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 $l(x)$, $0<x<1$, 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.

*

*I believe I might have a counter example, but it's far from clear.

*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.

*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.

*If these conditions fail, are there stronger conditions which will
make it work?

 A: 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 $\alpha(t)$.)
