Getting rotation vector from spherical coordinates angles derivatives I have a unit vector q that is represented in spherical coordinates:
\begin{equation}
q=
\begin{bmatrix}
    \sin(\theta)\sin(\varphi) \\
    \sin(\theta)\cos(\varphi)\\
    \cos(\theta)
\end{bmatrix}
\end{equation}
What I need to get is a transformation that goes from $\;\biggl\{\dfrac{\partial\theta}{\partial t};\dfrac{\partial\varphi}{\partial t}\biggr\}$ to $w$. Where $w$ is the rotation vector of q around the origin.
I know that $\dfrac{\partial q}{\partial t}=w\times q$ but as the cross product does not have an inverse i cannot isolate $w$.
 A: Given
$$
{\bf q} = \left[ {\matrix{
   {\sin \theta \sin \phi }  \cr 
   {\sin \theta \cos \phi }  \cr 
   {\cos \theta }  \cr 
 } } \right]
$$
then
$$
d{\bf q} = \left[ {\matrix{
   {\cos \theta \sin \phi }  \cr 
   {\cos \theta \cos \phi }  \cr 
   { - \sin \theta }  \cr 
 } } \right]d\theta  + \left[ {\matrix{
   {\sin \theta \cos \phi }  \cr 
   { - \sin \theta \sin \phi }  \cr 
   0  \cr 
 } } \right]d\phi 
$$
and of course, being $\bf q$ unitary, we will have:
$$
{\bf q} \cdot d{\bf q} = 0
$$
So $|d \bf q|$ represents the rotation angle, which is on the plane containing
$\bf q$ and $\bf q+d\bf q$ and any linear combination of them, thus also $\bf q$ and $d \bf q$, which are normal to each other.
Therefore their cross product is normal to the rotation plane and is equal to
$$
\eqalign{
  & {\bf w} = {\bf q} \times d{\bf q} = \left( {\left| {\bf q} \right|\left| {d{\bf q}} \right|\sin \left( {\mathop {{\bf q},d{\bf q}}\limits^ \wedge  } \right)} \right)\;{\bf n}
 = \left| {d{\bf q}} \right|\;{\bf n} =  {d\alpha} \; {\bf n} \cr 
  &  = \left[ {\matrix{
   {\sin \theta \sin \phi }  \cr 
   {\sin \theta \cos \phi }  \cr 
   {\cos \theta }  \cr 
 } } \right] \times \left[ {\matrix{
   {\cos \theta \sin \phi }  \cr 
   {\cos \theta \cos \phi }  \cr 
   { - \sin \theta }  \cr 
 } } \right]d\theta  + \left[ {\matrix{
   {\sin \theta \sin \phi }  \cr 
   {\sin \theta \cos \phi }  \cr 
   {\cos \theta }  \cr 
 } } \right]\times \left[ {\matrix{
   {\sin \theta \cos \phi }  \cr 
   { - \sin \theta \sin \phi }  \cr 
   0  \cr 
 } } \right]d\phi  =   \cr 
  &  = \left[ {\matrix{
   { - \cos \phi }  \cr 
   {\sin \phi }  \cr 
   0  \cr 
 } } \right]d\theta  + {1 \over 2}\left[ {\matrix{
   {\sin \phi \sin (2\theta )}  \cr 
   {\cos \phi \sin (2\theta )}  \cr 
   {\cos (2\theta ) - 1}  \cr 
 } } \right]d\phi  \cr} 
$$
where the sign of $\alpha$ is according to the Right-Hand Rule (bringing "starting" v. to "ending" v.).
