How can I find the projection of the vector (2,2,1) on the plane that has the equation z=x-y?
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.
How do I find the normal vector of the plane?
The plane's equation can be written x - y + z = 0, so a normal vector is (1, -1, 1) (the coefficients of the variables).
so the projection of the vector (2,2,1) onto the plane (z=x-y) is
How did you arrive at that? I got a different answer.
I subtracted the vectors, guess it's more complicated than that?
Yes, it's more complicated than that. You have to go back to Muzza's original response and DO what he said there!
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.
This is quite a puzzle but thanks for helping, I think I've figured this out.
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