Rotation of a point in 3d space I'm trying to rotate a point around a single axis of a 3D system.
Given $P=\begin{pmatrix}
101 \\
102 \\
103
\end{pmatrix}
$,
And the rotation matrix formula for rotation around the X axis only, I get:
$Rx(\psi=90^\circ)=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & 0
\end{pmatrix}$
The rotation yields 
$Rx*P =\begin{pmatrix}101 \\ -103 \\ 102\end{pmatrix}$
...but I expected a 90 degree rotation around $x$ would mirror the $y$ coordinate, yielding $\begin{pmatrix}101 \\ -102 \\ 103\end{pmatrix}$. By now I'm utterly confused by the various rotation conventions, and I'd be grateful for any help in clearing up where I went wrong with this simple operation.
Thanks!
 A: The word "mirror" should be a hint: the transformation
$$\begin{pmatrix}
x\\ y\\ z\end{pmatrix}\mapsto \begin{pmatrix}
x\\ -y\\ z\end{pmatrix}$$is  a reflection, not a rotation.
The result you're getting is correct. To get an intuition for it, try considering a simpler example: rotating the plane $90^\circ$ counterclockwise around the origin turns $(1,0)$ into $(0,1)$, not $(-1,0)$.
A: The matrix multiplication you performed is correct... the expectation sentence is wrong.
Rotations and reflections are different, in general. Rotations are generally taken to be orientation preserving rigid motions, and it turns out that any matrix for such a rotation has determinant 1. Reflections, on the other hand, are orientation reversing, and they have to have determinant -1. 
If it helps, we can think about plane geometry for a second. If you have a triangle drawn on a sheet of paper and a point marked as the origin, you can rotate this paper in place around the point on the paper. To perform a reflection you have to flip the paper over! This is the difference between a rotation and reflection in the 2-d plane.

Exercises to convince yourself the computation you did is correct:
1) You can confirm that the reflection you were performing was $\begin{bmatrix}1&0&0\\0&-1&0\\0&0&1\end{bmatrix}$, and that the determinants of this matrix and your original matrix are -1 and 1 respectively. This shows the two operations are definitely of different characters.
2) Look at the "shadow" of the point in the $y-z$ plane as the sun shines down the $x$ axis. (I mean $(102,103)$). If you rotate that 90 degrees around the $x$ axis, you can see that it lands in the second $y-z$ quadrant. Of course, this point is the shadow of the original point after rotation. Draw a triangle between the old shadow, the new shadow and the origin, and verify that it's a right triangle. 
3) It would also be a good exercise to try a 45 degree rotation instead. Do a 45 degree rotation in the $y-z$ plane, and check to see where the point $(0,100,100)$ goes. It will certainly not be $(0,-100,100)$ :)
