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by addle_brains on March 19th, 2009

addle_brains

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Help answer this question below.

Prove that for any 2 vectors, U and V, we get ||U+V||^2 + ||U-V||^2 = 2(||U||^2 + ||V||^2). I have proven this with simple algebra. But what geometric fact does it prove?

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Answers. 1 helpful answer below.

  • by iwnit on March 19th, 2009

    iwnit

    This can be proven using the law of cosines.

    |U+V|^2 = |U|^2 + |V|^2 +2*|U|*|V|*cosß
    |U-V|^2 = |U|^2 + |V|^2 -2*|U|*|V|*cosß

    Now just add the two to get your equality.

    This means that in a random parallelogram, the sum of the squares of the diagonals is the sum of the squares of the sides. This is the parallelogram law.

    Further information:
    http://en.wikipedia.org/wiki/Law_of_Cosines

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