distance from point to plane (plane was created from 3d point data)

Let's say : a plane N contain a set of 3d coordinate data (A).

A=[x1 y1 z1
   x2 y2 z2
   x3 y3 z3
   ........]

plane N is determined by normal vector (n) and center coordinate (P0).

I have a point X(x0,y0,z0) which is located outside plane N. How can I find perpendicular distance from point X to plane N?

Note: I know how to find the normal vector(n) and P0 from the set of data A.

Answer 1

[John D'Errico]

You need to understand what the equation of a plane tells you.

A plane is defined by a point on the plane (P0), and the normal vector to the plane(N). Thus any point on the plane X satisfies the constraint

dot(X-P0,N) = 0

If the normal vector has unit length, so it is normalized to have norm(N)==1, then the solution to your problem is trivial.

The distance to the plane is then simple. It is just:

dot(X-P0,N)

If a point lies on the plane, then the distance to the plane is 0. And that is embodied in the equation of a plane that I gave above!

Finally, you might recognize that the above dot product is simply computed using the function dot, but even more simply written as a matrix multiply, if you have more than one point for which you need to compute this distance.

[ha ha]

If we have a set of point

X=[x0 y0 z0
   x1 y1 z1
   x2 y2 z2]

----->dist=abs((X-P0)*n);

[John D'Errico]

Yes ... and that is the dot product that I referred to. As I said in my last paragraph, you can use just a matrix multiply. Abs handles the issue that a point can be on one side of the plane or the other. So the sign tells you where the point is relative to the normal vector.

I wrote it using the dot function to make the idea of a dot product clear.

[Joerg Fricke-Schmidt]

@John D'Errico: How exactly do I have to multiply a matrix with multiple points with P0 and N of the surface? I would like to have distances calculated for multiple points gathered in may X, Y, Z matrix.

Thanks a lot!