4) and perpendicular to

Find the distance between the points (2, -3, 5) and (7, 5, -2).

Show that the points (-1, 2, 5), (1, 1, 6) and (0, 5, 6) form a right triangle.

Show that the points (1, -1, 2), (3, -2. 3) and (5, -3, 4) are collinear.

3.2 Lines

Directed value of a line segment: For a line segment PQ, directed values (l, m, n) of PQ, where P is (x₁, y₁, z₁) and Q is (x₂, y₂, z₂) is defined by

(PQ) = (l,m,n) = (x₂ - x₁, y₂ - y₁, z₂ - z₁)

Then the length of the segment PQ is

PQ = √ (x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)² = √ l² + m² + n²

Real numbers and points on the line: For every real number k, there is a point R on the line PQ with respect to point P and k vice versa.

Case 1. 0 ≤ k ≤ 1

Point R is between P and Q such that PR = kPQ.

Draw PA, RC and QB perpendicular to xy-plane: AB is projection of PQ on xy-plane.

Draw PT ∥ AB and RS ∥ AB.

fig 3.7

Then ^TS ⁄ _TQ = ^PR ⁄ _PQ
^{z - z₁}⁄ _{z₂ - z₁} = k
z = z₁ + k(z₂ - z₁)
z = z₁ + kn
Similarly, by drawing the projection of PQ on yz-plane, we have
^{x - x₁} ⁄ _{x₂ - x₁} = k
x = x₁ + k(x₂ - x₁)
x = x₁ + kl
and by drawing the projection of PQ on a zx-plane, we have
^{y - y₁} ⁄ _{y₂ - y₁} = k
y = y₁ + k(y₂ - y₁)
y = y₁ + km

Case 2. k > 1

Point R is after Q such that PR = kPQ.

Draw PT ∥ AC and QS ∥ AC. Then we have
fig 3.8

^TR⁄ _TS = ^PR⁄_PQ
^{z - z₁} ⁄ _{z₂ - z₁} = k
z = z₁ + k(z₂ - z₁)
z = z₁ + kn

Case 3. k < 0.

Point R is before P such that RP = -kPQ.
Draw PT ∥ CB and RS ∥ CB. Then we have
fig 3.9

^ST ⁄ _TQ = ^RP⁄ _PQ
^{z₁ - z} ⁄ _{z₂ - z₁} = -k
z = z₁ + k(z₂ - z₁)
z = z₁ + kn

Similarly, for Case 2 and Case 3,
^{x - z₁} ⁄ _{x₂ - x₁} = k
x = x₁ + k(x₂ - x₁)
x = x₁ + kl

^{y - y₁} ⁄ _{y₂ - y₁} = k
y = y₁ + k(y₂ - y₁)
y = y₁ + km
Therefore the coordinates of point R on the line PQ with respect to the point P and a real number k are

(x, y, z) = (x₁ + kl, y₁ + km, z₁ + kn)

This equation is called coordinate form of the equation of line PQ and k is called a parameter.
In general, if a segment of a line through (x, y, z) has directed values (l, m, n) which are not equal to 0, then the equation of the line can be written as

^{x - x₁} ⁄ _l = ^{y - y₁} ⁄ _m = ^{z - z₁} ⁄ _n

Since
(PR) = (x₁ + kl - x₁, y₁ + km - y₁, z₁ + kn - z₁)
= (kl, km, kn)

and R is any point on the line PQ with (PQ) = (l, m, n), we may define directed values of the line PQ as (kl, km, kn) where k is a real number.

Example 2.

Given P(1, 2, 3) and Q(3, 6, 5), find the coordinates of point R(x, y, z) on the line PQ with rspect to the point P and the following parameters.

(a) k = ¹ ⁄ ₂ (b) k = 2 (c) k = -2

Solution

Given P(1, 2, 3) and Q(3, 6, 5), we have (PQ) = (l, m, n) = (2, 4, 2)

(a) k = ¹⁄ ₂
(x, y, z) = (1 + ¹⁄ ₂(2), 2 + ¹ ⁄ ₂(4), 3 + ¹ ⁄ ₂(2) ) = (2, 4, 4)

(b) k = 2
(x, y, z) = (1 + 2(2), 2 + 2(4), 3 + 2(2) ) = (5, 10, 7)

(c) k = -2
(x, y, z) = (1 + (-2)(2), 2 + (-2)(4), 3 + (-2)2 ) = (-3, -6, -1)

Example 3.
Given P(-1, 2, 3) and Q(3, 5, -2), determine whether or not the following points are on the line PQ. If the point is on the line PQ, Find the corresponding parameter with respect to the point P.

