On a real projective plane $\mathbb{P}^2$, say we have two parallel lines; namely, $2x+y=0$ and $4x+2y+1=0$. What would be the equations of the projective lines, and how to find the point of their intersection?
Thanks a lot!
[Math] How to get the intersection of two parallel lines on projective plane $\mathbb{P}^{2}$
algebraic-geometryprojective-geometryprojective-space
Related Solutions
This is not true in ordinary plane geometry, and so it cannot be proved.
It is true, sort of, in a different form of geometry known as projective geometry, however.
As a quick intuitive introduction to projective geometry, imagine that you're standing on the ordinary Euclidean plane. Your head is about 2 meters above the plane, so when you look down you see whatever is drawn on the plane, stretching out to the horizon. Details on the plane right where you stand look large to you; the same details a long distance away will look small to you and be seen very close to the horizon.
Now it's a common enough experience that if we draw to parallel infinite lines on a plane, when we look at them from a point above the plane, it will look as if they meet at the horizon. We can decide to consider the points on the horizon line "equally real" as points on the plane. The horizon then becomes the "line at infinity" and parallel lines in the plane actually do meet at a point on the line at infinity. Then any two lines always meet. Every line in the plane meets the line at infinity at a point determined by its direction; two lines in the plane with different directions intersect in the plane itself, and two lines in the plane with the same direction both meet the line at infinity at the same point, and therefore meet there.
So far there's a strange thing about points on the horizon: they are at the "edge of the world", with plane below them but nothing above them. This is sort of untidy, and there are two ways to fix that. One is simply to decide to draw stuff on the sky. That leads to spherical geometry, an ancient and venerated area of study that is the foundation of astronomy. But it's not what today's lecture is about.
In projective geometry, we're standing on the Euclidean plane, but we're in a virtual reality constructed by a careless and/or lazy programmer. When we look in any direction, our VR helmet computes the infinite line through our head in the direction we're looking, and figures out where that infinite line intersects the plane below us. Whatever is at the plane there is what we see. So when we look at the sky, what we see is the plane behind our head! Only when we look in a perfectly horizontal direction does this procedure not work, but we posit that there's nevertheless some points to look at there, which form a "line at infintity" as before. Because we're not distinguishing between points in front of us and points behind us, there are only 180° of horizon all in all; if we turn 180° we will be looking at the same points at infinity.
(A different way of looking at it is to imagine that we took a copy of the ordinary Euclidean plane and lifted it up to hover 2 meters above our head, and then turned it by 180° about a vertical axis. Note that since we're looking at the sky plane from below, things in the sky will be the mirror image of the same thing if the turn around and look down at them instead).
In projective geometry any two different lines have exactly one point in common. If they are non-parallel lines in the original plane they will cross once on the ground, and we will also see that crossing in the sky -- but these are just two images of the same point in the projective plane. Two parallel lines will cross exactly once at the line at infinity -- again we see two images of that crossing when we turn around, but they are by definition the same point. And any line in the plane will cross the line at infinity once.
The main thing that makes this cool is that the line at infintity now has no special properties. If we tilt our head and forget which way was up and down, there is no way we can deduce from the geometrical properties of what we see which direction the "true" horizon lies. Projective geometry is about properties of figures that are invariant of rotating our view of the world. It is also about properties of figures that don't change as we walk around on the plane, or move our head closer to (or farther from) the plane. So it has no concept of scale or distance either.
These movements both preserve lines (a line stays a line in our view when we tilt our head and/or walk around), so "line" is a projective concept. They don't preserve circles -- if we stand above the center of a circle, it will look circular, but when we walk away from it it starts to look like an ellipse instead. So "circle" and "ellipse" are not projective concepts. On the other hand, surprisingly, "non-degenerate conic section" is a projective concept, and all non-degenerate conics are equivalent. An ellipse is simply a circle seen from a distance. A parabola is a circle that is tangent to the line at infinity. A hyperbola is a circle or ellipse that intersects the line at infinity twice.
Suppose that your eye is at the origin, and the "canvas" on which you draw is in front of your eye, and is the $z = 1$ plane. Then the point $(2, 3, 5)$ in space will project, in your "drawing" or "seeing" of the world, to the point $(2/5, 3/5, 1)$ in the $z = 1$ plane. In general, any point $(x, y, z)$ will project to $(x/z, y/z, 1)$.
Now consider two parallel lines; the first one consisting of all points of the form $(1, t, 4)$ and the second consisting of points $(3, t, 2)$. The projections of these to the drawing plane will consist of points of the form $(1/4, t/4, 1)$ and $(3/2, t/2, 1)$, respectively, i.e., they'll still be parallel.
Now look at the lines $(-1, 0, t)$ and $(1, 0, t)$. These project to $(-1/t, 0, 1)$ and $(1/t, 0, 1)$, which "meet" at the point $(0, 0, 1)$ when $t$ goes to $\infty$.
Does that help at all?
Best Answer
There are various approaches, but it is common to go to homogeneous coordinates. So the projective lines have homogeneous equations $2x+y=0$ and $4x+2y+z=0$. They meet where $z=0$ and $2x+y=0$, so at $(1,-2,0)$.
Because we are using homogeneous coordinates, each component of $(1,-2,0)$ can be multiplied by the same non-zero constant.
Remark: I do not know what notation is used in your course, so used standard old-fashioned notation. Your version may use equivalence classes. If so, it should not be difficult to translate.