# Determinant of vector of AXB for 3-D

dpa
Hi all,

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

thank you

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• Screenshot from 2012-09-23 13:19:16.png
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DonAntonio
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

dpa
I am talking abou this proof:

Thank You.

#### Attachments

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

dpa
Thank You.