Axis of rotation of composition of rotations (Artin's Algebra) Say $R_1$, $R_2$ are rotations in $\mathbb{R}^3$ with axes and angles $(v_1,\theta_1), (v_2,\theta_2)$ respectively. Since $SO_3$ is a group, we have that $R_2 \circ R_1$ is a rotation with some axis $v$. Is there a geometric way of finding $v$? This is problem 4.5.10 in Artin's Algebra.
My attempts have included looking at $v$ as a differentiable function of $\theta_1$ and $\theta_2$, writing out the corresponding matrix equations, staring at a wall, and guessing. Not sure where to go from here.
 A: A rotation around an axis is a composition of two symmetries (reflections) wrt planes which intersect at this line at half the angle. If your two rotations are represented as $R_i=S_{i1}\circ S_{i2}$ so that each plane $S_{12}$ and $S_{21}$ contains both axes, then $S_{12}=S_{21}$, and $R_1\circ R_2=S_{11}\circ S_{22}$.
A: $\def\\#1{{\bf#1}}$We can find the axis of $R_2R_1$ by writing each of the rotations as a product of two reflections.  We may assume without loss of generality that $\\v_1$ and $\\v_2$ are unit vectors, and are independent (if they are scalar multiples of each other then $\\v_1=\pm\\v_2$ and the problem is easy).  Define the following vectors:
$$\eqalign{
  \\w&=\\v_1\times\\v_2\cr
  \\x&=\\w\times\\v_1\cr
  \\w_1&=\\w\cos(\theta_1/2)+\\x\sin(\theta_1/2)\cr}$$
(suggestion: draw them in order to visualise what we are doing) and let
$$S_1=(I-2\\w\\w^T)(I-2\\w_1\\w_1^T)\ .$$
Then $S_1$ is a product of two reflections in $\Bbb R^3$ and is therefore a rotation.  In fact, we shall show that $S_1$ is the given rotation $R_1$.  This takes a bit of algebra: it is helpful to note that we have a lot of perpendicular vectors here, and in particular
$$\\w\cdot\\v_1=\\x\cdot\\v_1=\\w_1\cdot\\v_1=0\ .$$
Also note that if $\\a,\\b,\\c$ are column vectors of the same size then
$$\\a\\b^T\\c=\\a(\\b\cdot\\c)=(\\b\cdot\\c)\\a\ .$$
Now we have
$$(I-2\\w_1\\w_1^T)\\v_1=\\v_1-2(\\w_1\cdot\\v_1)\\w_1=\\v_1$$
and so
$$S_1\\v_1=(I-2\\w\\w^T)\\v_1=\\v_1-2(\\w\cdot\\v_1)\\w=\\v_1\ ;$$
therefore the rotation $S_1$ has axis $\\v_1$.  Moreover,
$$\eqalign{S_1\\x
  &=(I-2\\w\\w^T)(I-2\\w_1\\w_1^T)\\x\cr
  &=(I-2\\w\\w^T)(\\x-2\\w_1\sin(\theta_1/2))\cr
  &=\\x-2\\w_1\sin(\theta_1/2)+4\\w\cos(\theta_1/2)\sin(\theta_1/2)\cr
  &=\\x\cos\theta_1+\\w\sin\theta_1\ .\cr}$$
Since $\\v_1,\\x,\\w$, in that order, form a right-handed system of axes, this shows that the angle of rotation is $\theta_1$.
Now define also
$$\eqalign{
  \\y&=\\w\times\\v_2\cr
  \\w_2&=\\w\cos(\theta_2/2)-\\y\sin(\theta_2/2)\cr}$$
and let
$$S_2=(I-2\\w_2\\w_2^T)(I-2\\w\\w^T)\ .$$
The same sort of calculations as above show that $S_2=R_2$.  Hence
$$\eqalign{R_2R_1=S_2S_1
  &=(I-2\\w_2\\w_2^T)(I-2\\w\\w^T)(I-2\\w\\w^T)(I-2\\w_1\\w_1^T)\cr
  &=(I-2\\w_2\\w_2^T)(I-2\\w_1\\w_1^T)\ ;\cr}$$
the fact that the middle terms cancel is because they give the product of a reflection with itself; alternatively we can do the algebra to show that the product is the identity.
All this gives $R_2R_1$ as the product of reflections with axes $\\w_1$ and $\\w_2$, and this is a rotation with axis $\\w_1\times\\w_2$.
Thus, the required axis is $\\w_1\times\\w_2$, where $\\w_1$ and $\\w_2$ are given by the above equations.
