# Projection of 5 skew lines

Given five skew lines, is it possible to find a point $P$ and a plane $\pi$ such that the projections of the five lines from $P$ onto $\pi$ intersect in the same point $Q$? [editet: rewritten clearly, it wasn't clear that the 5 lines are given a priori]

-

[[Edited to revert an error I noticed right after posting and which Ross Millikan & joriki subsequently plugged up.]]

If the five lines are fixed a priori then you generally won't be able to find a point and plane to make that possible. It's equivalent to finding a line $L$ that passes through all five given ones. Let's say we're working in $\mathbb{R}^m$, and parametrize each of the five lines with $\{ \vec{n}_i + t \vec{p}_i | t\in \mathbb{R} \}$, indexing $i=1,2,3,4,5$. Each vector has $m$ components. Say the desired line $L$ has parametrization $\vec{a}+t\vec{b}$; without loss of generality we can fix $\|\vec{a}\|=1$ and hence affix it to the $(m-1)$-dimensional sphere, likewise we can fix $\vec{b}$ to be orthogonal to $\vec{a}$ (namely, it is the displacement from the origin), so that it belongs in the $(m-1)$-dimensional hyperplane perpendicular to $\vec{a}$, thereby making them jointly $2(m-1)$-dimensional objects with that number of variables needed to represent them.

Then we are tasked with the system of equations

$$\vec{n}_i + t_i \vec{p}_i = \vec{a} + s_i \vec{b}, \quad i=1,2,3,4,5$$

Notice that this amounts to $5m$ (second-degree polynomial) equations in $2(5)+2(m-1)=2m+8$ unknowns, therefore it is overdetermined for $m\ge 3$. This means it is only a fortuitous case when such a line $L$ can be found.

An alternative, geometric reasoning is as follows: selecting any three of the given lines will determine a hyperbolic paraboloid (set of all points of all lines that go through all three), and it is easily possible for one of the other two lines to not even touch this surface. Even if one does, almost surely it will only intersect the surface a finite number of times, and then almost surely the fifth line would not touch any of these few points. So ultimately we find that it is in general impossible to find a line passing through any three of the given lines that also passes through the other two.

-
Actually it's $2m+10$ unknowns so my first line of reasoning looks to be invalid in dimension $m=3$ as it currently stands... –  anon Jul 12 '11 at 12:17
so, the answer to the question could be positive? –  zar Jul 12 '11 at 12:32
@anon: you lose one unknown as you can force $\vec{a}$ to be perpendicular to $\vec{b}$ by a shift of the $s_i$. This still doesn't rule out $m=3$ –  Ross Millikan Jul 12 '11 at 12:46
In fact you lose two, since the length of $\vec{b}$ doesn't matter. That makes $14<15$, so it's not possible in general. –  joriki Jul 12 '11 at 12:50
@joriki: good point. It seemed like it shouldn't be possible. –  Ross Millikan Jul 12 '11 at 17:53