(a) (1, ⁷ ⁄ ₂, ¹ ⁄₂) (b) (7, 8, -7)
(c) (-5, -1, 8) (d) (7, 8, -2)
Solution
Given P(-1, 2, 3) and Q(3, 5, -2), we have (PQ) = (l, m, n) = (4, 3, -5). The equation of the line PQ is
^{x + 1} ⁄ ₄ = ^{y - 2} ⁄ ₃ = ^{z - 3} ⁄ _-5

(a) If (x, y, z) = (1, ⁷ ⁄ ₂, ¹ ⁄₂), then
^{x + 1} ⁄₄ = ^{1 + 1} ⁄₄ = ¹ ⁄₂
^{y - 2} ⁄₃ = ^{⁷ ⁄₂ -
2} ⁄₃ = ¹ ⁄₂
^{z - 3} ⁄_-5 = ^{¹ ⁄₂ - 3} ⁄_-5 = ¹ ⁄₂
and so ^{x + 1} ⁄₄ = ^{y - 2} ⁄₃ = ^{z - 3} ⁄_-5 for (x, y, z) = (1, ⁷ ⁄₂, ¹ ⁄₂).

Therefore the point (1, ⁷ ⁄₂, ¹ ⁄₂) is on the line PQ, corresponding parameter is ¹ ⁄₂.

(b) If (x, y, z) = (7, 8, -7), then
^{x + 1} ⁄₄ = ^{7 + 1} ⁄₄ = 2
^{y - 2} ⁄₃ = ^{8 - 2} ⁄₃ = 2
^{z - 3} ⁄_-5 = ^{-7 - 3} ⁄_-5 = 2
and so ^{x + 1} ⁄₄ = ^{y - 2} ⁄₃ = ^{z - 3} ⁄_-5 for (x, y, z) = (7, 8. -7)
Therefore the point (7, 8, -7) is on the line PQ, corresponding parameter is 2.

(c) If (x, y, z) = (-5, -1, 8), then
^{x + 1} ⁄₄ = ^{-5 + 1} ⁄₄ = -1
^{y - 2} ⁄₃ = ^{-1 - 2} ⁄₃ = -1
^{z - 3} ⁄_-5 = ^{8 - 3} ⁄_-5 = -1
and so ^{x + 1} ⁄₄ = ^{y - 2} ⁄₃ = ^{z - 3} ⁄_-5 for (x, y, z) = (-5, -1, 8).
Therefore the point (-5, -1, 8) is on the line PQ, corresponding parameter is -1.

(d) If (x, y, z) = (7, 8, -2), then
^{x + 1} ⁄₄ = ^{7 + 1} ⁄₄ = 2
^{y - 2} ⁄₃ = ^{8 - 2} ⁄₃ = 2
^{z - 3} ⁄_-5 = ^{-2 - 3} ⁄_-5 = 1
and so ^{x + 1} ⁄₄ = ^{y - 2} ⁄₃ ≠ ^{z - 3} ⁄_-5 for (x, y, z) = (7, 8, -2)
Therefore the point (7, 8, -2) is not on the line PQ.

Example 4.
Given P(2, 1, 3) and Q(6, -5, 3), determine whether or not the following points are on the line PQ. If the point is on the line PQ, find the corresponding parameter with respect to the point P.

(a) (4, -2, 3) (b) (-2, 7, 3)
(c) (10, -11, 3) (d) (1, 1, 3)

Solution
Given P(2, 1, 3) and Q(6, -5, 3), we have (PQ) = (l, m, n) = (4, -6, 0).
Then the coordinates of the point (x, y, z) on the line PQ are
(x, y, z) = (2 + 4k, 1 - 6k, 3)
This means that the line PQ is on the plane z = 3.
(a) If (x, y, z) = (4, -2, 3), then
(4, -2, 3) = (2 + 4k, 1 - 6k, 3)
we have k = ¹ ⁄₂. Therefore the point (4, -2, 3) is on the line PQ with corresponding parameter ¹ ⁄₂.

