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Help with a proof.

  1. Oct 3, 2011 #1
    Hi, I am having trouble with this proof, I am wondering what to do because the way I have attempted it is incorrect.

    I want to prove

    ||x-y|| = ||x|| ||y|| ||x bar - y bar||

    where

    x and y are vectors in Rn

    and u bar is defined by u/(||u||^2)


    so the question asks to prove this analytically, I couldn't figure out how to do so, the only attempt I made was mathematically by expanding the right side using the formula for norm and trying to simplify it to the left. Anyway that results in a mess and I was wondering where to start.
     
  2. jcsd
  3. Oct 3, 2011 #2

    Ben Niehoff

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    I would draw a picture and do this geometrically. Draw two rays emanating from the origin; these rays are the directions of x and y. Choose a point on each ray to represent each vector. Then ||x-y|| is the length of the line segment that goes between these two points. Can you think of another line segment you can draw between the two rays that obviously has the same length as ||x-y||?
     
  4. Oct 3, 2011 #3
    thanks for the tip. I've started doing it geometrically but I'm still stuck on the right side of the proof and how it equates to the left. The only other ray with the same length should be lly-xll = llx-yll, but I'm still unsure how to end up there.
     
  5. Oct 3, 2011 #4

    Ben Niehoff

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    You should have a triangle OXY, where O is the origin and X and Y are the endpoints of the vectors x and y. Then ||x-y|| is the length of XY, i.e. the side opposite the angle O.

    Now take this picture, and reflect it about the line that bisects angle O. Overlay this on top of your original picture. How do you express the new legs OX' and OY' in terms of the original OX and OY? Then consider that it should be obvious that the legs X'Y' and XY have the same length.
     
  6. Oct 3, 2011 #5
    For intuition, drawing a picture is fine, but for proof, I would use algebraic approach. This formula solves it all:

    [tex]
    \|A + B\|^2 = \|A\|^2 + \|B\|^2 + 2A\cdot B
    [/tex]
     
  7. Oct 3, 2011 #6

    Deveno

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