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Math Help - Proof with cross product

  1. #1
    a.a
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    Proof with cross product

    Proove that the magnitude of vector a cross b is equal to :
    sqrt. ((a dot b) (b dot b) - (a dot b)^2)
    I have no idea how to go about proveing this...:S
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  2. #2
    Eater of Worlds
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    It would appear you mean LaGrange's identity.

    ||a\times{b}||^{2}=||a||^{2}||b||^{2}-(a\cdot{b})^{2}

    ||a\times{b}||^{2}=(a_{2}b_{3}-a_{3}b_{2})^{2}+(a_{3}b_{1}-a_{1}b_{3})^{2}+(a_{1}b_{2}-a_{2}b_{1})^{2}

    ||a||^{2}||b||^{2}-(a\cdot{b})^{2}=(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})(b_  {1}^{2}+b_{2}^{2}+b_{3}^{2})-(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^{2}

    Now, just multiply out the right sides and verify the equality.
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