# Problem about $\mathbb{P}^3(K)$

Show that four skew lines in $\mathbb{P}^3$ have two transversals in common.

I know that exist a quadric which contains three of the four lines....but i'm stuck

EDIT:

If the skew lines are $a,b,c,d$. Let $Q$ the quadric such that $a,b,c \subset Q$. Then $d \cap Q=\text{{two point}}$? Why?

• – MvG May 17 '14 at 13:24
• I have already showed it – Mark May 17 '14 at 13:30