Proof of the shortest distance from a point to a plane formula?

How can you prove that, for a plane with equation ax+bx+cz=d and a point (alpha, beta, gamma) the shortest distance from the point to the plane is given by

(aalpha+b beta+cgamma-d)/sqrt(a^2+b^2+c^2)

1 Answer
Feb 10, 2018

Quite easy.

Given the plane

Pi-> << p, vec n >> = d

with

p = (x,y,z)
vec n = (a,b,c)

and the point

p_0 = (alpha, beta, gamma) the line

L-> p = p_0 + lambda vec n

is normal by construction to Pi and passes by p_0

The intersection point p_1 = Pi nn L is obtained by solving

<< p_0+lambda vec n, vec n >> = d for lambda giving

lambda = (d - << p_0, vec n >>)/norm(vec n)^2 and the intersection point is

p_1 = p_0 + ((d - << p_0, vec n >>)/norm(vec n)^2) vec n

and now the sought distance is given by

norm(p_1-p_0) = abs(d- << p_0, vec n >>)/norm(vec n) or

norm(p_1-p_0) = abs(alpha a+beta b+gamma c-d)/sqrt(a^2+b^2+c^2)