calculus

show that if a, b & c are in v3, then (a cross b) dot [(b cross c)×(c cross a)] = [a dot (b cross c)]^2

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  1. If 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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