transforming $(A,B,C)$ to $(0, 0, 1)$ by rotations I'm trying to reflect the "world" through a specified plane $p:Ax+By+Cz=0$.
I know how to reflect the "world" through the $xy$-plane, so I want to rotate $p$ in the $3$ axes ($x,y,z$-axes) so it will be contained in the $xy$-plane, which means its normal will be $(0,0,1)$.
That's why I want transform $p$'s normal to $(0,0,1)$.
I'm trying to solve this using the three rotations with known transformations:
Rotation about the $z$-axis for example:
\begin{pmatrix}
\cos A & -\sin A & 0 \\ 
\sin A & \cos A &  0 \\
  0   &   0    &  1
\end{pmatrix}
So, using only this kind of matrix (for the $x$- and $y$-axes too) I would like to transform $(A,B,C)$ to $(0,0,1)$.
Thank you so much for helping.
 A: First rotate around $X$ with angle $-\arctan(B/C)$ to cancel $B$
$$(A,B,C)\to(A,0,C'),$$
Then rotate around $Y$ with angle $-\arctan(A/C')$ to cancel $A$
$$(A,0,C')\to(0,0,C'').$$
A: What you want to do is rotate your vector $(A,B,C)^T$ three times, around each axis, and find the appropriate angles of rotation. You can write:
$$\begin{pmatrix}0\\0\\1\end{pmatrix}=R_X\cdot R_Y\cdot R_Z\cdot\begin{pmatrix}A\\B\\C\end{pmatrix}$$
$$\begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}1&0&0\\0&\cos X&-\sin X\\0&\sin X&\cos X\end{pmatrix}\begin{pmatrix}\cos Y&0&\sin Y\\0&1&0\\-\sin Y&0&\cos Y\end{pmatrix}\begin{pmatrix}\cos Z&-\sin Z&0\\\sin Z&\cos Z&0\\0&0&1\end{pmatrix}\begin{pmatrix}A\\B\\C\end{pmatrix}$$
Now you can multiply these matrices, and you should end up with a system of three equations with three unknowns, namely the angles $X$, $Y$ and $Z$.
A: For a detailed algorithm you look here:
http://inside.mines.edu/fs_home/gmurray/ArbitraryAxisRotation/
This is what @Yves suggested, in a more detailed way.
Thank you all!