(b) If (x, y, z) = (-2, 7, 3), then
(-2, 7, 3) = (2 + 4k, 1 - 6k, 3)

We have k = -1. Therefore the point (-2, 7, 3) is on the line PQ with corresponding parameter -1.
(c) If (x, y, z) = (10, -11, 3), then
(10, -11, 3) = (2 + 4k, 1 - 6k, 3)
We have k = 2. Therefore the point (10, -11, 3) is on the line PQ with corresponding parameter 2.

(d) If (x, y, z) = (1, 1, 3), then
(1, 1, 3) = (2 + 4k, 1 - 6k , 3)
There is no value k that satisfies this condition. Therefore the point (1, 1, 3) is not on the line PQ.

Exercise 3.2

Given P(3, 1, 5) and Q(-3, 7, -2), find the coordinates of point R(x, y, z) on the line PQ with respect to the point P and the following parameters.

¹ ⁄ ₂

Given P(-2, 1, 3) and Q(4, 4, -3), determine whether of not the following points are on the line PQ. If the point is on the line PQ, find the corresponding parameter with respect to the point P.

Given P(3, 2, -1) and Q(4, 2, 5), determine whether or not the following points are on the line PQ. If the point is on the line PQ, find the corresponding real number with respect to the point P.

⁷ ⁄ ₂

Find the points of intersection of the line joining the two points (2, 4, 5) and (3, 5, -4) with the following planes.

3.3 Parallel, Skew and Perpendicular Lines

Parallel lines:

If P is (x₁, y₁, z₁) and Q is (x₂, y₂, z₂) with directed values, (PQ) = (l, m, n) = (x₂ - x₁, y₂ - y₁, z₂ - z₁) then segment PQ and segment OA where A = (l, m, n) are parallel and PQ = OA as shown in the given figure.

So line PQ and line OA are parallel if they have same directed values. But if directed values of a line are given by (l, m, n), then (kl, km, kn) are also directed values of the line. Therefore line PQ and line OA are parallel if their directed values multiples of each other by a real number k. Any line parallel to the line OA is also parallel to the line PQ, so that

Two lines are parallel if and only if their directed values are multiples of each other by some real number.

For example, if line PQ has a directed values (2, 3, 5) and line RS has a directed values (4, 6, 10) then PQ ∥ RS.

Skew Lines: In space, there are pair of lines that are neither parallel nor intersect. These pairs of lines are called skew lines.

Example 5.

Given P(2, 1, 3), Q(6, -5, 4), R(2, 3, 4) and S(-1, 5, 1), determine whether the lines PQ and RS are parallel or skew or intersect.

Solution
For P(2, 1, 3) and Q(6, -5, 4),
(PQ) = (4, -6, 1)

For R(2, 3, 4)and S(-1,5,1),
(RS) = (-3, 2, -3) are not parallel.
Since ⁴ ⁄ _-3 ≠ - ⁶ ⁄ ₂ ≠ ¹ ⁄ _-3, directed values of PQ are not multiple of RS.
So the two lines are not parallel.

If a point (x, y, z) is on the line PQ and RS, then

x = 2 + 4s	x = 2 - 3t
y = 1 - 6s	y = 3 + 2t
z = 3 + s	z = 4 - 3t

for real numbers s and t.
Then we have the following system of three equations

2 + 4s	= 2 - 3t
1 - 6s	= 3 + 2t
3 + s	= 4 - 3t

Solving the first two of these equations, we have s = - ³ ⁄ ₅ and t = ⁴ ⁄ ₅. But
3 + (- ³ ⁄ ₅) ≠ 4 - 3( ⁴ ⁄ ₅)
these values of s and t do not satisfy the last equation.
So the system of equations has no solution and hence the given lines do not intersect each other. Therefore the given lines are skew.

