Find the Glide Line and Glide Vector This is a question from a homework assignment.  I have been spinning my wheels for some time and could really use a hint.  Here is the set up.  Let $t_a$ denote the isometry of translation by a vector $a$, let $\rho_{\theta}$ denote the isometry of rotation through the origin by $\theta$, and let $r$ denote the isometry of reflection about the horizontal axis.  The question asks

Determine the glide line and glide vector for the glide reflection $t_a \rho_{\theta} r$.

Say the line $L$ makes an angle of $\theta$ with the horizontal axis (not necessarily passing through the origin).  I want to change coordinates so that $L$ is horizontal.  To do this, I conjugate by $\rho_{- \theta/2}$.  The isometry $t_a \rho_{\theta} r$ in this new coordinate system now becomes
$$ \rho_{- \theta / 2} t_a \rho_{\theta} r \rho_{- \theta / 2}^{-1} = t_{\rho_{- \theta / 2}(a)} r. $$
Now, by the definition of a glide reflection, I know that $a$ was parallel to $L$.  Since the line of reflection of $\rho_{\theta} r$ makes an angle of $\theta / 2$ with the horizontal, this implies $a' = \rho_{- \theta / 2}(a)$ is parallel to the line of reflection of $\rho_{\theta} r$.  However, I do not think $a'$ is the glide vector.  I worked out an example using $\theta = 90^{\circ}$ and $a = (0, 4)^t$.  I found that the glide vector is $(2, 2)^t$, and the glide line is $\{ (0, 2) + t(1, 1) : t \in \mathbb{R} \}$.  However, my work up to this point implies that the glide vector is $(2 \sqrt{2}, 2 \sqrt{2})^t$.  I feel confident that I worked the example correctly, so the issue must be in the argument.  Any hints of suggestions would be greatly appreciated.
 A: First change coordinates by rotation through $- \frac{\theta}{2}$.  In this new coordinate system, the isometry is $t_a r$ (where $a = (a_1, a_2)^t$), representing a reflection about the line $\text{span} (0, \frac{a_2}{2})^t$ followed by a translation by $(a_1, 0)^t$.  Thus, the glide line and glide vector in the original coordinates is found by rotating through $\frac{\theta}{2}$.  That is, the glide line is spanned by the vector
$$ \begin{bmatrix}
\cos \frac{\theta}{2} & - \sin \frac{\theta}{2} \\
\sin \frac{\theta}{2} & \cos \frac{\theta}{2}
\end{bmatrix} \begin{bmatrix}
0 \\
\frac{a_2}{2}
\end{bmatrix} = \begin{bmatrix}
- \frac{a_2}{2} \sin \frac{\theta}{2} \\
\frac{a_2}{2} \cos \frac{\theta}{2}
\end{bmatrix}, $$
and the glide vector is
$$ \begin{bmatrix}
\cos \frac{\theta}{2} & - \sin \frac{\theta}{2} \\
\sin \frac{\theta}{2} & \cos \frac{\theta}{2}
\end{bmatrix} \begin{bmatrix}
a_1 \\
0
\end{bmatrix} = \begin{bmatrix}
a_1 \cos \frac{\theta}{2} \\
a_1 \sin \frac{\theta}{2}
\end{bmatrix}. $$
