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Determinant of vector of AXB for 3-D

  1. Sep 23, 2012 #1


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    Hi all,


    This is a beginning step in proving aXb=|a||b|sin(theta)

    thank you

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  2. jcsd
  3. Sep 23, 2012 #2
    Assuming that expression with a's is the determinant, then yes: it must be the whole thing squared.

  4. Sep 23, 2012 #3


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    I am talking abou this proof:


    Thank You.

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  5. Sep 23, 2012 #4


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    Hey dpa.

    These brackets refer to the norm and the norm of a vector in R^n is simply is the sum of the components squared. [This norm is simply the ||x-0|| for some vector x]

    Also you can't use this to prove what AXB is, but you can prove that for some directional normal vector n^ then you can prove that AXB = nhat*length where length is |AXB|.
  6. Sep 23, 2012 #5


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    Thank You.
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