3-D Rotation Matrices

If we consider two vectors in $\Re ^3$, $\vec{v_1} = (1,0,1)$ and $\vec{v_2} = (0,1,1)$ (represented as (3,0,3) and (0,3,3) for clarity) and operate on them with a linear transformation that preserves their length but rotates them $\frac{\pi}{2}$ radians, we can see that our $\vec{v_1}' = \vec{v_2}$ and our $\vec{v_2}' = \vec{v_1}$ with a negative x component.

Based on the analysis in the paper in question, we are now able to assign a "rotation vector" as our transformation of $\vec{v_1}$.

(1)
\begin{align} T_{\theta}(\vec{v_1}) = \begin{bmatrix} \cos{\theta}\\ \sin{\theta} \\ 0 \end{bmatrix} \end{align}

It therefore follows that our transformation of $\vec{v_2}$ can be shown as the transformation of $\vec{v_1}$ transformed by a rotation of 90 degrees. We must, however, include a column in matrix $T_{\frac{\pi}{2}}$ because our Z component is preserved regardless of the rotation.

(2)
\begin{align} T_{\theta}(\vec{v_2}) = T_{\frac{\pi}{2}}T_{\theta}(\vec{v_1}) \\ T_{\theta}(\vec{v_2}) = \begin{bmatrix} 0&-1&0\\1&0&0\\0&0&1 \end{bmatrix} \begin{bmatrix} \cos{\theta}\\ \sin{\theta} \\ 0 \end{bmatrix} \\ = \begin{bmatrix} -\sin{\theta} \\ \cos{\theta} \\ 0 \end{bmatrix} \end{align}

Compiling all of these vectors into an A matrix that we can call our rotational transformation,

(3)
\begin{align} A = \begin{bmatrix} \cos{\theta} & -\sin{\theta} &0\\ \sin{\theta} & \cos{\theta} & 0 \\ 0& 0 & 1 \end{bmatrix} \end{align}

which looks very similar to the two dimensional rotation matrix, but this one preserves the sign and value of the z component which makes sense if the vectors are rotating about the z axis.

page revision: 0, last edited: 05 May 2015 06:08