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How can I find a fourth line $L$ that intersects three given lines $L_1$, $L_2$, $L_3$ in 3D space?

We can assume that $L_1$, $L_2$, $L_3$ are in "general position", so no two of them are coplanar, etc.

I'm not even sure that three lines is enough to uniquely define $L$, actually. If three lines is not enough, how many do I need?

The question is related to this one. Specifically, see idea #4 in my list of suggested approaches. It requires finding a line that intersects with a few given ones.

Edit:

Apparently, I need four lines, not three, to uniquely define $L$. So, how can I construct a fifth line that interesects four given ones?

I found this paper, and this one, but they are both difficult for me to read. Surely there must have been solutions before 2008, and, if so, I'm hoping that these are easier to understand.

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No, the $L$ will still not be unique, in the generic situation. –  Ted Shifrin Dec 15 '13 at 3:58

2 Answers 2

There is a one-parameter set of such lines, and their union is a ruled surface. The classic question from enumerative geometry is to ask (at most, over $\Bbb R$) how many lines meet four lines in general position. (These sorts of questions are best asked in projective space. But in Euclidean space by "general position" you rule out any sort of parallelism of planes and lines, etc.)

Here's a hint on how to get started: $L_1$ and an arbitrary point on $L_2$ determine a plane, and $L_3$ intersects that plane in a unique point.

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Thanks very much. If the four-line question is classical, can you point me to an easy-to-read reference, please. Hopefully something that's not too abstract. I know nothing about projective geometry. I live in a Euclidean space :-) I'm interested in computing the common line, so a constructive process would be welcome. –  bubba Dec 15 '13 at 3:26
    
I actually have a hands-on treatment of this in the abstract algebra book I wrote, but it is all set in projective space. I would suggest taking three general skew lines: say the $x$-axis, the line $(0,1,0) + t(1,0,1)$, and the line $(1,2,2)+s(1,0,2)$. Find the ruled surface explicitly (i.e., equation). Then take a general fourth line and see where it intersects that ruled surface. Now figure out, from the nature of the ruled surface, exactly what lines meet all four. If you get stuck in the middle, I'll help. :) BIG HINT: That ruled surface will be DOUBLY ruled. –  Ted Shifrin Dec 15 '13 at 3:52
    
I did some numerical examples. I set up parametric equations for the lines, and wrote equations to express the intersections. I found that my equations often had only complex solutions. In any case (real or complex), there were always two distinct solutions. Is this what you would expect? –  bubba Dec 15 '13 at 14:27
    
Well, that's the problem with the real number system. For the three lines I gave you, you get a familiar surface. Some lines in R3 do not intersect it (hence your complex roots), but lots do. And, yes, the surface is a quadric, so the general line will meet it in either 2 or 0 points. –  Ted Shifrin Dec 15 '13 at 14:44

Note that lines are determined by 4 numbers (a point with three coordinates which can be nirmalized so that one coordinate is 0, and a direction with 2 coordinates up to scaling). Note that requiring a line to intersect another given line gives a single equation (e.g. for the x-axis, the requirement is that y=0 when z=0 or vice versa), cutting the number of free coordinates down by one. So to get down to a single line one would expect to require it to intersect four lines.

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Well, to a finite number of lines ... :) –  Ted Shifrin Dec 15 '13 at 3:18

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