Page 215 - A Course in Linear Algebra with Applications

7.1: Scalar Products in Euclidean Space 199

We are now in a position to calculate the shortest distance
I from the point (XQ, yo, z 0) to the plane. Let (x, y, z) be a,
point in the plane, and write

x \ x 0
X=\ y\ and Y = \ y 0
z I \z 0

Then I is simply the scalar projection of XQ — X on N, as may
be seen from the diagram below:

(*> y. z)

Therefore
T
\(X -X) N\
0
IliVll
Now

T
(X 0 - X) N = a(x 0 -x) + b(y Q - y) + c(z 0 - z)
= ax 0 + by 0 + cz 0 - d :

for ax + by + cz = d since the point (x, y, z) lies in the plane.
Thus we arrive at the formula

\ax 0 + fa/o + cz 0 - d\
/ = 2 2 2
Va + b + c