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If I have three lines $a,$ $b$ and $c$ in the euclidean 3D space, which are pairwise non-coplanar, is there always a fourth line $x$, that intersects theses three lines?

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Label the three lines $\ell_1$, $\ell_2$, and $\ell_3$. They cannot intersect for then they would be co-planar. Since lines $\ell_1$ and $\ell_2$ do not intersect, they have points where they are closest and the line $m$ connecting those two points will be perpendicular to both $\ell_1$ and $\ell_2$. In fact, perpendicular to this line $m$ are two parallel planes: $\mathcal{P}_1$ which contains $\ell_1$ and $\mathcal{P}_2$ which contains $\ell_2$.

Now pick any point $A$ of line $\ell_3$ not in either plane. The point $A$ together with the line $\ell_2$ defines a plane $\mathcal{Q}$ that contains them. This plane, since it intersects plane $\mathcal{P}_2$, must intersect the parallel plane $\mathcal{P}_1$. Moreover, the line $r$ formed by the intersection of planes $\mathcal{Q}$ and $\mathcal{P}_1$ is parallel to line $\ell_2$. Note that lines $r$ and $\ell_1$ are both on $\mathcal{P}_1$ and cannot be parallel. If they were, $\ell_2$ would also be parallel to $\ell_1$, and $\ell_1$ and $\ell_2$ would, therefore, be co-planar. So, lines $r$ and $\ell_1$ must intersect at some point $B$. If you connect points $A$ and $B$ with a line, it will connect point $A$ on $\ell_3$ passes through $\ell_2$ and connect to $B$ on $\ell_1$.

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This proof is the easiest to understand for me. Thank you. –  FUZxxl Feb 23 '11 at 16:28
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The answer is yes. Three lines in general position determine a unique hyperboloid of one sheet (since they determine a ruling, or a one parameter family of lines on a hyperbola). But there is a second ruling of the same hyperboloid; every line in this ruling meets the original three lines. So Ross is correct, there is a one parameter family of solutions (in the shape of a hyperboloid).

If I'm ever not up to my eyeballs in homework, I'll make a picture of this and post it.

NOTE: Questions of this sort can be addressed in generality using the Schubert calculus, which is roughly a part of enumerative geometry.

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I think so, but have not finished the proof. I would argue that if you pick two of the lines you can find an affine transformation to make the them (x,0,0) and (0,y,1) where x and y parameterize the lines. The general line through these points is (x-xt,yt,t) where t parameterizes this line. The third line is (u+u's, v+v's, w+w's). So to get an intersection, you have three equations in four unknowns: x,y,t,s. There ought to be a one-parameter family of solutions unless something goes wrong. You have to show that non-coplanar is enough to make sure nothing goes wrong.

Added: the equations are $\begin {align} x-xt&=u+u's \\ yt&=v+v's \\ t&=w+w's \end{align}$

where $u,u',v,v',w,w'$ are the fixed parameters of the third line. So you can pick $s$ at will, calculate $t$ from $w, w'$, calculate $y$ from $v, v'$ as long as $t \neq 0$ and calculate $x$ from $u, u'$ as long as $t \neq 1$. This gives only two values of $s$ that are not acceptable. So we have the promised one dimensional family of solutions.

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Actually, this was not homework or something else. I thought about this in context of another problem and found out, that this can be false if the lines might be coplanar. But in this case I can't think of any reason why there should be no such fourth line. This is a good idea for a proof. I'm going to think about it. –  FUZxxl Feb 12 '11 at 15:44
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