# Projection of a point on the plane defined by 3 other points.

1. Aug 8, 2013

### Pejeu

Is there already a solution to this available somewhere?

Am I not googling it right?

I did come upon solutions for this with a point and plane defined by way of vectors and normals but not points.

Have I stumbled upon a blank I can fill? Is there room here for a math paper?

My approach is trilateration and averaging of the symmetric (with respect to the plane) solutions. Then simplifying the result.

Naturally, I need only calculate the solution for one coordinate axis. The rest will be analogues.

Last edited: Aug 8, 2013
2. Aug 9, 2013

### Simon Bridge

Welcome to PF;
You are correct - this is something that is pretty routine.
The three points define a plane - then you define what you mean by "projection of a point" and perform the associated transformation.

3. Aug 9, 2013

### Pejeu

Thank you.

I think a paper is in order, though.

Yes, I know the 3 points define the plane. I want to compute the orthogonal projection of the given 4th point on the plane defined by the other 3 given points.

Knowing all 4 points I can compute the respective distances from the one to the 3.

Then I can compute by trilateration the 2 points (of which one is the very point I need to project and the other is its symmetric with respect to the plane defined by the other 3 given points) that satisfy those 3 distances in space from their afferent reference points (an intersection of 3 spheres).

Except I don't actually do that but instead average the solutions together before I compute them as I'm interested in their mean, not their actual selves. Their mean will give me the mid-way point, which lies smack-dab on the plane.

Then I simplify the result. It should be something analogous to this:

I just don't want to be a dunce and go to the trouble of writing a paper on a subject one has already been written about. But I just can't find anything resembling what I'm after. I searched on arxiv too.

And I have to do this by hand, MathCAD and Mathematica couldn't solve the system for me.

They fail by memory overrunning.

I also tried a couple of free online alternatives that failed the same way.

Last edited: Aug 9, 2013
4. Aug 9, 2013

### HallsofIvy

Given three points, the equation of the plane they determine can be found by looking at the vectors from one point to the other two. That is, if the three points are $(x_0, y_0, z_0)$, $(x_1, y_1, z_1)$, and $(x_2, y_2, z_2)$ then two vectors in the plane are $\vec{A}= (x_1- x_0)\vec{i}+ (y_1- y_0)\vec{j}+ (z_1- z_0)\vec{k}$ and $\vec{B}= (x_2- x_0)\vec{i}+ (y_2- y_0)\vec{j}+ (z_2- z_0)\vec{k}$. The cross product of those two vectors, $A\times B= C_x\vec{i}+ C_y\vec{j}+ C_z\vec{k}$ is perpendicular to the plane and the equation of the plane is $C_x(x- x_0)+ C_y(y- y_0)+ C_z(z- z_0)= 0$. If (a, b, c) is some fourth point, then the line $x= a+ C_xt$, $y= b+ C_yt$, $z= c+ C_xt$ is the line perpendicular to the plane through (a, b, c). Replacing x, y, and z in the equation of the plane by those gives $C_x(C_xt+ a- x_0)+ C_y(C_yt+ b- y_0)+ C_z(C_zt+ c- z_0)= 0$ can be solved for t and then (x, y, z) can be calculated.

5. Aug 9, 2013

### Simon Bridge

The exercises you describe are standard in undergraduate algebra courses - so you won't find them in recent academic journals. You should have been looking in text books, not arxiv et al.
HallsofIvy has illustrated the basic method.

Try search terms like "find the closest point on a plane"

6. Aug 9, 2013

### Enigman

I did it in high school.....

7. Aug 9, 2013

### Simon Bridge

OTOH: learning to write something like this up formally is good exercise for writing actual papers.