1 dof parametrization of a rotation matrix I found the following identity:
${\bf v}( \alpha ) = \cos(\alpha)\frac{{\bf x}+{\bf y}}{\| {\bf x}+{\bf y} \|}    + \sin(\alpha)\frac{{\bf x}\times{\bf y}}{\| {\bf x} \times {\bf y} \|}$ (1)
where  ${\bf v}( \alpha )$ is a 1dof parametrization of the rotation $R \in SO(3)$ such that ${\bf y}=R {\bf x}$ and $\times $ is the cross product. The corresponding rotation angle $\gamma $  for a given $\alpha$ can be found using
$\gamma =\arccos\left( \frac{{\bf p}_x^T {\bf p}_y }{\|{\bf p}_x\|\|{\bf p}_y\|} \right)$
where ${\bf p}_x={\bf x} -({\bf v}^T{\bf x}){\bf v}$ and  $p_y={\bf y} -({\bf v}^T{\bf y}){\bf v}$ are projections of the vector ${\bf x}$ and ${\bf y}$ onto the plane whose normal is the rotation axis ${\bf v}$. The  rotation $R$ can be computed using the exponential map using ${\bf v}$ and $\gamma$.
How can I prove identity (1)?
Br,
Nicolas
 A: A rotation, by definition, keeps the euclidean norm (length) of the vectors. To pass from 
$\mathbf{x}$ to $\mathbf{y}$ you need in general a rotation and a scale factor.
$$ \frac{\mathbf{y}}
{{\left\| \mathbf{y} \right\|}} = R\;\frac{\mathbf{x}}
{{\left\| \mathbf{x} \right\|}}\quad  \Rightarrow \quad \mathbf{y} = \left( {\frac{{\left\| \mathbf{y} \right\|}}
{{\left\| \mathbf{x} \right\|}}} \right)R\;\mathbf{x} $$
Apart from that, your scheme is correct.  In fact, putting
$$ \mathbf{t} = \frac{{\mathbf{x} + \mathbf{y}}}
{{\left\| {\mathbf{x} + \mathbf{y}} \right\|}}\quad \mathbf{n} = \frac{{\mathbf{x} \times \mathbf{y}}}
{{\left\| {\mathbf{x} \times \mathbf{y}} \right\|}}\quad \mathbf{v}\left( \alpha  \right) = \cos \left( \alpha  \right)\mathbf{t} + \sin \left( \alpha  \right)\mathbf{n}
$$
we have that
$\mathbf{t} $ is a unit vector, co-planar with $\mathbf{x} $ and  $\mathbf{y} $;
$\mathbf{n} $ is the unit normal vector to the plane of $\mathbf{x}$ , $\mathbf{y}$;
$\mathbf{v}\left( \alpha  \right)$ is a unit vector, in the plane $\mathbf{t}$, $\mathbf{n}$.
Then the angle of rotation around $\mathbf{v}$ that brings $\mathbf{x} $ onto  $\mathbf{y} $
is the dihedral angle between the planes $\mathbf{x} $, $\mathbf{v}$  and $\mathbf{y} $, $\mathbf{v}$.
That can be computed in multiple ways.
One way is that you adopted of computing the components of $\mathbf{x} $ and $\mathbf{y} $
orthogonal to $\mathbf{v} $, and then taking the angle between them.
Another way would be to compute the angle between $ \mathbf{x} \times \mathbf{v}$ and $\mathbf{y} \times \mathbf{v}$ (angle between the normals to the planes).
Finally note that if you apply your scheme starting with  $ \frac{\mathbf{x}}{{\left\| \mathbf{x} \right\|}} $ and $ \frac{\mathbf{y}}{{\left\| \mathbf{y} \right\|}} $, in place of 
$\mathbf{x} $ and $\mathbf{y} $, then these unit vectors will be symmetric with respect to the plane containing $\mathbf{t} $, $\mathbf{v} $,$\mathbf{n} $.
