Get the rotation matrix from two vectors Given $v=(2,3,4)^t$ and $w=(5,2,0)^t$, I want to calculate the rotation matrix (in the normal coordinate system given by orthonormal vectors $i,j$ and $k$) that projects $v$ to $w$ and to find out which is the rotation axis.
I Started by calculating the vector product $v \times w$, which I need in order to calculate the angle between the two vectors as $\varphi=arcsin({||v \times w|| \over ||v||||w||})$. This equals to $arcsin(\frac{3\sqrt{65}}{29})$. 
I now that the rotation in the plane spanned by v and w is given by $A=\pmatrix{cos(arcsin(\frac{3\sqrt{65}}{29}) & - \frac{3\sqrt{65}}{29} \\ \frac{3\sqrt{65}}{29} & cos(arcsin(\frac{3\sqrt{65}}{29}))}$ and $v \times w$ ist he rotation axis, but I'm lost when it comes to determining the rotation matrix in 3D.
 A: Try writing the matrix $A_{B_2}$ in terms of the basis $B_2 = \{\hat v, \hat w, \widehat{v \times w}\}$.  This is very similar to the matrix for the plane that you've already written.
Then, think about if you know a way to change from the standard basis $B_1 = \{\hat i, \hat j, \hat k\}$ to this basis.  That is, let $P_{1 \to 2}$ be a linear map such that $P_{1 \to 2}(\hat i) = \hat v$, $P_{1 \to 2}(\hat j)= \hat w$, and so on.  Then the matrix representation of $P$ can be used to find the matrix representation $A_{B_1}$ (in the standard basis) like so:
$$A_{B_1} = P^{-1}_{1 \to 2} A_{B_2} P_{1 \to 2}$$
You should feel comfortable with the idea that $P^{-1}_{1 \to 2}$ is just the change of basis that goes from $B_2 \to B_1$, which is what you need.
Think about how you would write the change of basis matrix $P_{1 \to 2}$ and its inverse.  Remember that this is an orthogonal matrix, and as such, its inverse is equal to its transpose (greatly simplifying things).  All that should be left once you know that matrix is to do the matrix multiplications:  a tedious process by hand, but that's all.
