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

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ha ha on 7 Dec 2017
Edited: ha ha on 26 Apr 2018
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. John D'Errico on 7 Dec 2017
Edited: John D'Errico on 7 Dec 2017
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.
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John D'Errico on 15 Dec 2017
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.