Is this true:
“Suppose that a line is chosen from within a plane, and the plane is rotated 180° clockwise around this line. This will mirror the plane, reflecting its points such that the points leftward from the line become rightward from it and vice-versa. But suppose that the rotation is done counter-clockwise instead. Then the same thing happens, and there is no difference.”
Yes – it’s absolutely true. Here’s why:
-
A 180° rotation about a line in the plane acts like a mirror reflection across that line.
- Take for example the horizontal $x$–axis in the $xy$–plane. If you rotate the plane 180° about that axis (in three‑dimensional space), any point $(x,y,0)$ goes to $(x,-y,0)$. That is exactly the effect of reflecting across the $x$–axis.
- More generally, for any line $L\subset P$, the restriction to $P$ of the half‐turn about $L$ is the involution “reflect across $L$.”
-
“Clockwise” versus “counter‑clockwise” makes no difference at 180°.
- Angles are taken modulo 360°, so rotating by $+180°$ or by $-180°$ lands you in the same place:
$$ R_{L}(+180°)\;=\;R_{L}(-180°). $$
- In other words, a half‑turn is its own inverse, and the sign convention for direction drops out.
-
Orientation flip on the plane.
- In three dimensions the rotation by 180° about a line lying in a plane reverses the plane’s “sidedness” (it sends a chosen normal vector to its negative). That’s exactly what a reflection in the line does: it reverses orientation in the plane.
So rotating $P$ by 180° around $L$ is the same map as reflecting $P$ across $L$, and whether you spin “clockwise” or “counter‑clockwise” your half‑turn is identical.
No comments:
Post a Comment