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, and $$ \eqalign{ & {\bf w} = {\bf q} \times d{\bf q} = \left( {\left| {\bf q} \right|\left| {d{\bf q}} \right|\sin \left| {d{\bf q}} \right|} \right)\;{\bf n} = \left| {d{\bf q}} \right|^{\,2} \;{\bf n} = \cr & = \left[ {\matrix{ {\sin \theta \sin \phi } \cr {\sin \theta \cos \phi } \cr {\cos \theta } \cr } } \right] \times \left[ {\matrix{ {\cos \theta \sin \phi } \cr {\cos \theta \cos \phi } \cr { - \sin \theta } \cr } } \right]d\theta + \left[ {\matrix{ {\sin \theta \sin \phi } \cr {\sin \theta \cos \phi } \cr {\cos \theta } \cr } } \right]\left[ {\matrix{ {\sin \theta \cos \phi } \cr { - \sin \theta \sin \phi } \cr 0 \cr } } \right]d\phi = \cr & = \left[ {\matrix{ { - \cos \phi } \cr {\sin \phi } \cr 0 \cr } } \right]d\theta + {1 \over 2}\left[ {\matrix{ {\sin \phi \sin (2\theta )} \cr {\cos \phi \sin (2\theta )} \cr {\cos (2\theta ) - 1} \cr } } \right]d\phi \cr} $$