# 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

Staff Emeritus
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.