This is a beginning step in proving aXb=|a||b|sin(theta)
Assuming that expression with a's is the determinant, then yes: it must be the whole thing squared.
I am talking abou this proof:
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|.
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