Minimum Distance From A Point To A Plane Tutorial

by bkenwright@xbdev.net

Both methods generate the same solution, but its how you think about solving it....one method uses a parametric line view of it, the other things more in terms of planes and the plane equation.

...Using Method 1....

o Step 1.

Lets make some assumptions first, n is our plane normal, and its normalised.  So |n|=1. 

    P1 to P0 = P1P0 = P0-P1

And P1 is a point on our plane.

d is the shortest distance from our point P0 to the plane

o Step 2.

This gives us:

    d = n dot |P0 - P1|

o Step 3.

Now d is a distance, its a value, not a vector.  So we still need to determine the point Px on the plane.  We use our n direction value to know which way where going, and combine it with an amplitude so we know how far to go.  (Not forgetting P0 as our starting point)

    Px = P0 + d*n

where:

    d = n dot |P0-P1|

....Method 2....

For this second method, we sort of work with a non-normalised plane vector and the plane equation.

As before, our point P0, and we want to find the closest distance to the plane.  Our plane is defined by P1, a ponit on the plane (x,y,z) and the planes normal n.

Now for a plane, any point on the plane must satisfy the plane equation, which is:

A x + B y + C z + D = 0

The minimum distance is then the absolute value of:

   ( A x0 + B y0 + C z0 + D ) / sqrt( A² + B² + C² )

But ackk...what is that D value?

D is the distance from the plane to the origin.

    minimum distance = d = (A (x0 - x1) + B (y0 - y1) + C (z0 - z1)) / sqrt(A² + B² + C²)

                                      = ( A x0 + B y0 + C z0 + D ) / sqrt(A²+B²+C²)

                                      = [  (n dot P0) - (n dot P1) ] / sqrt(A²+B²+C²)

                                      = n dot (P0 - P1) /  sqrt(A²+B²+C²)

We don't have a normalised normal n for the second method example, which is why we divide by |n|.