My task is: "Describe rotation $S \circ R$ by axis and angle, where $R$ is rotation around $(0,1,1)$ by 90 degrees, and $S$ is rotation around $(1,-1,0)$ by 90 degrees." I should use quaternion multiplication to solve that. How do I do that? The way I see it now, I represent $S$ with $e^{\pi/4 (j+k)}$ and $R$ with $e^{\pi/4 (i-j)}$, and then multiply them, using the quaternion rules for that: $e^{\pi/4 (j+k)} \cdot e^{\pi/4 (i-j)} = (cos (\pi /4) + (j+k)sin(\pi /4))\cdot(cos (\pi /4) + (i-j)sin(\pi /4)) = \dots = 1 + i +1/2 j$, which should correspond to rotation by $0$ degrees around vector $(2,1,0)$, but I guess something is wrong with my answer.
• Can you clarify the order in $S \circ R$? Am i right in assuming its $R$ followed by $S$ ? – vnd Feb 9 '16 at 14:08
The quaternions for rotations $R$ and $S$ are: $$t_R=\cos \frac{\pi}{4}+\frac{j+k}{\sqrt 2}\sin \frac{\pi}{4}$$ $$t_S=\cos \frac{\pi}{4}+\frac{i-j}{\sqrt 2}\sin \frac{\pi}{4}$$ The product rotation $S \circ R$ would then be given by $t_Rt_S$, which is easily calculated by the rules of quaternion multiplication to be $$t_Rt_S=\frac{3}{4} +\frac{\sqrt 7}{4}\left[\frac{(\sqrt 2 +1)i + j + (\sqrt 2 - 1)k}{\sqrt 7}\right].$$ The rotation angle is therefore $2\arctan \frac{\sqrt 7}{3}$ and the rotation axis is in the square brackets of the above expression.
• Can you explain a bit more about how you got $arctan$ here? – Dark Archon Feb 9 '16 at 15:17
• a quaternion of the form $\cos \theta + u \sin \theta$ operates by conjugation on $R^3$ to cause a rotation of $2\theta$ about the axis $u$ in $R^3$. Comparing one has $\cos \theta = 3/4$ and $\sin \theta = \sqrt 7 /4$. – vnd Feb 9 '16 at 15:21