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Is $Qa \wedge Qb = \pm Q(a \wedge b)$, where $a$ and $b$ are two unitary vectors in $E^3$ and $Q$ is an orthogonal matrix ??? Thanks

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Look near the bottom of this section at Wikipedia. Does that help? –  Dylan Moreland Sep 2 '11 at 17:39
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Does $Q(a\wedge b)$, as opposed to $Q(a\times b)$, have a conventional definition? I'm not sure how the naive meaning for the cross product in 3 dimensions would generalize to exterior products in general. –  Henning Makholm Sep 2 '11 at 18:11
    
@Henning Oh, good point. Also, to me the generalization of $\times$ relates to the Hodge star. –  Dylan Moreland Sep 2 '11 at 18:15

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up vote 2 down vote accepted

I assume you mean the standard 3-dimensional cross product $\times$. Then, yes it holds for an orthogonal matrix. We can see this from the fact, that

$\mathbf{Qa}\times\mathbf{Qb}=(\det\mathbf{Q})\mathbf{Q}^{-T}(\mathbf{a}\times\mathbf{b})$

and the fact, that for an orthogonal matrix

$\mathbf{Q}^{-T}=\mathbf{Q}^{T^T}=\mathbf{Q},\qquad\qquad\qquad \det\mathbf{Q}=\pm 1$

But you can also reason this geometrically, by understanding the cross product of two vectors as the vector orthogonal to both of them. In this context an orthogonal matrix is equivalent to either a rotation or a reflection (depending on the determinant being 1 or -1). So the vector orthogonal to a roatated/reflected pair of vectors is the same as the vector orthogonal to the original pairs and then rotated/reflected, though when reflecting the cross product directly the sign changes (because when reflecting the original vectors we reflect two times).

The vectors needn't even be unitary, I think.

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