Angle created by tilting a plane. I have a math problem involving tilting planes and the getting the angle of the normal from the starting point normal to the new normal after the plane is tilted.
I have a plane $x-y$ with a normal $z$. The plane tilts, lets say $+4$ degrees along the $x$, and $+18$ degrees along the $y$. 
How do I solve for the angle between the original normal to the plane and the new normal to the new plane as created by the $x$ and $y$ angular changes? 
Also how do I calculate the direction angle (as in a compass) that the new normal $z$ projects onto the original plane? This doesn't seem like it should be hard but it has me puzzled.
Thanks
 A: You can use the well-known Rodriguez rotation formula (https://en.wikipedia.org/wiki/Rodrigues%27_rotation_formula), which states
$
\textbf{v}' = \textbf{v}\cos\theta + (\textbf{r}\times\textbf{v})\sin\theta + \textbf{r}(\textbf{v}\cdot\textbf{r})(1-\cos\theta)
$
where $\textbf{v}$ is the original vector, $\textbf{r}$ is the vector about which the rotation on angle $\theta$ is performed, and $\textbf{v}'$ is the vector after rotation.
Now, by applying this formula twice (first, for example, for the rotation on angle $\alpha$ about $x-$axis and then on angle $\beta$ around $y-$axis), we get
$\textbf{k}'' = \textbf{i}\cos\alpha\sin\beta -\textbf{j}\sin\alpha + \textbf{k}\cos\alpha\cos\beta$.
Now, to find the angle between $\textbf{k}$ and $\textbf{k}''$, you can take the dot-product of them
$\cos \angle(\textbf{k},\textbf{k}'') = \textbf{k}\cdot\textbf{k}'' = \cos\alpha\cos\beta$.
The compass direction is given by the vector $\textbf{k}''_{||} = \textbf{i}\cos\alpha\sin\beta -\textbf{j}\sin\alpha$.
Now, let consider specific example of $\alpha = 4^\circ$ and $\beta = 18^\circ$. In this case, $\cos \alpha = 0.9976$, $\sin\alpha = 0.07$, $\cos\beta = 0.951$, and $\sin\beta = 0.309$. As a result, combining all these numbers together, we get 
$\angle(\textbf{k},\textbf{k}'') = 18.42^\circ$,
and the compass direction with respect to the the original $x-$axis is $-\tan^{-1}(\tan\alpha\sin\beta) = 1.24^\circ$.
Hope this helps.
