The angle between two unit vectors is not what I expected Ok imagine a vector with only X and Z components that make a 45 degree angle with the positive X axis. It's a unit vector. Now also imagine a unit vector that has the same direction as the positive x axis. Now imagine rotating both of these around the Z axis. 
I expect the angle between these vectors to still be 45 degrees. But it's not. If you don't believe me look here. Angle between two 3D vectors is not what I expected.
Another way to think about it is to draw a 45 degree angle between two lines on a piece of paper. Now stand the paper up, and rotate the paper. The angle between the lines are still 45 degrees.
Why is my way of thinking wrong?
 A: If you start with the vectors $(1,0,0)$ and $(1/\sqrt{2},0,1/\sqrt{2})$, and rotate both by $45^\circ$ about the $z\text{-axis}$, then you end up with $(1/\sqrt{2},1/\sqrt{2},0)$ and $(1/2,1/2,1/\sqrt{2})$.  The second point is not $(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})$ as you imagined.  If you think about it, the $z\text{-coordinate}$ cannot be changed by this rotation.  If the $z\text{-axis}$ is vertical, and the $x\text{-}y$ plane is horizontal, then the height of the point above the plane is not changed by rotation about the $z\text{-axis}$.  The height remains $1/\sqrt{2}$, and the length of the horizontal coordinate remains $1/\sqrt{2}$ as well.  That would not be the case if the final vector were what you thought it was.
A: The angle remains the same.
Let $x_1 = \frac{1}{2}(1, 0, 1)^T$, $x_2 = (1,0,0)^T$.
$\arccos (x_1^T x_2) = \frac{\pi}{4}$.
Now let $Q_\theta = \begin{bmatrix}\cos \theta & -\sin \theta & 0\\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1  \end{bmatrix}$. This represents a rotation of $\theta$ around the $z$ axis.
Note that $Q_\theta^T Q_\theta = I$. Then $\arccos ((Q_\theta x_1)^T x_2) = \arccos (x_1^T Q_\theta^T Q_\theta x_2) = \arccos (x_1^T x_2) = \frac{\pi}{4}$.
