Prove $R(a \times b) = Ra \times Rb$ given $R \in \mathcal{SO}(3)$ and $a,b \in \mathbb{R}^3$. Here, $R$ is a proper rotation matrix, and $(\times)$ is the cross-product.
I already found 3 ways of proving this, all have problems, and I am requesting an elegant approach.
(1) Direct calculation, $R = R_{\mathbf{u},\theta}$ has an explicit expression, where $\mathbf{u}$ is the vector about which the rotation of $\theta$ is carried out. It is essentially possible to calculate both sides and compare. Inelegant.
(2) Using antisymmetric matrices: $Ra \times Rb=S(Ra)Rb=RS(a)R^\top Rb=RS(a)b=R(a\times b)$.
 My issue with this is that the equality I am trying to prove, is used to prove $RS(a)R^\top=S(Ra)$. And so using this feels circular.
(3) $Ra \times Rb=\|Ra\|\|Rb\|\sin(Ra,Rb)\mathbb{\hat{u}_1}$ and $a \times b=\|a\|\|b\|\sin(a,b)\mathbb{\hat{u}_2}$.

Here, essentially $\|Ra\|$ should equal $\|a\|$ since $R$ only affects orientation.

Because the relative orientation does not change, the $\sin(\cdot)$ term should be equal.

Likewise, $\mathbb{\hat{u}_1}$ and $\mathbb{\hat{u}_2}$ intuitively I know are equal but I am having a hard time expressing it.

Lastly, I have no idea how to bridge $(a \times b)$ to $R(a \times b)$.

I intuitively see it, and perhaps $\det R = 1$ might be useful, but I feel it is hard to write.
Please, a fourth approach is welcome, and insight is always appreciated.
 A: Recall that the cross product $a\times b$ is characterized by the property that 
$$
\det(x,a,b)=\langle x,a\times b\rangle, \qquad \forall x\in\mathbb{R}^3.
$$
Now let $R\in\mathcal{SO}(3)$. Then by using the fact that $R^{\mathsf{T}} = R^{-1}$, we get
$$
\langle x, R(a \times b) \rangle
= \langle R^{\mathsf{T}}x, a \times b \rangle
= \langle R^{-1}x, a \times b \rangle
= \det(R^{-1}x, a, b).
$$
Then, utilizing the assumption $\det(R) = 1$,
$$
= \det(R) \det(R^{-1}x, a, b)
= \det(x, Ra, Rb)
= \langle x, Ra \times Rb \rangle.
$$
Finally, since $\langle x, R(a \times b) \rangle = \langle x, Ra \times Rb \rangle$ holds for any $x\in\mathbb{R}^3$, the desired identity follows.

Addendum. A similar argument shows that, for any invertible $3\times 3$ real matrix $T$,
$$ T(a \times b) = \frac{1}{\det T}(T T^{\mathsf{T}})( Ta \times Tb). $$
A: The simplest approach I see is the following,
$$\langle R(a\times b), Ra\rangle = (Ra)^TR(a\times b)$$
$$=a^T(a\times b)=\langle a\times b, a\rangle =0$$
Similarly, 
$$\langle R(a\times b), Rb\rangle = 0$$
Hence, it follows that 
$$R(a\times b)= k( Ra\times Rb)$$
for some non-zero real number $k$. Now, observe that,
$$\frac{1}{k}||R(a\times b)||^2$$
$$\langle R(a\times b), Ra\times Rb \rangle = det (R(a\times b), Ra,Rb)$$
$$=(det R)(det(a\times b,a,b))$$
$$=||a\times b||^2$$
$$=||R(a\times b)||^2$$
Required result follows.
A: I don't know whether this is really different from what you had in mind, but I would explain this as follows. It has a bit of Lie algebra air to it.
Let $R_{\vec{a},t}$ be the right-handed rotation about the axis $\vec{a}$ by the angle $t|\vec{a}|$. These form a 1-parameter group under composition, basically just add the angles of rotation as the axis is constant. The Lie algebra trick is to study the derivatives of such 1-parameter groups at $t=0$. The reason why cross products often appear when studying rotations in $\Bbb{R}^3$ is the following.

For all vectors $\vec{b}\in\Bbb{R}^3$ we have
  $$\frac d{dt}R_{\vec{a},t}\vec{b}\big\vert_{t=0}=\vec{a}\times\vec{b}.$$ In other words, cross product by $\vec{a}$ is an infinitesimal generator for the group of rotations about the axis $\vec{a}$.