Finding the measure of ∠PAQ : Consider P(x₁, y₁, z₁), A(a, b, c) and Q(x₂, y₂, z₂),

(AP)	= (l₁, m₁, n₁) = (x₁ -a, y₁ - b, z₁ - c)
(AQ)	= (l₂, m₂, n₂) = (x₂ - a, y₂ - b, z₂ - c)
(PQ)	= (l₃, m₃, n₃)
	= (x₂ - x₁, y₂ - y₁, z₂ - z₁)
	= (l₂ - l₁, m₂ - m₁, n₂ - n₁)

(AP)² + (AQ)² - (PQ)²
= l₁² + m₁² + n₁² + l₂² + m₂² + n₂² - (l₂ - l₁)² - (m₂ - m₁)² - (n₂ - n₁)²
= 2(l₁l₂ + m₁m₂ + n₁n₂)
By the law of cosines,

cos ∠PAQ	= ^{(AP)² + (AQ)² - (PQ)²} ⁄ _{2 . AP . AQ}
	= ^{2 (l₁l₂ + m₁m₂ + n₁n₂)} ⁄ _{2 . AP . AQ}
	= ^{l₁l₂ + m₁m₂ + n₁n₂} ⁄ _{√ l₁² + m₁² + n₁² √ l₂² + m₂² + n₂²}

If l₁l₂ + m₁m₂ + n₁n₂ = 0, then cos ∠PAQ = 0 and hence ∠PAQ = 90°.

Perpendicular lines: Two lines are perpendicular if and only if they are intersect and
l₁l₂ + m₁m₂ + n₁n₂ = 0
for any directed values (l₁, m₁, n₁) and (l₂, m₂, n₂) of the line.

Exampole 6.
Given P(0, 0, 1), Q(3, 6, 4), R(0, 3, 1) and S(3, 0, 4), show that lines PQ and RS are perpendicular.

Solution
Given P(0, 0, 1) and Q(3, 6, 4), we have (PQ) = (3, 6, 3)
Given P(0, 3, 1) and Q(3, 0, 4), we have (RS) = (3, -3, 3)
And we have 3(3) + 6(-3) + 3(3) = 0
If a point (x, y, z) is on the line PQ and RS, then

x = 0 + 3s	x = 0 + 3t
y = 0 + 6s	y = 3 - 3t
z = 1 + 3s	z = 1 + 3t

for real numbers s and t. Then we have the following system of three equations

3s	= 3t
6s	= 3 - 3t
1 + 3s	= 1 + 3t

Solving the first two of these equations, we have s = ¹ ⁄ ₃ and t = ¹ ⁄ ₃. The two lines intersect each other and the point of intersection is (1, 2, 2). Hence, the two lines are perpendicular to each other.

Example 7.
Find the equation of the line passing through the point (-4, 7, -3) and perpendicular to the line
(x, y, z) = (3 + 2k, -1 + 3k, 1 - k)
Find also the point of intersection of two lines.

Solution
Directed value of the given lines are (2, 3, -1).
Directed value of the required lines are
(-4 - (3 + 2k), 7 - (-1 + 3k), -3 - (1 - k) )
= ( -7 - 2k, 8 - 3k, -4 + k)
For some real number k.
If the two lines are perpendicular, then

2 (-7 - 2k) + 3 (8 - 3k) + (-1) (-4 + k)	=	0
-14 - 4k + 24 - 9k + 4 - k	=	0
14 k	=	14
k	=	1

So directed values of the required line are (-9, 5, -3) and the equation of the line is
(x, y, z) = (-4 - 9t, 7 + 5t, -3 - 3t)
The point of intersection is
(x, y, z) = (5, 2, 0)

Exercise 3.3

Find cos ∠PAQ for the followings.

Determine whether the line PQ and RS are parallel or skew or intersect. If PQ and RS intersect, are they perpendicular?

Find the equation of the line passing through the point (8, -1, -10) and perpendicular to the line
(x, y, z) = (1 + 2k, 2 - k, 3 - 7k)
Find also the point of intersection of two lines.

3.4 Planes

A plane is determined by three points which are not on the same line.
Let P(x, y, z) be a point on the plane through A, B, C. Since A, B, C are not on the same line, line segment joining any two points will intersect each other. Let AB intersect AC at A. Draw a line through P parallel to AC. This line will meet AB at P (x₁ + sl₁, y₁ + sm₁, z₁ + sn₁) for some parameter s.

