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

  1. Oct 16, 2004 #1
    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. jcsd
  3. Oct 17, 2004 #2
    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.
  4. Oct 17, 2004 #3
    How do I find the normal vector of the plane?
  5. Oct 17, 2004 #4
    The plane's equation can be written x - y + z = 0, so a normal vector is (1, -1, 1) (the coefficients of the variables).
  6. Oct 17, 2004 #5
    so the projection of the vector (2,2,1) onto the plane (z=x-y) is
  7. Oct 17, 2004 #6
    How did you arrive at that? I got a different answer.
  8. Oct 17, 2004 #7
    I subtracted the vectors, guess it's more complicated than that?
  9. Oct 17, 2004 #8


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    Yes, it's more complicated than that. You have to go back to Muzza's original response and DO what he said there!
  10. Oct 17, 2004 #9


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    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.
  11. Oct 18, 2004 #10
    This is quite a puzzle but thanks for helping, I think I've figured this out.
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