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

Question

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)#

4483 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-10T18:14:06

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)#