# Intersection of 4 lines in projective space

I was reading about projective geometry and I came across this exercise:

Let $L_1$, $L_2$, and $M$ be three lines of $\mathbb{P}_\mathbb{R}^3$ such that $L_1 \cap M = \emptyset$, $L_2 \cap M = \emptyset$ and $L_1 \cap L_2 = \{P\}$ (a point). Show how to find all lines of $\mathbb{P}_\mathbb{R}^3$ meeting $L_1$, $L_2$, and $M$.

I am confused about how to proceed. So far the definition that I know of a line in projective space is as the span of two points: if $P = [x_0:...:x_n]$ and $Q = [y_0:...:y_n]$ then $<PQ> = \{[ \lambda x_0 + \mu y_0:...:\lambda x_n + \mu y_n] | (\lambda,\mu) \in \mathbb{R}^2\setminus\{(0,0)\} \}$. Is there any better definition that I could use, or an alternative way to visualise this problem? (I got this problem from an old past paper so it is likely that I may not have covered enough material)

Any help would be very appreciated, thank you very much!

• One thing that might help you get started is that if two lines intersect, then they’re coplanar, so any other line that intersects $L_1$ and $L_2$ must also lie in that plane. – amd Apr 20 '18 at 23:02
• @amd, this is not true. Consider the $x-y$ plane in $\mathbb{A}^3(\mathbb{R})$ with the two lines $L_1$ and $L_2$ as $x$ and $y$ axis, respectively. Then $z$ axis intersects $L_1$ and $L_2$ at the origin $(0,0,0)$ while it doesn't lie in the $x-y$ plane. Concisely speaking, the point of intersection with $L_1$ and $L_2$ may be same. – user2902293 May 14 '18 at 15:48
• @user2902293 Fair point. This adds a second pencil of lines to the solution. – amd May 17 '18 at 6:10

[Updated to include the second pencil of lines pointed out by user2902293.].

I’m going to use the language of joins and meets, which I hope you can translate into whatever formalism is used by the material that you’re studying.

$L_1$ and $L_2$ define a unique plane $\Pi$ that can be described as their join, i.e., all non-zero linear combinations of the two lines. All lines on $\Pi$ intersect both $L_1$ and $L_2$, therefore $M$ does not lie on $\Pi$.

There are two possibilities for a line that intersects $L_1$ and $L_2$: it is either coincident with them, passing through their intersection $P$, or its intersections with them are distinct and it is coplanar with them. In the first case, for every point $Q$ on $M$ we have a line $\mathbf{\text{Join}}[P,Q]$ that intersects $L_1$, $L_2$ and $M$. In the second case, for a line in $\Pi$ to intersect $M$, it must include the intersection of $M$ with this plane, which is a single point that does not lie on either $L_1$ or $L_2$. Therefore, all of the lines of $\mathbf{\text{Join}}[L_1,L_2]$ that pass through $\mathbf{\text{Meet}}[M,\mathbf{\text{Join}}[L_1,L_2]]$ also meet the criteria of the problem.

For a slightly different way to describe the second set of lines, we can use the fact that $M$ defines a pencil of planes $\Pi_\alpha$ that all contain $M$. We know from above that $\mathbf{\text{Join}}[L_1,L_2]$ is not one of these planes so $\mathbf{\text{Meet}}[\Pi_\alpha,\mathbf{\text{Join}}[L_1,L_2]]$ is a line that passes through the intersection of $M$ and the plane of $L_1$ and $L_2$.

• So the plane $\Pi$ is equal to $\langle L_1, L_2 \rangle$? – The Coding Wombat Mar 28 at 17:49
• And for the second case, could I visualize this by having two lines in the same plane in a cross shape, and a third line perpendicular to these first two lines, but not intersecting them? So if the first lines were on a sheet of paper, the third line would go through the paper? – The Coding Wombat Mar 28 at 17:58
• @TheCodingWombat For the first comment: correct. For the second, that’s a good mental picture to start from, but since we’re working in a projective space instead of a Euclidean space, you also have to consider lines $M\parallel\Pi$—they, too, intersect $\Pi$ in a single point. – amd Mar 28 at 19:04