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In 3 dimensional space, given a line segment $S$ which has a line $l(0)$ passing through one endpoint at right angles and another, $l(1)$, passing through the other endpoint at right angles, such that $l(0)$ is not parallel to $l(1)$, prove that the lines passing through each point $l(x)$, $0<x<1$, will also be perpendicular to $S$ on the assumption that none of the lines through $S$ intersect. It is assumed that the $l(x)$, $0\le x \le 1$, are distributed continuously.

  1. I believe I might have a counter example, but it's far from clear.
  2. Note that this is true for a ruling line of the one-sheeted hyperboloid (or for those of a hyperbolic paraboloid) and its intersection of its dual ruling lines.
  3. Of course, this condition could be weakened to the two end lines intersecting $S$ at any pair of congruent angles as long as their concave sides are facing the same directions.
  4. If these conditions fail, are there stronger conditions which will make it work?
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  • $\begingroup$ @MishaLavrov, No. this is in 3 dimensional space. I'm correcting my statement to make that clear. $\endgroup$ Commented Aug 31, 2021 at 4:23
  • $\begingroup$ In three dimensional space the statement fails. $\endgroup$
    – subrosar
    Commented Aug 31, 2021 at 4:26
  • $\begingroup$ In 3 dimensional space, $\ell(0)$ and $\ell(1)$ practically don't constrain $\ell(x)$ for $0<x<1$. For example if $S$ is the segment from $(0,0,0)$ to $(1,0,0)$, we could have segments through $\ell(0)$ and $\ell(1)$ lie in the $(x,y)$-plane while the segments through $\ell(x)$ all lie in the $(x,z)$ plane, making any angle of your choice with $S$. $\endgroup$ Commented Aug 31, 2021 at 4:31
  • $\begingroup$ Yes. I succumbed to the fallacy that people can read my mind. I edited my statement. In any case, my statement in (3) was flat-out wrong and I've withdrawn it; thanks for the correction. $\endgroup$ Commented Aug 31, 2021 at 4:38
  • $\begingroup$ @MishaLavrov, if I add the condition of continuity, your objection about the case where $l(0)$ parallel to $l(1)$ seems to vanish. $\endgroup$ Commented Aug 31, 2021 at 5:51

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A counterexample: suppose that the line segment is on the $x$-axis, starting at $(0,0,0)$ and ending at $(\pi,0,0)$. Let $\alpha(t) = \frac\pi4 + \frac12|t - \frac \pi2|$. Through the point $(t,0,0)$, draw the line with unit direction vector $(\sin \alpha(t), \cos t \cos \alpha(t), -\sin t \cos \alpha(t))$.

This vector is orthogonal to the vector $(0, \sin t, \cos t)$, as is the vector $(1,0,0)$, so the entire line we draw lies in the plane with equation $(x,y,z) \cdot (0, \sin t, \cos t) = 0$. As $t$ varies, these planes also vary, repeating with period $\pi$, and they only intersect on the $x$-axis. So as long $t_1 \not\equiv t_2 \pmod \pi$, the line through $(t_1,0,0)$ and the line through $(t_2,0,0)$ cannot intersect; they're skew.

This leaves us with complete freedom to choose the angle that the line makes with the $x$-axis, which we take advantage of: the angle is actually $\alpha(t)$. This is $\frac \pi2$ when $t=0$ or $t=\pi$, but not otherwise.

(I originally wanted to make $\alpha(t) = t + \frac\pi2$, which would be nicer, but the problem we run into is that $\alpha(t)$ should never be $0$. In that case, we end up drawing a line through the line segment, intersecting all the other lines. There's still probably a nicer choice of $\alpha(t)$.)

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