transforming $(A,B,C)$ to $(0, 0, 1)$ by rotations

Question

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.

Answer 1

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 $A$

$$(A,0,C')\to(0,0,C'').$$

Answer 2

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 $Z$.

Answer 3

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!