Theorem
Let $\vR$ be a rotational $3\times 3$ matrix, i.e., a real matrix with $\det \vR = 1$ and $\vR^{-1} = \vR^T$ . Then for all vectors $\vu,\vv$ in $\mathbb{R}^3$ , $$ \vR \cdot (\vu\times \vv) = (\vR\cdot \vu)\times (\vR\cdot \vv).$$

Proof. Let us first fix some right hand oriented orthonormal basis in $\mathbb{R}^3$ . Further, let $\{u^1,u^2,u^3\}$ and $\{v^1,v^2,v^3\}$ be the components of $\vu$ and $\vv$ in that basis. Also, in the chosen basis, we denote the entries of $\vR$ by $R_{ij}$ . Since $\vR$ is rotational, we have $R_{ij} R_{kj} = \delta_{ik}$ where $\delta_{ik}$ is the Kronecker delta symbol. Here we use the Einstein summation convention. Thus, in the previous expression, on the left hand side, $j$ should be summed over $1,2,3$ . We shall use the Levi-Civita permutation symbol $\varepsilon$ to write the cross product. Then the $i$ :th coordinate of $\vu\times \vv$ equals $(\vu\times \vv)^i = \varepsilon^{ijk} u^j v^k$ . For the $k$ th component of $(\vR\cdot \vu)\times (\vR\cdot \vv)$ we then have \begin{eqnarray*} ((\vR\cdot \vu)\times (\vR\cdot \vv))^k &=& \varepsilon^{imk} R_{ij} R_{mn} u^j v^n \\ &=& \varepsilon^{iml} \delta_{kl} R_{ij} R_{mn} u^j v^n \\ &=& \varepsilon^{iml} R_{kr} R_{lr} R_{ij} R_{mn} u^j v^n \\ &=& \varepsilon^{jnr} \det \vR\, R_{kr} u^j v^n. \end{eqnarray*}The last line follows since $\varepsilon^{ijk} R_{im} R_{jn} R_{kr} = \varepsilon^{mnr}\varepsilon^{ijk} R_{i1} R_{j2} R_{k3} = \varepsilon^{mnr} \det \vR$ . Since $\det \vR = 1$ , it follows that \begin{eqnarray*} ((\vR\cdot \vu)\times (\vR\cdot \vv))^k &=& R_{kr} \varepsilon^{jnr} u^j v^n\\ &=& R_{kr} (\vu\times \vv)^r \\ &=& (\vR\cdot \vu\times \vv)^k \end{eqnarray*}as claimed.