Determinant of vector of AXB for 3-D

  • Thread starter dpa
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  • #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
 

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  • #2
606
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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
Assuming that expression with a's is the determinant, then yes: it must be the whole thing squared.

DonAntonio
 
  • #3
dpa
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I am talking abou this proof:

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


Thank You.
 

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  • #4
chiro
Science Advisor
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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|.
 
  • #5
dpa
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Thank You.
 

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