Prove a formula for the distance between a plane determined by three points and a point

by RoRi

May 30, 2016

If a plane is determined by the points $A,B,C$ prove that the distance from a point $Q$ to this plane is given by
$\frac{|(Q-A) \cdot (B-A) \times (C-A)|}{\lVert (B-A) \times (C-A) \rVert}.$
Using part (a) compute the distance in the case
$Q = (1,0,0), \quad A = (0,1,1), \quad B = (1,-1,1), \quad C = (2,3,4).$

Proof. We know the distance from a plane containing a point $P$ to a point $Q$ not on the plane is given by the formula
$\frac{(Q-P) \cdot N}{\lVert N \rVert}.$

Since the plane through $A,B,C$ is the set of points

$M = \{ A + s(B-A) + t(C-A) \}$

we have $N = (B-A) \times (C-A)$ . Thus, the distance from $Q$ to $M$ is

$d = \frac{|(Q-A) \cdot (B-A) \times (C-A)|}{\lVert (B-A) \times (C-A) \rVert}.$
For the given points $Q,A,B,C$ we have
$d = \frac{|((1,0,0) - (0,1,1)) \cdot ((1,-1,1) - (0,1,1)) \times ((2,3,4) - (0,1,1))|}{\lVert ((1,-1,1) - (0,1,1)) \times ((2,3,4) - (0,1,1)) \rVert} = \frac{|-6+3-6|}{9} = 1.$