Recall the rules of quaternion multiplication are given by
$$\begin{matrix} ij=k, & ik=-j, & jk=i \\ ji=-k, & ki=j, & kj=-i. \end{matrix}$$
If $i,j,k$ are interpreted as the standard basis vectors in $\R^3$, then multiplication corresponds precisely to the cross product! I've always wondered why, and I finally managed to find a satisfactory explanation.
I was directly inspired by wy.22/6/2021 where I derived $\Aut(Q_8)$. In particular I had the idea of interpreting the product $ijk$ as the $3\times3$ matrix $I=(i\ j\ k)$, and interpreting the equality $ijk=-1$ as saying that $I$ has positive orientation. (It is weird that $-1$ should correspond to positive orientation, but it doesn't lead to any major problems.)
In general, if $a,b,c$ is a permutation of $i,j,k$ then the sign of the product $abc$ is equal to (the opposite of) the orientation of the matrix $(a\ b\ c)$. This is because every permutation can be decomposed into transpositions, and transpositions flip both the sign of the product $abc$ and the orientation of $(a\ b\ c)$. For instance, $ij=-ji$ so $jik=1$, and indeed $(j\ i\ k)$ has negative orientation. Therefore, for $abc=-1$ to be true, we must have $c=a\times b$, so that by the property of cross product $(a\ b\ c)$ has positive orientation.
This argument easily generalises if we allow $a,b,c$ to be signed, i.e. take values in $\{\pm i,\pm j,\pm k\}$.