# Composition of two rotations in $\mathbb{R}^3$ around skewed axis.

Prove that the composition of two rotations in $$\mathbb{R}^3$$, through skewed axis $$l,m$$ (this means there is no plane that contains both) and angles $$\alpha ,\beta$$ : $$\rho_{m,\beta} \circ \rho_{l,\alpha}$$ is a screw translation, i.e. the composition of a rotation with a translation through a vector paralel to the axis, also note that they commute: $$\rho_{n,\gamma} \circ T_u=T_u \circ \rho_{n,\gamma}$$ with $$n = P + \mathcal{L}\{u\},\quad P \in \mathbb{R}^3$$.

I've tried using the fact that one can decompose a rotation in a composition of two reflections about hiperplanes such that their intersection is the axis of rotation and their angle (oriented) is half of the angle of rotation, in mathematical notation: $$\rho_{l,\alpha} =R_{\mathcal{H}_1} \circ R_{\mathcal{H}_2}$$ and $$\rho_{m,\beta} =R_{\mathcal{H}_3} \circ R_{\mathcal{H}_4}$$ with $$\mathcal{H}_1 \cap \mathcal{H}_2 = l,\mathcal{H}_3 \cap \mathcal{H}_4 = m$$ and $$\angle_{or}(\mathcal{H}_2 ,\mathcal{H}_1)=\alpha /2,\angle_{or}(\mathcal{H}_4 ,\mathcal{H}_3)=\beta /2$$. Note that for each rotation the choice of one of the planes is arbitrary with the only condidtion being that contains the axis. Using this fact yields: $$\rho_{m,\beta} \circ \rho_{l,\alpha} =R_{\mathcal{H}_3} \circ R_{\mathcal{H}_4} \circ R_{\mathcal{H}_1} \circ R_{\mathcal{H}_2}$$ Im having trouble in the choice of the planes for each rotation such that the desired isometry (screwed translation) appears.