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February 27, 2015

February 27, 2015

Posted by **matt** on Wednesday, September 14, 2011 at 12:53pm.

- calculus -
**Steve**, Thursday, September 22, 2011 at 10:27amIf you look for some list of identities for vector products, such as wolfram: cross product you will find the following:

A.(BxC) = det(ABC)

(AxB)x(CxD) = det(ABD)C - det(ABC)D

If we adapt this formula to our problem, we need to set

A = b

B = c

C = c

D = a

Now, we plug and chug:

bxc x cxa = det(bca)c - det(bcc)a

Now, a determinant is zero if two rows are the same. So, we can toss out he last term on the right, and we are left with just

bxc x cxa = det(bca)c

AxB.C = C.AxB so,

axb . det(bca)c = det(bca)c . axb

= det(bca) * det(cab)

When we swap columns, the detrminant changes sign. Swapping columns twice, we get

det(bca) * det(cab) = det(abc) * det(abc) = (a.bxc)^2

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