There is a unique quadric through three disjoint lines

There is a classical exercise that three disjoint lines in $\mathbb{P}^3$ are contained in a quadric surface $Q$. The existence is trivial. Every quadric in $\mathbb{P}^3$ is determined by nine coefficients and if three distinct points of a line lie on $Q$ then the whole line lie on $Q$. Thus we obtain the system of 9 linear conditions on $Q$. But why such a quadric is unique? To be more precise, why these linear equations are linearly independent?

Solution 1:

There exists exactly one quadric containing three mutually disjoint lines in $\mathbb P^3$:
Indeed you may assume that these lines are

$$L_1: x_0=x_1=0\quad \quad L_2:x_2=x_3=0\quad\quad L_3: x_0=x_2 ,\:x_1=x_3$$

and then the unique quadric containing these three line $L_i$ has the equation

$$x_0x_3-x_1x_2=0$$

This is easy to check by writing that the polynomial

$$ax_0^2+bx_0x_1+\dots+jx_3^2$$

vanishes on the $L_i$'s.

Solution 2:

Mutually skew lines $a,b$ and $c$ lie in the same ruling, and the quadric is the union of the lines of the other ruling. There exists exactly one line containing a given point $p \in a$ and intersecting $b$ and $c$, so the quadric is unique as the union of all the lines intersecting with $a,b$ and $c$.