What is the minimum number of rotations about axes in a plane that can describe an arbitrary rotation in 3D? I'm trying to decompose an arbitrary rotation of a 3D sphere into a series of rotations about any axes that lie in the equatorial plane. 
I know I can factor the arbitrary rotation into three rotations in the equatorial plane using Euler angle formulas. Can I factor into two rotations if I allow any choice of axis in the equatorial plane? If it exists, is this factorization unique?
 A: My answer is that you can. 
If you know about quaternions, then you know that a rotation about a unit vector $v \in \mathbb{R}^3$ through an angle $\theta$ can be accomplished by regarding elements $w = (a, b, c) \in \mathbb{R}^3$ as purely quaternionic elements $ai + bj + ck$, and then mapping $w \mapsto uwu^{-1}$ where $u = \cos(\frac{\theta}{2})\cdot 1 + \sin(\frac{\theta}{2})v$. Such $u$ are precisely quaternions of unit norm. 
Let us denote the rotation above by $R_u$ (so $R_u(w) := uwu^{-1}$). Notice that a composition of two rotations $R_u \circ R_v$ is just $R_{u v}$; this follows by associativity of quaternionic multiplication. 
So we can translate your problem into the following: for any rotation $R_w \in SO(3)$ given by a nonzero quaternion $w$, show that there exist quaternions $u, v$ in the linear span of $1, i, j$ such that $u v = w$. For these $u, v$ describe rotations about vectors in the equatorial plane spanned by $i, j$. 
Now this is not difficult. Let us write $u = a + bi + cj$ and $v = a' + b'i + c'j$. We compute 
$$(a + bi + cj)(a' + b'i + c'j) = aa' - bb' - cc' + (ab' + a'b)i + (ac' + a'c)j + (bc' - b'c)k$$ 
and so given a nonzero quaternion $w = p + qi + rj + sk$, our task is to cook up parameters $a, b, c, a', b', c'$ such that 
$$p = aa' - bb' - cc'$$ 
$$q = a b' + a'b$$ 
$$r = ac' + a'c$$ 
$$s = bc' - b'c$$ 
In fact, let's make our life easier and simply set $b = 0, b' = 1$. Then $a = q$, $c = -s$, and we can solve for $a', c'$ in the linear system 
$$p = qa' + sc'$$ 
$$r = -sa' + qc'$$ 
provided the determinant $q^2 + s^2$ is nonzero, i.e., provided one of $q, s$ is nonzero. If both $q, s = 0$, then of course $w = p + rj$ corresponds to a rotation about the vector $j$ which is already in the equatorial plane, so there was nothing to do in this case. 
Once you have solutions $u = a + bi + cj, v = a' + b'i + c'j$ to the equation $uv = w$ in your hands, the rotations $R_u, R_v$ are rotations about the vectors $bi + cj$ and $b'i + c'j$, respectively. If you want the rotation angles, then normalize $u$ and $v$ (i.e. divide by their norms $(a^2 + b^2 + c^2)^{1/2}$ and $(a')^2 + (b')^2 + (c')^2)^{1/2}$) and use the fact that $R_u = R_{\frac{u}{\|u\|}}$. You can then read off the desired angles by writing e.g. $\frac{u}{\|u\|} = \cos(\frac{\theta}{2}) + \sin(\frac{\theta}{2})\frac{bi + cj}{(b^2 + c^2)^{1/2}}$, so for instance $\cos(\frac{\theta}{2}) = \frac{a}{(a^2 + b^2 + c^2)^{1/2}}$. 
By the way, this solution also shows that the decomposition is in general non-unique. For example, we could have also chosen $c = 0, c' = 1$ and arrive at quaternionic solutions. 
A: The answer is no, because when letting $v$ be an arbitrary direction about which to rotate by an angle $\rho$, that rotation can only then be realized with a rotation around an axis in the xy-plane if $v^Tz=0$; that shows that at least three rotations are necessary  


*

*rotate about $v\times z$ to bring $v$ into the $xy$-plane   

*rotate about the image of $v$ in the $xy$-plane  

*rotate the image of $v$ back around $v\times z$ to bring the image back to the original direction of $v$ 

