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- Thread starter dpa
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- #2

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Hi all,

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

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- #4

chiro

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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

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Thank You.

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