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Vector Rhombus Proof

  1. Dec 3, 2006 #1
    1. The problem statement, all variables and given/known data
    Prove that the diagonals of a paraelogram are perpendicular iff the parallelogram is a rhombus.

    2. Relevant equations

    a (dot) b = 0

    3. The attempt at a solution
    This is how I started:

    By definition, a rhombus is a quadrilateral with all sides equal in length. So this means that if I have two vectors, a and b that form the corner of a rhombus, then that means that the magintude of a and b are equal. By inspection of a diagram of this vector problem, I found that (a+b) (dot) (a-b) = 0 iff the magintude of a and b are equal.

    This is great, however, it will not fly because I cannot just say "by inspection of the diagram" right. How can I put this in words that will make my proof make sense?

    Thanks
     
  2. jcsd
  3. Dec 3, 2006 #2
    The diagonals of the parallelogram are precisely a+b and a-b. if you are talking about proving your equation above, multiply it out, keeping in mind:

    [tex] (a+b)\cdot (a-b) = a\cdot a - b\cdot b + b\cdot a - a\cdot b[/tex]

    edit: fixed my mistake
     
    Last edited: Dec 3, 2006
  4. Dec 3, 2006 #3
    AWESOME!! Thank you so much for you help. That was a lot easier than I thought. So, after multiplying it out, I came up with a²-b²=0. So this is true iff a² = b². Thanks for your help!
     
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