Does SO(3) preserve the cross product?

Let $g\in SO(3)$ and $p,q\in\mathbb{R}^3$. I wondered whether it is true that $$g(p\times q)=gp\times gq$$

I am not sure how to prove this. I guess I will use at some point that the last row $g_3$ of $g$ can be obtained by $g_3=g_1\times g_2$.

But I assume there is an easier proof than writing everything out.

You may use the scalar triple product formula $r \cdot (p\times q)=\det(r,p,q)$ to prove that $$ gr \cdot (gp\times gq)=gr \cdot g(p\times q)\tag{1} $$ ($=\det(r,p,q)$) for any vector $r$. Since $g$ is invertible, if $(1)$ holds for every vector $r$, we must have $gp\times gq=g(p\times q)$.

Two ways of seeing it more directly:

• The cross product of two vectors can be expressed in terms of the norms of the vectors and the angle between them, and those properties are preserved by rotations. (Update: As mephistolotl mentioned in the comments, you also need the fact that rotations preserve the orientation. Otherwise, you'll get a vector of correct length, but different direction.)
• Numerically, you would need to show $\epsilon_{ijk} O_{jl} O_{km} =O_{ip}\epsilon_{plm}$, which follows from orthogonality and the fact that $\det O=1$.

Since both expressions $g(p\times q)$ and $g(p)\times g(q)$ are linear in each of the variables $p,q$, it suffices to check the equality for the nine cases $p,q=e_1,e_2,e_3$, i.e. when $p,q$ are basis vectors. This method is quite often employed when dealing with equality of multilinear functions.