Line intersecting three (or four) given lines 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.
 A: 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.
A: To build up on Ted Shifrin's comment, assuming you have four skew lines $L_1$, $L_2$, $L_3$ and $L_4$, a costructive and an easy way to find a fifth line that intersects all the four lines would be as follows:
Construct two ruled surfaces (hyperboloids or hyperbolic paraboloids) uniquely defined by $L_1$, $L_2$, $L_3$ and $L_1$, $L_2$, $L_4$ that form the reguli of the quadrics $H_{123}$ and $H_{124}$. Find out the intersection between those two quadrics which definitely contains $L_1$ and $L_2$. Along with these two lines, the intersection must also contain two other lines (from the complementary reguli of the quadrics), say $M$ and $N$ that intersect all four given lines. Here's an example:
$$L_1=(-3,-5,-2)+ t(1,0,0)$$
$$L_2=(-3,-5,2)+ t(0,1,0)$$
$$L_3=(3,5,0)+ t(0,0,1)$$
$$L_4=(3,10,0)+ t(1,0,1)$$
$$H_{123}=4\,xy-10\,xz+6\,yz-60$$
$$H_{124}=4\,xy-15\,xz+4\,yz+15\,{z}^{2}-10\,x+4\,y-25\,z-130$$
The intersection of these two quadrics yields 4 lines (I did it by finding the Gröbner basis of the ideal $\langle H_{123}, H_{124} \rangle$ using lexicographic ordering and factoring the first polynomial, there are other ways to do it and almost all computational environments have a Gröbner basis command if you want to check it out):
$$L_1: z+2=y+5=0 $$
$$L_2: -z+2=x+3=0 $$
$$M: x+ \left( -\frac {1}{10}\,\sqrt {73}-{\frac {7}{10}} \right) y+\frac {1}{2}\,\sqrt {73}+
\frac {1}{2}= 5\,z\sqrt {73}+10\,\sqrt {73}+8\,y-55\,Z-70=0 $$
$$N: x+ \left( \frac {1}{10}\,\sqrt {73}-{\frac {7}{10}} \right) y-\frac {1}{2}\,\sqrt {73}+\frac {1}{2}= 5\,z\sqrt {73}+10\,\sqrt {73}-8\,y+55\,Z+70=0 $$
where $M$ and $N$ intersect $L_i, i=1,2,3,4$.
A: 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.
A: I occasionally need to actually write down the formulas for the lines in $\mathbb{R}^4$ which meet two given lines. Here is how I do it.
Make a projective change of coordinates so that $L_1$ is the $z$-axis and $L_2$ is the intersection of the plane at infinity with the $xy$ plane. Parametrize $L_3$ and $L_4$ as $(a z+ b, c z + d, z)$ and $(a' z'+ b', c' z' + d', z')$. 
The line through $(a z+ b, c z + d, z)$ and $(a' z'+ b', c' z' + d', z')$ will meet $L_2$ if and only if $z=z'$, and will meet $L_1$ if and only if the vectors $\left[ \begin{smallmatrix} a z +b \\ c z + d \end{smallmatrix} \right]$ and $\left[ \begin{smallmatrix} a' z +b' \\ c' z + d' \end{smallmatrix} \right]$ are proportional. In other words, we want
$$\det \begin{bmatrix} az+b & a' z + b' \\ cz+d & c' z+d' \\ \end{bmatrix} = 0.$$
This is a quadratic equation which can be solved for $z$. 
