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Since n !
0n ! 1A, B, C2, P 0 P 1 1x1 x 0 , y 1 y 0 , z 1 z 0 2
0 A 2 B 2 C 2
d @PR ! @ 1A, B, C2 1x 1 x 0 , y 1 y 0 , z 1 z 0 2
A 2 B 2 C 2
or
and
d Ax 1 Ax 0 By 1 By 0 Cz 1 Cz 0
A 2 B 2 C 2
Since P 1 1x 1 , y 1 , z 1 2 is a point on Ax By Cz D 0,
Ax 1 By 1 Cz 1 D 0 and Ax 1 By 1 Cz 1 D.
Rearranging the formula,
d Ax 0 By 0 Cz 0 Ax 1 By 1 Cz 1
A 2 B 2 C 2
Therefore,
d Ax 0 By 0 Cz 0 D
A 2 B 2 C 2
d 1Ax 0 By 0 Cz 0 D2
A 2 B 2 C 2
Since the distance d is always positive, the formula is written as
d 0Ax 0 By 0 Cz 0 D 0
A 2 B 2 C 2
Distance from a Point P 0 ( x 0 , y 0 , z 0 ) to the Plane with Equation
Ax By Cz D 0
In R 3 , d 0Ax 0 By 0 Cz 0 D 0 , where d is the required distance between the
A 2 B 2 C 2
point and the plane.
EXAMPLE 1
Calculating the distance from a point to a plane
Determine the distance from S11, 2, 42 to the plane with equation
8x 4y 8z 3 0.
Solution
To determine the required distance, we substitute directly into the formula.
Therefore, d
08112 4122 8142 3 0
045 0
8 2 142 2 8 2 12
45
12 3.75
The distance between S11, 2, 42 and the given plane is 3.75.
NEL
CHAPTER 9 543