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Determinant of vector of AXB for 3-D

  1. Sep 23, 2012 #1

    dpa

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    Hi all,

    attachment.php?attachmentid=51137&stc=1&d=1348399197.png

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

    thank you
     

    Attached Files:

    dpa, Sep 23, 2012
  2. jcsd
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  3. Sep 23, 2012 #2

    DonAntonio

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    Assuming that expression with a's is the determinant, then yes: it must be the whole thing squared.

    DonAntonio
     
    DonAntonio, Sep 23, 2012
  4. Sep 23, 2012 #3

    dpa

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    I am talking abou this proof:

    attachment.php?attachmentid=51141&stc=1&d=1348405375.png

    Thank You.
     

    Attached Files:

    dpa, Sep 23, 2012
  5. Sep 23, 2012 #4

    chiro

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    Science Advisor

    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|.
     
    chiro, Sep 23, 2012
  6. Sep 23, 2012 #5

    dpa

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    Thank You.
     
    dpa, Sep 23, 2012
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