Question

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)`

Answer 1

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)`