# 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 ?

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Consider what happens to a point on $d_1$. – Gerry Myerson Jan 29 '13 at 11:49
it is a fix point – Mathematician Jan 29 '13 at 15:52
It is a fixed point of $R_1$. So what happens when you apply the two compositions to it? – Gerry Myerson Jan 29 '13 at 23:32

$R_1(p)=p$ 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 $d_2$.

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The more one reflects on this problem, the simpler the answer gets :-) +1 – robjohn Jan 30 '13 at 23:19

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.

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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.

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But reflection in x axis commutes with reflection in y axis. Composed either way it gives the antipodal map $f(x,y)=(-x,-y)$. – coffeemath Jan 29 '13 at 14:16
Ah, yes. I factored out the $u_1\cdot u_2$ and forgot to consider what the case $u_1\cdot u_2=0$ gave. – robjohn Jan 29 '13 at 14:31
I think this can be done using another way by considering the two effects as to whether the lines meet or are parallel. +1 . – coffeemath Jan 29 '13 at 14:37
That looks good, too. +1 – robjohn Jan 29 '13 at 14:40