Composition of Reflections If R1 and R2 are two planar reflections corresponding to the lines d1 and d2. What is the necessary and sufficient condition on the lines d1 and d2 such that R1oR2=R2oR1 ?
 A: $R_1(p)=p$ if and only if $p$ is on $d_1$. 
If $p$ is on $d_1$, then from $R_1\circ R_2(p)=R_2\circ R_1(p)$ we deduce $R_2(p)$ is a fixed point of $R_1$, so $R_2(p)$ is on $d_1$. 
This says the reflection of the line $d_1$ in the line $d_2$ is the line $d_1$. A little thought about the geometry tells you this can happen if, and only if, $d_1$ is $d_2$, or $d_1$ is perpendicular to $d_2$. 
A: Let $d_i$ be $\{x:(x-p_i)\cdot u_i=0\}$ where $\|u_1\|=1$. Then reflection across $d_i$ would be
$$
r_i(x)=x-2(x-p_i)\cdot u_i u_i\tag{1}
$$
Composition yields
$$
\begin{align}
r_1\circ r_2(x)
&=(x-2(x-p_2)\cdot u_2 u_2)-2((x-2(x-p_2)\cdot u_2 u_2)-p_1)\cdot u_1u_1\\
&=x-2(x-p_1)\cdot u_1u_1-2(x-p_2)\cdot u_2u_2+4(x-p_2)\cdot u_2u_2\cdot u_1u_1\tag{2}
\end{align}
$$
and
$$
\begin{align}
r_2\circ r_1(x)
&=(x-2(x-p_1)\cdot u_1 u_1)-2((x-2(x-p_1)\cdot u_1 u_1)-p_2)\cdot u_2u_2\\
&=x-2(x-p_1)\cdot u_1u_1-2(x-p_2)\cdot u_2u_2+4(x-p_1)\cdot u_1u_1\cdot u_2u_2\tag{3}
\end{align}
$$
Equating these, we need
$$
u_1\cdot u_2(x-p_2)\cdot u_2u_1=u_1\cdot u_2(x-p_1)\cdot u_1u_2\tag{4}
$$
If $u_1\cdot u_2\ne0$, then $u_1=\pm u_2$ (they are parallel unit vectors). Since $u_2$ and $-u_2$ define the same line with $p_2$, let $u_1=u_2$. Thus, $(4)$ implies that
$$
(p_1-p_2)\cdot u_1=0\tag{5}
$$
Thus, $(x-p_1)\cdot u_1=(x-p_2)\cdot u_2$ and the lines must be the same.
If $u_1\cdot u_2=0$, then $(4)$ is satisfied trivially.
Therefore, it is necessary that the lines must be the same or perpendicular. Obviously, this is also sufficient.
A: If the lines $d_1,d_2$ are parallel at distance $d$ apart, then $R_2 \circ R_1$ (i.e. $R_1$ done first ) gives a translation through a vector of length $2d$ pointing from line $d_1$ toward line $d_2$ (and perpendicular to the two lines). So in this case one must have $d_1=d_2$ for the two compositions to be the same.
If the lines $d_1,d_2$ intersect, making the angle $\theta$, then $R_2 \circ R_1$ gives a rotation through the angle $2\theta$ in the direction of rotation from $d_1$ toward $d_2$. So in this case the two orders give the same map when $2\theta=-2\theta$ mod $2\pi$, which means $4\theta$ is $0$ mod $2\pi$ so that $\theta=\pi/2$ and the lines are perpendicular. 