In the figure, coordinates of any point (x, y, z) on the plane are

x	=	x₁ + sl₁ + tl₂
y	=	y₁ + sm₁ + tm₂
z	=	z₁ + sn₁ + sn₂ for some parameter t.

Let
a = m₁n₂ - m₂n₁
b = n₁l₂ - n₂l₁
c = l₁m₂ - l₂m₁ fig 3.4-2

Then
al₁ + bm₁ + cn₁ = 0
and al₂ + bm₂ + cn₂ = 0
Thus we have the following plane equation:
ax + by + cz = ax₁ + by₁ + cz₁
Let d = ax₁ + by₁ + cz₁, then we have the Cartesian form of the plane equation as

ax + by + cz = d

Since
al₁ + bm₁ + cn₁ = 0
and al₂ + bm₂ + cn₂ = 0
the line l with equation
^{x - x₁} ⁄ _a = ^{y - y₁} ⁄ _b = ^{z - z₁} ⁄ _c
is perpendicular to both of the lines AB and AC. So the line l is perpendicular to the plane ABC. Hence any line with directed values (ka, kb, kc), for some parameter k is perpendicular to the plane ABC.

fig 3.3

Example 8.

Find the equation of the plane containing A(1, 0, 1), B(3, 6, 4), and C(-2, 3, 1).

Solution
Given A(1, 0, 1) and B(3, 6, 4) we have
AB = (l₁, m₁, n₁)
= (2, 6, 3)
Given a(1, 0, 1) and C(-2, 3, 1) we have
AC = (l₂, m₂, n₂)
= (-3, 3, 0)

a	=	m₁n₂ - m₂n₁
	=	6(0) - 3(3)
	=	-9

b	=	n₁l₂ - n₂l₁
	=	3(-3) - 0(2)
	=	-9

c	=	l₁m₂ - l₂m₁
	=	2(3) - (-3)6
	=	6 + 18
	=	24

and

d	=	ax₁ + by₁ + cz₁
	=	-9(1) + (-9)0 + 24(1)
	=	15

Therefore the equation of the plane is
-9x - 9y + 24z = 15

Note that we can use any point on the plane to calculate the value of d.

Example 9.
Find the equation of the line passes through the point (-1, 3, 2) and perpendicular to the plane 3x - 2y - z = 3. Find the point of intersection of the line and the given plane.
solution
Directed values of the line perpendicular to the plane 3x - 2y - z = 3 are (3, -2, -1).
Then the equation of the line is
^{x - (-1)} ⁄ ₃ = ^{y - 3} ⁄ _-2 = ^{z - 2} ⁄ _-1
So coordinates of the points on the line are of the form
fig 3.4-4

x	=	-1 + 3s
y	=	3 + (-2)s
	=	3 - 2s
z	=	2 + (-1)s
	=	2 - s

If one of these point P is on the plane,
3 (-1 + 3s) - 2 (3 - 2s) - (2 - s) = 3
-3 + 9s - 6 + 4s - 2 + s = 3
14s = 14
s = 1
Therefore the point of intersection is (2, 1, 1)

Example 10.
Find the equation of the plane containing the point (-1, 3, 2) and parallel to the plane 3x - 2y - 3z = 2.
Solution
Directed values of the line perpendicular to the plane 3x - 2y - 3z = 2 are (3, -2, -3). This line is also perpendicular to the required plane. So the equation of the required plane is
3x - 2y - 3z = d
The point (-1, 3, 2) is on the plane. So
3(-1) - 2(3) - 3(2) = d
d = -15
The equation of the required plane is
3x - 2y - 3z = -15.
fig 3.2

Exercise 3.4

Find the equation of the plane containing

Find the equation of the line passing through the point (3, -2, -2) and perpendicular to the plane -2x + 3y - z = 4.
Find the point of intersection of the line and the plane.
Find the equation of the plane containing the point (2, 3, -1) and parallel to the plane -2x + y + 3z = 6.

3.5 Spheres

The distance between center (x₁, y₁, z₁) and any point (x, y, z) of a sphere is radius r.
√ (x - x₁)² + (y - y₁)² + (z - z₁)² = r

Therefore the equation of the sphere with center (x₁, y₁, z₁ and radius r is

(x - x₁)² + (y - y₁)² + (z - z₁)² = r²

Example 11.
Find the equation of the plane tangent to the sphere
(x - 2)² + (y - 1)² + (z + 1)² = 14
at the point (3, 4, 1).

