Let's suppose we have two rotations about two different axes represented by vectors $v_1$ and $v_2$: $R_1(v_1, \theta_1)$, $R_2(v_2,\theta_2)$.
It's relatively easy to prove that composition of these two rotations gives rotation about axis $v_3$ distinct from axes $v_1$ and $v_2$ .
Indeed
if for example $v_3=v_1$ then
$R_1(v_1, \theta_1) R_2(v_2,\theta_2)=R_3(v_1,\theta_3)$ leads to $R_2(v_2,\theta_2)=R_1^T(v_1, \theta_1)R_3(v_1,\theta_3)=R(v_1,\theta_3 -\theta_1)$ what gives $v_1=v_2$. ... Contradiction...
We see that composition of two rotations about different axes always generates a new axis of rotation.
The problem can be extended for condition of the plane generated by the axes.
Question:
Is it true that composition of two rotations generates the axis which doesn't belong to the plane which is constructed by the original axes of rotations ?
How to prove it ?
If the statement is not however true what are conditions for not changing a plane during the composition of rotations $ ^{[1]}$ ?
$ ^{[1]}$ It can be observed that even in the case of quite regular rotations the above statement is true
Let's take $Rot(z,\dfrac{\pi}{2})Rot(x,\dfrac{\pi}{2})= \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix} = Rot([1,1,1]^T, \dfrac{2}{3}\pi)$
or
$Rot(x, \pi )Rot(z, \pi )= \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \\ \end{bmatrix} = Rot( y, \pi)$
So I suppose it is generally true but how to prove it ?