Question

Projection of a point onto a plane?

Imagine that a wire frame cube sits in three dimensional space above a white piece of paper in the plane spanned by (1,2,2) and (2,1,2) and light shines orthagonal to the paper onto the cube casting a shadow onto the paper. The cube is centered at (-2,-2,3) and has corners at (-1,-1,4), (-1,-1,2), (-1,-3,4), (-1,-3,2). Find the coordinates of the shadows of the corners of the cube.

Answer 1

So first of all I have found the other four corners. they are (-3,-3,2) (-3,-3,4) (-3,-1,2) (-3,-1, 4). Now my problem is projecting them onto the column space of the plane.

Answer 2

just put the last coordinate 0 in the triplet (a,b,c) so you have (a,b,0) and you got projection on xy plane. With projections to other planes do the same with corresponding coordinate

Answer 3

Okay, so our first task is to find a projection matrix for the plane. This is a matrix P such that \(P=P^2\). Unfortunately, (1,2,2), (2,1,2) is not an orthonormal basis for that plane, which makes it a little more complicated. Probably the best way to proceed is to use Gram-Schmidt to find an orthonormal basis for that plane.

myko: That would work if this were an xy-plane or similar plane for which two unit vectors formed an orthonormal basis, but this plane is not of that type.

Answer 4

i ddon't read the compleat question, but it doesn't really matter if its ortogonal basis or not, and even less if it's ortonormal. The only change in my previous answer is to make first a coordinate change, so xy plane becomes (1,2,2), (2,1,2)

Answer 5

Yeah, you can do a change of basis for the entire cube. Or you can just use Gram-Schmidt so that the projection matrix is as easy as calculating \(AA^T\). Either method ought to work.

Answer 6

I think my problem here is I don't know how to set up a projection matrix.

Answer 7

the solutions I have do not use graham schmidt I don't think.

Answer 8

Well, this is just the method that seemed to make sense to me. I haven't actually done this type of problem before. There might be another easier method.

Answer 9

it gives me \[\mathbb{P}c_1=\frac{1}{17}\left[\begin{matrix}13 & -4 &6 \\ -4 & 13&6\\6&6&8\end{matrix}\right]\ \left[\begin{matrix}-1\\ -1\\4\end{matrix}\right]\]for the first projection(projection of (-1,-1,-4). I have no idea where that first matrix comes from. assuming it is something divided by a length?

Answer 10

Ugh. I tried my method using Gram-Schmidt and it just got really messy and then didn't work.

Answer 11

Oh, wait, no I guess it did work, because I got that exact same matrix that you have there.

Answer 12

And, crap, I already deleted my whole explanation. Argh!

Basically though, I did gram-schmidt which gave me this basis matrix: \[
\begin{bmatrix}
1/3 & 10/(3 \sqrt{17}) \\ 2/3 & -7/(3 \sqrt{17})\\ 2/3 & 2/(3 \sqrt{17})
\end{bmatrix}
\]Then I just multiplied that matrix by its transpose and it gave me the projection matrix you have there.

Answer 13

After that it's as easy as multiplying the projection matrix by each vector and you're done. The messy part is all just doing gram-schmidt.

Answer 14

Wolfram will do Gram-Schmidt for you: http://www.wolframalpha.com/input/?i=orthogonalize+%7B%7B1%2C2%2C2%7D%2C%7B2%2C1%2C2%7D%7D though it is good exercise to do it yourself.

Answer 15

I'll admit that I'm still confused because that projection matrix does not seem to be idempotent, and they're supposed to be. Weird.

Answer 16

Thanks a lot. I'm still confused because I can't see how this process you described will produce a 3x3 matrix

Answer 17

See Strang's lecture
http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-15-projections-onto-subspaces/

The idea is you have 2 vectors a1 and a2 that serve as a basis for a plane.
A= [a1 a2] and we are interested in the column space of matrix A

We have a point b that we project onto the C(A). Point p in C(A) is the point in the plane closest to b. The key idea is that the error vector e is perpendicular to plane, and
p+e= b