Projection of the vector (2,2,1) on the plane

1. Oct 16, 2004

gunnar

How can I find the projection of the vector (2,2,1) on the plane that has the equation z=x-y?

Last edited: Oct 17, 2004
2. Oct 17, 2004

Muzza

First project it on the plane's normal vector (call the resulting vector x), then the vector you're searching for is (2, 2, 1) - x.

3. Oct 17, 2004

gunnar

How do I find the normal vector of the plane?

4. Oct 17, 2004

Muzza

The plane's equation can be written x - y + z = 0, so a normal vector is (1, -1, 1) (the coefficients of the variables).

5. Oct 17, 2004

gunnar

so the projection of the vector (2,2,1) onto the plane (z=x-y) is
(1,3,0)?

6. Oct 17, 2004

Muzza

How did you arrive at that? I got a different answer.

7. Oct 17, 2004

gunnar

I subtracted the vectors, guess it's more complicated than that?

8. Oct 17, 2004

HallsofIvy

Yes, it's more complicated than that. You have to go back to Muzza's original response and DO what he said there!

9. Oct 17, 2004

mathwonk

to project v onto w != 0, find c such that v-cw is perpendicular to w, i.e. solve the equation w.(v-cw) = 0. i.e. solve w.v = c w.w, i.e. let c = w.v/w.w. then the projection is cw, I think, but i am geting confused looking at my fingers and doing this in my head.

anyway you get v = cw + (v-cw), where the first term is parallel to w and the second term is perpendicular to w, and so surely that does it.

10. Oct 18, 2004

gunnar

This is quite a puzzle but thanks for helping, I think I've figured this out.