Equation of a regulus How does one compute the equation for the points on a regulus given the equations of 3 mutually skew lines that define it?  I saw the definition of a regulus in a geometry book but I had trouble computing its equation given 3 skew lines, and searching the internet does not seem to help. 
 A: I presume this argument is written up in some of the classic books, most likely Griffiths and Harris' Principles of Algebraic Geometry, but I don't have access to it right now, so instead let me sketch a solution below.
For simplicity, let me answer the question in the case when the three lines have particularly simple descriptions; you can then adapt the argument to the general case. So let's assume the lines have the form
$$ \begin{align*} L_1 &= [X,Y,0,0]; \\
L_2 &= [0,0,Z,W]; \\
L_3 &= [X,Y,-X,-Y]. \end{align*} $$
(Indeed, by applying projective transformations you can take any pair of skew lines to $L_1$ and $L_2$ respectively, but in general $L_3$ will be more complicated. That won't change the argument much, though.)
Now take a point $p=[a,b,c,d]$ in projective space, let's see what the condition is on the coordinates so that there is a line through $p$ touching all these lines. We can assume that $p$ doesn't lie on any of the lines.
Assume there is such a line, and suppose it intersects $L_1$ in the point $q=[e,f,0,0]$. Then a general point on this line has coordinates of the form
$$ [sa+te,sb+tf,sc,sd] $$
for some $s$ and $t$, not both zero. Evidently this must intersect $L_2$ in the point $[0,0,c,d]$, so there must be values of $s$ and $t$ such that
$$ sa+te=0=sb+tf.$$
Juggling this a bit, we get the relation $af=eb$, which is to say that $q=[a,b,0,0]$. That is, there is a unique line through $p$ that intersects both $L_1$ and $L_2$, namely the line $\Lambda_p$ joining $[a,b,c,d]$ to $[a,b,0,0]$. It's not hard to see that $\Lambda_p$ is defined by the equations 
$$aY=bX, \, cW=dZ.$$
Finally, we want to determine when such a line intersects $L_3$: to do this, use the fact that on $L_3$ we have $Z=-X,\, W=-Y$. Plugging these into the previous pair of displayed equations we get the equations
$$aY=bX, \, cY=dX$$
which are consistent precisely when $ad-bc=0$. So we end up with a quadric relation on the coordinates of $p$: that is, the points lying on these lines sweep out a quadric surface in $\mathbf{P}^3$.  
