cross product and RHR Edit: I really like the approach taken by $\mathbb{R}^n$ in the comments below. I posted this late last night and didn't get to ask him why $\bf R(a \times b) = Ra \times Rb$ for any rotation matrix $\bf R$. I'm not sure how to prove that: along with trying to evaluate the determinant directly, I looked in a linear algebra book and tried using the "norm-preserving" property of rotation matrices to evaluate $|\bf R(a \times b) - R(a) - R(b)|$ $^2$, but didn't succeed. How do you do this?

In one of my courses, a professor briefly summarized properties of the cross product at the start of last class. I realized that the assertion
$$ \bf a \times (b + c) = a\times b + a\times c$$
was actually surprising to me. Clearly this is fundamental (if you don't accept this, you can't derive a way of computing the cross product), but the proof is a bit slippery.
If you start from $\bf{ a \times  b}$ $:= \bf \hat n |a||b|$ $ \sin\theta$, where $ \bf \hat n$ comes from the right hand rule, then if you can prove as a lemma $\bf a \cdot (b \times c) = (a \times b) \cdot c$, then there's a neat little proof which I found here. But the only argument I've seen for the needed lemma is about the volume of a parallelepiped, which only convinces me that $\bf | a \cdot (b \times c)| = |(a \times b) \cdot c|$. 
I think I prefer the approach in one of my textbooks, which starts by defining the cross product by determinants - so, distributivity holds - and proves most of the needed properties. But it wusses out at a crucial point: "it can be proven that the orthogonal vector obtained from this matrix obeys the Right Hand Rule". 
Could somebody either prove that lemma, or the textbook claim? (Preferably the latter.)
 A: Typically, distributivity is assumed (more broadly, we assume the product is linear in each of its arguments, or that it is bilinear).  Without this, even if one defines the action of the cross product on basis vectors, one would be hard-pressed to talk about its action on arbitrary vectors, as these are built up linearly from basis elements.
In short, one uses bilinearity to build up other results, not vice versa.
Furthermore, note that if $\hat n$ comes from a "left-hand rule" instead, none of the logic is really affected.  As long as the cross product is internally consistent, there's no issue.  The determinant approach accomplishes this by imposing an orientation, but this fundamentally gets one of the punchlines of linear algebra backwards.  Using determinants to talk about cross products is backwards.  One should instead use cross products to infer things about determinants.
Or better yet, not use cross products at all, but that's a whole 'nother argument.  Using wedge products, for instance, the equivalent of $a \cdot (b \times c) = (a \times b) \cdot c$ is $(a \wedge b) \wedge c = a \wedge (b \wedge c)$, which is trivially true because wedge products are associative, and there's nothing to prove.
