Chapter 3

Analytical Solid Geometry

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
al₂ + bm₂ + cn₂ = 0
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 = 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 = (l₁, m₁, n₁)
= (-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.

Exercise 3.4
Answers
1. Find the equation of the plane containing
(a) A(2, -5, 4), B(-5, 2, 4) and C(-2, 3, -1)
Solution
Given A(2, -5, 4) and B(-5, 2, 4), we have

AB	=	(l₁, m₁, n₁)
	=	(x₂ - x₁, y₂ - y₁, z₂ - z₁)
	=	(-5 - 2, 2 - (-5), 4 - 4)
	=	(-7, 7, 0)

Given A(2, -5, 4) and C(-2, 3, -1), we have

AC	=	(l₂, m₂, n₂)
	=	(x₂ - x₁, y₂ - y₁, z₂ - z₁)
	=	(-2 - 2, 3 - (-5), -1 - 4)
	=	(-4, 8, -5)

a	=	m₁n₂ - m₂n₁
	=	7(-5) - 8(0)
	=	-35

b	=	n₁l₂ - n₂l₁
	=	0(-4) - (-5)(-7)
	=	-35

c	=	l₁m₂ - l₂m₁
	=	-7(8) - (-4)7
	=	-56 + 28
	=	-28

and

d	=	ax₁ + by₁ + cz₁
	=	-35(2) + (-35)(-5) + (-28)(4)
	=	-70 + 175 - 112
	=	-7

Therefore the equation of the plane is
-35x - 35y - 28z = -7

(b) A(4, 2, -3), B(1, -2, 4) and C(-1, 0, 3)

Solution
Given A(4, 2, -3) and B(1, -2, 4), we have

AB	=	(l₁, m₁, n₁)
	=	(x₂ - x₁, y₂ - y₁, z₂ - z₁)
	=	(1 - 4, -2 - 2, 4 - (-3))
	=	(-3, -4, 7)

Given A(4, 2, -3) and C(-1, 0, 3), we have

AC	=	(l₂, m₂, n₂)
	=	(x₂ - x₁, y₂ - y₁, z₂ - z₁)
	=	(-1 - 4, 0 - 2, 3 - (-3))
	=	(-5, -2, 6)

a	=	m₁n₂ - m₂n₁
	=	-4(6) - (-2)7
	=	-24 + 14
	=	-10

b	=	n₁l₂ - n₂l₁
	=	7(-5) - 6(-3)
	=	-35 + 18
	=	-17

c	=	l₁m₂ - l₂m₁
	=	-3(-2) - (-5)(-4)
	=	6 - 20
	=	-14

and

d	=	ax₁ + by₁ + cz₁
	=	-10(4) + (-17)2 + (-14)(-3)
	=	-40 - 34 + 42
	=	-32

Therefore the equation of the plane is
-10x - 17y - 14z = -32

2. 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.

solution
Directed values of the line perpendicular to the plane -2x + 3y - z = 4 are (-2, 3, -1).
Then the equation of the line is
^{x - x₁} ⁄ _a = ^{y - y₁} ⁄ _b = ^{z - z₁} ⁄ _c
^{x - 3} ⁄ _-2 = ^{y - (-2)} ⁄ ₃ = ^{z - (-2)} ⁄ _-1
^{x - 3} ⁄ _-2 = ^{y + 2} ⁄ ₃ = ^{z + 2} ⁄ _-1
^{x - 3} ⁄ _-2 = ^{y + 2} ⁄ ₃ = ^{z + 2} ⁄ _-1 = s

^{x - 3} ⁄ _-2 = s	and	^{y + 2} ⁄ ₃ = s	and	^{z + 2} ⁄ _-1 = s
x - 3 = -2s,		y + 2 = 3s,		z + 2 = -1s

So coordinates of the points on the line are of the form

x	=	3 - 2s
y	=	-2 + 3s
z	=	-2 - s

If one of these point P is on the plane -2x + 3y - z = 4,
-2 (3 - 2s) + 3 (-2 + 3s) - (-2 - s) = 4
-6 + 4s - 6 + 9s + 2 + s = 4
-10 + 14s = 4
14s = 14
s = 1

∴ x = 3 - 2s = 3 - 2(1) = 1
y = -2 + 3s = -2 + 3(1) = 1
z = -2 - s = -2 - 1 = -3
Therefore the point of intersection is (1, 1, -3)

fig 3.4-6

3. Find the equation of the plane containing the point (2, 3, -1) and parallel to the plane -2x + y + 3z = 6.
Solution
Directed values of the line perpendicular to the plane -2x + y + 3z = 6 are (-2, 1, 3).
This line is also perpendicular to the required plane. So the equation of the required plane is
-2x + y + 3z = d
The point (2, 3, -1) is on the plane. So
-2(2) + 3 + 3(-1) = d
-4 + 3 - 3 = d
d = -4
The equation of the required plane is
-2x + y + 3z = -4.
fig 3.4-7