The other fact I would use is the observation that

For all rotations $R$ we have the conjugation relation
  $$ R\circ R_{\vec{a},t}\circ R^{-1}=R_{R\vec{a},t}.$$ 
  In other words, to rotate a vector $\vec{b}$ about the vector $R\vec{a}$ we can first apply $R^{-1}$, to move the scene to be about rotations about $\vec{a}$, then rotate about $\vec{a}$ by the prescribed angle, and then as the last step rotate the axis back to $R\vec{a}$.

We have the obvious identity
$$
(R\circ R_{\vec{a},t}\circ R^{-1})R\vec{b}=R(R_{\vec{a},t}\vec{b}).\qquad(*)
$$
And your claim follows from equating the derivatives of both sides of $(*)$ at $t=0$.

Don't know whether this was buried in one of your proofs, or whether you see too much handwavium in it. 
A: Here's a component-wise proof. Note that I'm using the Einstein convention: If an index appears twice in a product, it is summed over. For example
$$(Rv)_i = R_{ij}v_j$$
actually means
$$(Rv)_i = \sum_{j=1}^3 R_{ij} v_j$$
because $j$ appears twice in $R_{ij}v_j$.
Also in case you don't know those symbols,
$$\delta_{ij} = \begin{cases} 1 & i=j\\ 0 & \text{otherwise} \end{cases}$$
and
$$\epsilon_{ijk} = \begin{cases}
  1 & ijk \in \{123,231,312\}\\
 -1 & ijk \in \{132,213,321\}\\
  0 & \text{otherwise}
\end{cases}.$$
Then we can easily check that
$$(a\times b)_i = \epsilon_{ijk}a_j b_k.$$
The equation we want to prove is
$$R(a\times b)=Ra\times Rb.$$
Now since $R$ is invertible, we can multiply both sides with $R^{-1}$ and get an equivalent equation:
$$a\times b=R^{-1}(Ra\times R_b)$$
Thus we get:
\begin{aligned}
&&(a\times b)_i &= (R^{-1}(Ra\times Rb))_i\\
\iff&&(a\times b)_i &= (R^{-1})_{ij}(Ra\times Rb)_j\\
\iff&&(a\times b)_i &= R_{ji}(Ra\times Rb)_j && \text{orthogonality of $R$}\\
\iff&&\epsilon_{ikl}a_kb_l &= R_{ji} \epsilon_{jmn}(Ra)_m(Rb)_n\\
\iff&&\epsilon_{ikl}a_kb_l &= R_{ji} \epsilon_{jmn}R_{mk}a_kR_{nl}b_l\\
\iff&&\epsilon_{ikl}a_kb_l &= (\epsilon_{jnm} R_{ji} R_{mk} R_{nl}) a_k b_l
\end{aligned}
Thus we have to prove that
$$\epsilon_{ikl} = \epsilon_{jnm} R_{ji} R_{mk} R_{nl} \tag{*}$$
Now $\epsilon$ is completely characterized by the following two properties:


*

*$\epsilon_{123} = 1$

*Exchange of two indices changes the sign.
So let's check both properties for the right hand side of the above equation (note that the free indices are $i$, $k$, and $l$):


*

*$ \epsilon_{jmn} R_{j1} R_{m2} R_{n3} \stackrel?= 1$
The left hand side is the determinant of $R$, which by assumption is $1$, so this
equation is indeed true.

*Exchange of two indices: Let's without loss of generality exchange the last two indices, $§k$ and $l$.
Clearly due to the commutativity of multiplication, we have
$$\epsilon_{jnm} R_{ji} R_{ml} R_{nk} = \epsilon_{jnm} R_{ji} R_{nk} R_{ml}$$
Now the indices of $\epsilon$ in this term are summed over, therefore we can just
rename them without changing anything. In particular, we exchange the names $m$
and $n$:
$$\epsilon_{jnm} R_{ji} R_{nk} R_{ml} = \epsilon_{jnm} R_{ji} R_{mk} R_{nl}$$
Now we exchange the last two indices of the $\epsilon$ factor:
$$\epsilon_{jnm} = -\epsilon_{jmn}$$
Putting all together, we then get
$$\epsilon_{jnm} R_{ji} R_{ml} R_{nk} = -\epsilon_{jmn} R_{ji} R_{mk} R_{nl}$$
Thus exchanging $k$ and $l$ indeed changes the sign.
Thus we see that equation $(*)$ indeed holds, and thus $R(a\times b)=Ra\times Rb$.
