Let $a$ and $b$ be two unitary vectors in $\mathbb E^3$, and let $Q$ be an orthogonal matrix. Does the following hold?
$Qa \wedge Qb = \pm Q(a \wedge b)$
Let $a$ and $b$ be two unitary vectors in $\mathbb E^3$, and let $Q$ be an orthogonal matrix. Does the following hold?
$Qa \wedge Qb = \pm Q(a \wedge b)$
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.