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

In summary, the conversation discusses finding the orthogonal projection of a given point onto a plane defined by three other points. The method involves defining the equation of the plane using vectors and finding the line perpendicular to the plane through the given point. This exercise is standard in undergraduate algebra courses and can be found in textbooks. The importance of writing this up formally is also emphasized as it can serve as good practice for writing academic papers.
  • #1
Pejeu
27
1
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.

Thanks for any advice.
 
Last edited:
Mathematics news on Phys.org
  • #2
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
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:

docs.google

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:
  • #4
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 [itex](x_0, y_0, z_0)[/itex], [itex](x_1, y_1, z_1)[/itex], and [itex](x_2, y_2, z_2)[/itex] then two vectors in the plane are [itex]\vec{A}= (x_1- x_0)\vec{i}+ (y_1- y_0)\vec{j}+ (z_1- z_0)\vec{k}[/itex] and [itex]\vec{B}= (x_2- x_0)\vec{i}+ (y_2- y_0)\vec{j}+ (z_2- z_0)\vec{k}[/itex]. The cross product of those two vectors, [itex]A\times B= C_x\vec{i}+ C_y\vec{j}+ C_z\vec{k}[/itex] is perpendicular to the plane and the equation of the plane is [itex]C_x(x- x_0)+ C_y(y- y_0)+ C_z(z- z_0)= 0[/itex]. If (a, b, c) is some fourth point, then the line [itex]x= a+ C_xt[/itex], [itex]y= b+ C_yt[/itex], [itex]z= c+ C_xt[/itex] 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 [itex]C_x(C_xt+ a- x_0)+ C_y(C_yt+ b- y_0)+ C_z(C_zt+ c- z_0)= 0[/itex] can be solved for t and then (x, y, z) can be calculated.
 
  • #5
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 textbooks, not arxiv et al.
HallsofIvy has illustrated the basic method.

Try search terms like "find the closest point on a plane"
 
  • #6
Simon Bridge said:
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 textbooks, not arxiv et al.
HallsofIvy has illustrated the basic method.

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

I did it in high school...
 
  • #7
OTOH: learning to write something like this up formally is good exercise for writing actual papers.
 

1. What is the projection of a point on a plane?

The projection of a point on a plane is the closest point on the plane to the given point. It is the intersection of a line drawn from the given point perpendicular to the plane.

2. How is the projection of a point on a plane calculated?

The projection of a point on a plane can be calculated using the dot product and cross product of vectors. The dot product is used to find the component of the vector that is parallel to the plane, and the cross product is used to find the component of the vector that is perpendicular to the plane. These components are then added to the original point to find the projection point on the plane.

3. Can the projection of a point on a plane be outside the defined plane?

No, the projection of a point on a plane will always lie on the defined plane. This is because the projection is the closest point on the plane to the given point, and any point outside the plane will be farther away.

4. What are the applications of projecting a point on a plane?

The projection of a point on a plane has various applications in fields such as computer graphics, engineering, and mathematics. It is used to determine the shortest distance between a point and a plane, to find the closest point on a plane for positioning objects, and to calculate reflections and shadows in computer graphics.

5. Can the projection of a point on a plane be negative or zero?

Yes, the projection of a point on a plane can be negative or zero. This can happen if the given point is located on the opposite side of the plane from the normal vector. In this case, the projection will be in the opposite direction of the normal vector, resulting in a negative value. Zero projection occurs when the given point lies on the plane itself, resulting in the projection point being the same as the given point.

Similar threads

Replies
4
Views
492
Replies
7
Views
3K
Replies
5
Views
2K
  • Topology and Analysis
2
Replies
38
Views
4K
  • Mechanical Engineering
Replies
11
Views
1K
Replies
2
Views
8K
Replies
1
Views
1K
  • General Math
Replies
1
Views
4K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
Back
Top