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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

3 Answers 3

up vote 2 down vote accepted

$R_1(p)=p$ 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

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