Solution
fig 3.5-2

Directed values of the line joining center C(2, 1, -1) of the sphere and the given point P(3, 4, 1) are
CP = (1, 3, 2).
Line CP is perpendicular to the tangent plane. Thus the equation of the plane is
x + 3y + 2z = d.
Since P(3, 4, 1) is on this plane, so we get
3 + 3(4) + 2(1) = d
d = 17
The equation of the plane is
x + 3y + 2z = 17.
Example 12.
Find the equation of the sphere with center (0, 1, 0) and touching the plane x - 2y + 2z + 5 = 0.

Solution
The equation of the line that passes through the center C(0, 1, 0) and perpendicular to the plane x - 2y + 2z + 5 = 0 is
^{x - 0} ⁄ ₁ = ^{y - 1} ⁄ _-2 = ^{z - 0} ⁄ ₂
^x ⁄ ₁ = ^{y - 1} ⁄ _-2 = ^z ⁄ ₂
So coordinates of the points on the line are of the form
x = s
y = 1 + (-2)s = 1 - 2s
z = 2s
If one of these points P is on the plane, then

s - 2(1 - 2s) + 2(2s) + 5	=	0
s - 2 + 4s + 4s + 5	=	0
9s	=	-3
s	=	- ¹⁄₃

Therefore the point of intersection is P(- ¹⁄₃, ⁵⁄₃, ^-2⁄₃). Thus we have the radius
CP = √ (- ¹⁄₃ - 0)² + (⁵⁄₃ - 1)² + (- ²⁄₃ - 0)²
= √ ¹⁄₉ + ⁴⁄₉ + ⁴⁄₉
= 1

Therefore the equation of the sphere is (x² + (y - 1)² + z² = 1.
Example 13.
Find the equation of a sphere that passes through the points (9, 0, 0), (3, 13, 5) and (11, 0, 10), given that its center lies on the yz-plane.
solution
The equation of the sphere with center (x₁, y₁, z₁) and radius r is
(x - x₁)² + (y - y₁)² + (z - z₁)² = r²
Distance from center (x₁, y₁, z₁) to each of the given points is

(9 - 0)² + (0 - y₁)² + (0 - z₁)²	=	r² (in yz-plane, x₁ = 0)
81 + y₁² + z₁²	=	r²	(1)
(3 - 0)² + (13 - y₁)² + (5 - z₁)²	=	r²
9 + (13 - y₁)² + (5 - z₁)²	=	r²	(2)
(11 - 0)² + (0 - y₁)² + (10 - z₁)²	=	r²
121 + y₁² + (10 - z₁)²	=	r²	(3)

From (1) and (3)
81 + y₁² + z₁² = 121 + y₁² + (10 - z₁)²
81 + y₁² + z₁² = 121 + y₁² + 100 - 20z₁ + z₁²
0 = 40 + 100 - 20z₁
z₁ = 7

From (1) and (2),
81 + y₁² + z₁² = 9 + (13 - y₁)² + (5 - z₁)²
81 + y₁² + 7² = 9 + 169 - 26y₁ + y₁² + (5 - 7)²
130 = 182 - 26y₁
y₁ = 2
Therefore x₁ = 0, y₁ = 2 and z₁ = 7 and substitute these values in equation (1), we get r² = 134.
The equation of the sphere is
(x - 0)² + (y - 2)² + (z - 7)² = 134.
Exercise 3.5

Find the equation of sphere with center C and radius r.

Check whether the given point P lies inside, outside or on a sphere.

Find the equation of the sphere on the join of (1, -1, 1) and (-3, 4, 5) as diameter.

Find the equation of the plane tangent to the sphere
(x + 2)² + (y - 1)² + (z + 3)² = 27
at the point (3, 2, -2).

Find the equation of the sphere with center (6, -7, -3) and touching the plane
4x - 2y - z = 17.

What is the equation of the sphere which passes through the points (3, 0, 2), (-1, 1, 1) and (2, -5, 4) and whose center lies on the plane 2x + 3y + 4z = 6 ?