# Condition for every line through a line segment $S$ to be perpendicular to $S$?

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 $$l(x)$$, $$0, will also be perpendicular to $$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?
• @MishaLavrov, No. this is in 3 dimensional space. I'm correcting my statement to make that clear. Commented Aug 31, 2021 at 4:23
• In three dimensional space the statement fails. Commented Aug 31, 2021 at 4:26
• 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 $S$. Commented Aug 31, 2021 at 4:31
• 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 In any case, my statement in (3) was flat-out wrong and I've withdrawn it; thanks for the correction. Commented Aug 31, 2021 at 4:38
• @MishaLavrov, if I add the condition of continuity, your objection about the case where $l(0)$ parallel to $l(1)$ seems to vanish. Commented Aug 31, 2021 at 5:51

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 $$\alpha(t)$$.)