Let $p,q $ be unit vectors in $\mathbb R^3$ with the Euclidean norm. Then it is known that there exists a rotation $f$ of $\mathbb R^3$ such that $f(p)=q$. It is easy to find two such rotation: one with rotation axis orthogonal to $p,q$ and the second with axis in the plane generated by $p,q$.
How can one prove that if $p+q\neq 0$ then the rotation axis of $f$ has to lie in the plane generated by $p\times q$ and $p+q$?
Let $u$ be some point on the rotation axis $L$, not the origin. Then $p$ and $q$ are equidistant from $u$. So $L$ lies within in the plane perpendicularly bisecting the segment $pq$. But this plane is through the origin ($|p|=|q|=1$) and is generated by $p+q$ and $p\times q$ as a vector space - both these vectors are in the plane, and clearly $p\times q$ is not in the plane through $0$, $p$ and $q$, so the vectors are linearly independent.
We have $ (p\times q)\times(p+q)=-p\times(p\times q)+(p \times q)\times q=-p<p,q>+q\|p\|^2+q<p,q>-p\|q\|^2=(q-p)+(q-p)<p,q>=(1+<p,q>(q-p), $ which is not zero, because $p\neq -q$. Thus
$ span\{p\times q, p+q\}=\{q-p\}^\bot. $
In order to show that the rotation's axis (=the set of fixed points of $f$) is contained in $span\{p\times q, p+q\}$ it suffices to show that for each $x\in \mathbb R^3$ if $f(x)=x$ then $<x,q-p>=0$.
Indeed, if $f(x)=x$ we have
$ <x,q-p>=<f(x),f(p)>-<x,p>=<x,p>-<x,p>=0. $
