Cross and Intersect

[Mr. Bland]

The coordinates of a line in two dimensions may be given in this form:

[MATH]ax + by + c = 0[/MATH]​

If [MATH]a[/MATH] and [MATH]b[/MATH] take some vector that is perpendicular to the line in question, then it is a simple matter to solve for [MATH]c[/MATH] using a point on the line, giving the line's coordinates.

Say you take two such lines, [MATH]ax + by + c = 0[/MATH] and [MATH]dx + ey + f = 0[/MATH], and find their point of intersection:

Spoiler

Solve for [MATH]x[/MATH]:

[MATH]aex + bey + ec = 0[/MATH][MATH]bdx + bey + bf = 0[/MATH][MATH]aex - bdx + ec - bf = 0[/MATH][MATH](ae - bd)x = bf - ec[/MATH][MATH]x = \frac{bf - ec}{ae - bd}[/MATH]​

Solve for [MATH]y[/MATH]:

[MATH]adx + aey + af = 0[/MATH][MATH]adx + bdy + cd = 0[/MATH][MATH]aey - bdy + af - cd = 0[/MATH][MATH](ae - bd)y = cd - af[/MATH][MATH]y = \frac{cd - af}{ae - bd}[/MATH]​

[MATH]x = \frac{bf - ce}{ae - bd}, y = \frac{cd - af}{ae - bd}[/MATH]​

If the lines are parallel, then [MATH]ae = bd[/MATH] and a division by zero results.

Now pretend we live in an alternate universe where the coordinates of these lines in two dimensions are actually the components of vectors in three dimensions and take their cross product:

[MATH]\begin{bmatrix}x\\y\\w\end{bmatrix} = \begin{bmatrix}a\\b\\c\end{bmatrix} \times \begin{bmatrix}d\\e\\f\end{bmatrix}= \begin{bmatrix}\textit{bf - ce}\\\textit{cd - af}\\\textit{ae - bd}\end{bmatrix}[/MATH]​

This winds up as the same operation as above, giving the position of the point of intersection in homogeneous coordinates such that [MATH]x' = \frac{x}{w}, y' = \frac{y}{w}[/MATH]. If the lines are parallel, then [MATH]w = 0[/MATH] and a division by zero results.

It's clear by following the algebra from the first example that it ultimately becomes the same operations as the cross product, but... why? Is there some relationship between two-dimensional lines and three-dimensional vectors? Or is it just a coincidence?

 

[Dr.Peterson]

The coordinates of a line in two dimensions may be given in this form:

[MATH]ax + by + c = 0[/MATH]​

Before I try to examine your question, I have to be sure of your premise. (I will admit that I am not familiar with the terminology of the page.)

The page you refer to does not say what you are saying here. You've just shown an equation; what they say is that the coordinates of the line whose equation is [MATH]ax + by + 1 = 0[/MATH] are (a, b). There is no c there!

The coordinates for your line (with c rather than 1) would therefore be (a/c, b/c).

I don't think your question is about coordinates of lines at all, but just about lines expressed in this standard form. Am I right?

 

[Mr. Bland]

You've just shown an equation; what they say is that the coordinates of the line whose equation is [MATH]ax + by + 1 = 0[/MATH] are (a, b). There is no c there!

The [MATH]c[/MATH] comes from the form buried a little further down the page:

The exclusion of lines passing through the origin [when using two coordinates] can be resolved by using a system of three coordinates (l, m, n) to specify the line in which the equation, lx + my + n = 0. Here l and m may not both be 0. In this equation, only the ratios between l, m and n are significant, in other words if the coordinates are multiplied by a non-zero scalar then line represented remains the same. So (l, m, n) is a system of homogeneous coordinates for the line.

On the topic of line-line intersection, Wikipedia further demonstrates that cross product can be used for this purpose. It was upon reading this some years ago that it was first brought to my attention that the two concepts are related.

I don't think your question is about coordinates of lines at all, but just about lines expressed in this standard form. Am I right?

I'm curious about the relationship between intersecting lines and cross product, if there is one. Where it stands, I can only see that the arithmetic output is the same regardless of whether solving algebraically or using cross product. It feels like the kind of thing that indicates a deeper connection, but I haven't seen any references discussing it.

 

[HallsofIvy]

It looks like you are talking about "projective coordinates", also called "homogeneous coordinates". https://en.wikipedia.org/wiki/Homogeneous_coordinates Any (x, y, z) where z is not 0 represents the point (x/z, y/z, 1) or (x/z, y/z) in Cartesian coordinates. One advantage of this is (x, y, 0) represents the "point at infinity" for any x, y.

 

[Mr. Bland]

I understand what homogeneous coordinates are (and in fact have mentioned them in every post I've made in this thread thus-far), but it's not about those...

A two-dimensional line in the form of [MATH]ax + by + c = 0[/MATH] can exist anywhere in the two-dimensional plane. It can be vertical (which doesn't work for the form [MATH]y = mx + b[/MATH]) and it can pass through the origin (which doesn't work for the form [MATH]ax + by + 1 = 0[/MATH]). Since multiplying these [MATH]a, b, c[/MATH] by some scalar will not change the position or orientation of the line, it is a system of homogeneous coordinates. However, these are homogeneous in the context of a line, not a point.

Cross product on vectors with three components is a strictly three-dimensional operation. For any two vectors of non-zero magnitude, the result of their cross product is a new vector whose direction is perpendicular to the plane shared by the other two and whose magnitude is the product of the magnitudes of the other two and the sine of the angle between them. It's an extraordinarily useful operation that is conveniently efficient to compute. However, it is still a "vectors in 3D thing".

When taking the homogeneous linear coordinates for two lines in two dimensions as the components of vectors in three dimensions, the vector cross product yields homogeneous coordinates for a point in two dimensions: the lines' point of intersection. A 3D vector operation on 2D lines gives a 2D point that solves a system of equations. This is uncanny. It's so interconnected that I feel like there must be some deeper relationship between the different concepts, but I can't say one way or the other whether there is. I brought it up in hopes that someone else might be able to shed some light on that connection.