Given one vector in two frames (with some constraints), what is the rotation matrix? I have a vector $g$ represented in frames A and B - $(g_{x_A}, g_{y_A}, g_{z_A})$ and $(g_{x_B}, g_{y_B}, g_{z_B})$. I know that the $y$-axes in both frames are antiparallel and that the frames share an origin. So the frames look like:

Given this information, how could I go about finding the rotation matrix between these two frames?
 A: Knowing that the origins are shared and $y_B$ is antiparallel to $y_A$ helps in that we can think of the rotation from $A$ to $B$ as first a rotation by $\pi$ about the $z_A$ axis, and then a single rotation by some angle $\theta$ in the shared (but flipped) $x-z$ plane that results.  
Let's represent the rotation from $A$ to $B$ with the rotation matrix $\mathbf{R}$ so that 
$$
\mathbf{g}_B = R \mathbf{g}_A
$$
or
$$
\begin{bmatrix}g_{x_B} \\ g_{y_B} \\ g_{z_B} \end{bmatrix} = \mathbf{R} \begin{bmatrix}g_{x_A} \\ g_{y_A} \\ g_{z_A} \end{bmatrix}
$$
We can decompose $R$ further into the two rotations we know must occur of $\pi$ about $z$ and then $\theta$ about $y_{B}$ (note: the negatives on the sines are swapped since it is really rotating by $-\theta$ relative to $A$).
$$
\mathbf{R} = 
\begin{bmatrix}
\cos(\theta) & 0 & -\sin(\theta) \\
0 & 1 & 0 \\
\sin(\theta) & 0 & \cos(\theta) 
\end{bmatrix}
\begin{bmatrix}
\cos(\pi) & -\sin(\pi) & 0 \\
\sin(\pi) & \cos(\pi) & 0 \\
0 & 0 & 1 
\end{bmatrix}
$$
or simplified a bit we have
$$
\mathbf{R} = 
\begin{bmatrix}
\cos(\theta) & 0 & -\sin(\theta) \\
0 & 1 & 0 \\
\sin(\theta) & 0 & \cos(\theta) 
\end{bmatrix}
\begin{bmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1 
\end{bmatrix}
= 
\begin{bmatrix}
-\cos(\theta) & 0 & -\sin(\theta) \\
0 & -1 & 0 \\
-\sin(\theta) & 0 & \cos(\theta) 
\end{bmatrix}
$$
So now you can write your rotation as
$$
\begin{bmatrix}g_{x_B} \\ g_{y_B} \\ g_{z_B} \end{bmatrix}  = 
\begin{bmatrix}
-\cos(\theta) & 0 & -\sin(\theta) \\
0 & -1 & 0 \\
-\sin(\theta) & 0 & \cos(\theta) 
\end{bmatrix}
\begin{bmatrix}g_{x_A} \\ g_{y_A} \\ g_{z_A} \end{bmatrix} 
$$
and multiplying we have
$$
\begin{bmatrix}g_{x_B} \\ g_{y_B} \\ g_{z_B} \end{bmatrix} =
\begin{bmatrix}
-g_{x_A} \cos(\theta) - g_{z_A} \sin(\theta) \\
-g_{y_A} \\
-g_{x_A} \sin(\theta) + g_{z_A} \cos(\theta) 
\end{bmatrix}
$$
Given your two vectors, you can solve this for $\theta$ and then plug it in to the rotation matrix $\mathbf{R}$ above. 
