# Determine lines intersecting four skew lines in $\mathbb{P}^3$

Let $l_1, l_2, l_3, l_4$ be four skew lines in a projective space $\mathbb{P}^3$ (meaning $l_i \cap l_j = \varnothing \;\forall i≠j$).

Let $R = \{ r : r \cap l_i ≠ \varnothing,\;i=1,...,4 \}$ be the set of all lines that intersect each one of the four skew lines.

Let $r, r' \in R$ and $p_i = r \cap l_i,\;p_i' = r' \cap l_i,\;i = 1,...,4$ be the points of intersection respectively.

Show that for the cross ratios it is true that:

a) $(p_1,p_2,p_3,p_4) = (p_1',p_2',p_3',p_4') \Rightarrow \#R = \infty$

b) $(p_1,p_2,p_3,p_4) ≠ (p_1',p_2',p_3',p_4') \Rightarrow R = \{r,r'\}$

Any help is very much appreciated!

To be honest i fail already at showing that $R$ is not the empty set and there are at least two lines intersecting all of the original four skew lines. Also i have the feeling that it might be necessary to understand why (geometrically) in the one case there are infinitely many lines and in the other just two - which i've tried but can not come up with the intuition.

The lines intersecting three skew lines form a regulus in three-space. This is a ruling of a non-singular quadric. A fourth line will in general meet the quadric in two points (over an algebraically closed field). In this case there will be exactly two lines meeting all four original lines. The other case is where the fourth line is tangent to the quadric. Either it meets in one point (when there is exactly one element in $R$) or it lies within the quadric (then $R$ is infinite).