In 3D, reflecting $\pi$ about a line is the same as reflecting about a plane followed by an enlargement

Question

Suppose I have a line through the origin $L$ and a point $P$. Rotating $P$ $\pi$ (rad) about $L$ is equivalent to reflecting $P$ in the plane through the origin to which $L$ is normal to a point $P'$ then stretching $P'$ scale factor $-1$ (i.e. $P'\rightarrow -P'$).

This is true and can be verified using matrix methods. I would like to know whether my following geometric reasoning is correct. I am also open to other (geometric) ideas to proving this, but if mine is correct I can't imagine anything being much simpler. I would also be interested in any opinions whether some of what I say, particularly beyond $x=y=d$, is redundant.

Let $Q$ be the point where $PP'$ intersects the plane $\Pi$(which is as was described previously). Then $OQP'$ and $P''P'P$ are similar because(by definition) $PQ=PP'$ and they share a right angle. Therefore $x=y=d$. The reflections we performed are preservative of length, that is $|OP|=|OP'|=|OP''|$. Let $S$ be the point where $L$ intersects $PP''$. Because $x=y$ and $|OP|=|OP'|$ and $\angle OLP''=\angle OLP = \frac{\pi}{2}$. the triangles are congruent and coplanar, i.e. reflections of each other.

Answer 1

Your proof looks good.

Here's another, that may or may not be simpler depending what geometric concepts are primitive: Let $\Pi_{1}$ be the plane containing $P$ and $L$, and let $\Pi_{2}$ be the plane containing $L$ and perpendicular to $\Pi_{1}$. Since

• The composition of reflections in perpendicular planes is the half-turn about their line of intersection,
• The product of reflections in three mutually-perpendicular planes through $O$ is scaling by $-1$,

we have (letting $\Pi$ also denote reflection in the plane $\Pi$) $$ P'' = -P' = \Pi_{1}\Pi_{2}\Pi P' = \Pi_{1}\Pi_{2} P. $$