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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.

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