I have been given the assignment to find an intersection between these two lines:
$$ G_1 =\left [ \begin{matrix} t & 1 \\ \end{matrix} \right ] \left [ \begin{matrix} -1 & 1 & 2 & 0 \\ 1 & -2 & 1 & -2 \\ \end{matrix} \right ] $$
$$ G_2 =\left [ \begin{matrix} t & 1 \\ \end{matrix} \right ] \left [ \begin{matrix} 2 & -1 & -1 & 0 \\ -1 & -1 & -1 & -1 \\ \end{matrix} \right ] $$
Those two lines are in 3D, in parametric form, with homogeneous coordinates.
The second matrix should be represented by following:
$$ \left [ \begin{matrix} B - A \\ A \\ \end{matrix} \right ] $$
where A and B are any two points on the line, A being the starting point and B-A being the vector that represents the direction of the line.
I know that, when talking about homogeneous coordinates of 3D points, when w=0, the point is in infinity, a neat way to denote that two lines don't have an intersection without having to divide by zero.
However, the two top rows in the matrices above are representing vectors. How is w=0 interpreted here and does it even make any sense to talk about homogeneous vectors?
As for finding the intersection, I could easily extract the points A and B from the matrix, divide them by w and then it gets really simple, so (I think) I do know how to solve the problem.
However, it seems that doing it that way defeats the purpose of homogeneous coordinates. Is there a better way to go about solving this?
First, let’s get the interpretation of $w=0$ out of the way. It’s more than just a nifty computational trick. In the projective plane, two lines always intersect in a single point: there’s no notion of “parallel” lines in projective geometry. You can think of the projective plane as a way of extending the Euclidean plane so that parallel lines do intersect: for each family of parallel lines, we add a “point at infinity” that’s their common intersection, and then add the line that consists of all of these points as the “line at infinity.” In the homogeneous coordinate system that you’re using, all of these points at infinity have $w=0$. A sometimes useful way to think about these homogeneous coordinate vectors is that when $w\ne0$, they correspond to points on the Euclidean plane, while when $w=0$, they correspond to directions in the plane. The same thing is going on in 3D, except that now all of those points at infinity form a plane at infinity, which also contains all of the lines in which parallel planes intersect. Note, too, that in 3D you can also have skew lines, so even with those points at infinity you’re no longer guaranteed that two lines will intersect.
Now, to compute the intersection of the two lines, one possibility is to use the dual matrices to the ones you have: instead of representing a line as the join of two points, you represent it as the meet of two planes. The rows of the matrix are the homogeneous vectors that represent those two planes. Essentially, you describe the line via a pair of implicit linear equations instead of parametrically as a linear combination of a pair of points.
Mechanically, this involves computing a pair of independent null vectors for each of the two matrices that you have, assembling them into a single matrix and then computing its null space. These are all fairly straightforward and simple calculations.
Why are we looking at null spaces for the solution? Those linear equations that we’re solving are homogeneous, so if they have a nontrivial solution, there will be an infinite number of them. In particular, if the two lines intersect at a single point, then the solution space to the system of four linear equations is going to be one-dimensional: a line in $\mathbb R^4$. That’s OK, though, since nonzero scalar multiples of a homogeneous coordinate vector represent the same point. Essentially, when we use homogeneous coordinates in 3D, we’re identifying points in $\mathbb R^3$ with lines through the origin in $\mathbb R^4$.
Alternatively, you could use the Plücker coordinates of the lines to compute their intersection, but you probably haven’t learned those yet. You can convert from the matrices that you have to Plücker coordinates by computing a bunch of $2\times2$ determinants and then use a special kind of multiplication to compute their intersection, but that’s a lot more work in this case than computing a few null spaces.