1. Sep 23, 2012 #1

   dpa

   

   Hi all,
   
   
   
   This is a beginning step in proving aXb=|a||b|sin(theta)
   
   thank you

    

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

   DonAntonio

   

   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

   

   I am talking abou this proof:
   
   
   
   Thank You.

    

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     Screenshot from 2012-09-23 15:01:33.png

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

   chiro

   

   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 = n^hat*length where length is |AXB|.

    

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

   dpa

   

   Thank You.

    

   dpa, Sep 23, 2012

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