I'm learning projective geometry and need help with the following problem :

Consider the affine plane $\mathbb{A}^2$ as the plane $\{z = 1\} \subset \mathbb{R}^3 - \{0\}$. $(1)$ Find the point of intersection $P \in \mathbb{RP}^2$ of the lines of $\mathbb{A}^2 : x + 4y = 3$ and $2x - 5y = -7$. $(2)$ Find the point of intersection $Q \in \mathbb{RP}^2$ of the lines of $\mathbb{A}^2 : 2x - 3y = 6$ and $-4x + 6y = 13$ (we want the homogeneous coordinates of $P$ and $Q$).

There are quite a few things which I don't understand here. To begin with, I'm going to share my understanding of the problem. Please do correct me if anything is incorrect.

First of all, an element of the projective plane $\mathbb{RP}^2$ is a line passing through the origin of $\mathbb{R}^3$. Therefore, the point $P$ we are looking for in $(1)$ is actually a line passing through the origin of $\mathbb{R}^3$. Since every "point" (i.e. line) of $\mathbb{RP}^2$ is entirely determine by homogeneous coordinates $[X: Y: Z]$ up to some nonzero scalar $k$, these are precisely the coordinates we are looking for.

Also, the affine plane $\mathbb{A}^2$ is just the $xy$-plane of $\mathbb{R}^3$ move up one unit in the $z$ direction. So if I'm sitting in the origin and look up in the $z$ direction, $\mathbb{A}^2$ is like a giant window where points are projected. Therefore the point $P$ I'm looking for must have homogeneous coordinates of the form $[X:Y:1]$. Am I thinking correctly until here?

One thing which I don't understand is why the lines $x+4y=3$ and $2x-5y=-7$ actually belongs in $\mathbb{A^2}$? If I use a graphing calculator I can actually see that those lines don't belong to the $z = 1$ plane yet the exercise seem to tell they actually belong to this plane.

Unfortunately I have no idea how to solve this problem. Possibly someone can clarify my misunderstandings and show me how to solve $(1)$ so I get try $(2)$ on my own.


A line in $\mathbb R^3$ can’t be represented by a single linear equation. If you plot, say $x+4y=3$ in $\mathbb R^3$, you should see a plane and not a line. If your calculator shows you a line, then it’s either not plotting in 3-d or it’s making some assumptions about the $z$-coordinates of the points that satisfy the equation. To find out how to plot this plane correctly, you’ll need to check the user’s manual.

When you embed $\mathbb A^2$ into $\mathbb R^3$ as the plane $z=1$, you tacitly add the latter equation to whatever planar equations you’re working with. So, using the same example as above, the line given by the equation $x+4y=3$ in $\mathbb A^2$ becomes the system of equations $$\begin{align}x+4y&=3\\z&=1\end{align}$$ in $\mathbb R^3$. The solution to this system consists of the line that is intersection of the two planes described by the individual equations, and it should be fairly obvious from that that this line does lie in the plane $z=1$.

As far as the homogeneous coordinates of the solutions go, any non-zero multiple of $[x:y:1]$ represents the same point, so that isn’t really going to help you find a solution. In fact, a very common way to solve this intersection problem will usually produce a homogeneous coordinate vector for which the last component is not zero.

I should think that whatever text or online materials you’re studying from would already have covered how to solve these problems, so I suggest going back over the material to see if you missed anything. However, you ought to be able to work this out yourself with a bit of thought.

If points in $\mathbb{RP}^2$ correspond to lines through the origin in $\mathbb R^3$, what might lines in $\mathbb{RP}^2$ map to? Consider a line in the $z=1$ plane. If you draw all of the lines through the origin that intersect this line, what do you get? A plane through the origin (less a line in the $x$-$y$ plane, but we can accommodate that, too). So, finding the intersection of a pair of lines in $\mathbb{RP}^2$ becomes finding the intersection of a pair of planes through the origin in $\mathbb R^3$. Unless the planes coincide, they intersect in a line through the origin, which maps to a point in $\mathbb{RP}^2$—just what we want.

What about that line in the $x$-$y$ plane that we left out? Imagine two planes that aren’t the $x$-$y$ plane and intersect in this line. Their intersections with the $z=1$ plane are a pair of parallel lines. In fact, the intersection of every plane other than the $x$-$y$ plane with $z=1$ is a family of parallel lines. This leads us to identify the line in the $x$-$y$ plane with the point at infinity at which all of these parallel lines (in $\mathbb{RP}^2$) intersect. If we include this point at infinity, there’s no longer a “hole” in the plane that models a line in $\mathbb{RP}^2$. Moreover, the $x$-$y$ plane contains all of these lines, so it must correspond to the line at infinity in this model of $\mathbb{RP}^2$.

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    $\begingroup$ I can't thank you enough for this answer. Your explanation was extremely helpful. I could not solve the exercise immediately after reading your answer (I had to go back to my notes) but your explanation helped me understand some very important notions of the projective plane which I previously did not understand. Interestingly for question $(2)$, the lines are actually parallel and they meet at a point at infinity with homogeneous coordinates $[3/2:1:0]$. For $(1)$, the lines intersect at the point $P$ with homogeneous coordinates $[-1:1:1]$. Thanks again for your help. $\endgroup$ – user347616 Sep 6 '17 at 21:27
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    $\begingroup$ @Elix Glad I could help. BTW, the “common way” to which I alluded is to take the cross product of the lines. Dually, the line through two points can be found by taking their cross product (all in homogeneous coordinates, of course). For the two questions, this is $[1:4:-3]\times[2:-5:7]=[13:-13:-13]$ and $[2:-3:-6]\times[-4,6,-13]=[75,50,0]$. It’s an instructive exercise to prove why these cross products work. $\endgroup$ – amd Sep 6 '17 at 22:01

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