Q: find closed-form formula for the intersection of two line segments in \(\bb{R}^2\) given by \((a,b)\) and \((c,d)\).

First, just find the intersection of the two lines. The methodical approach: write them both as parametric equations, solve the system.

Set equal to each other and rearrange:

This is a set of \(n\) equations for \(2\) unknowns, so in general it’s overdetermined and may not have a solution. This makes sense: it reflects the fact that two line segments in \(\bb{R}^{n \geq 3}\) may not intersect. The requirement that it has a solution is that \((ab)\) and \((cd)\) lay in the same plane. I believe (haven’t checked carefully) that this is equivalent to saying that \((1,a) \^ (1,b) \^ (1,c) \^ (1,d) = 0\), that is, their two lines have zero ‘volume’ (in projective \(n+1\) space). This expands to \((abcd) = 0\) and \((abc) - (abd) + (acd) - (bcd) = 0\). There might be a cleaner way though. Seems to work from trying one trivial example.) Also, trivially, \((b-a) \^ (d-c) \neq 0\) is required, since the two lines have to be not collinear.

todo: come up with a clean, compact formula for these requirements?

Assuming that there is a solution, we can solve this equation in many ways, since any two coordinates that aren’t trivial should give us a valid system of equations for \((t,s)\). One mindless approach is to contract with the two coefficients in turn to create two scalar equations:

Or we can do it by eliminating variables Cramer’s rule, which is really just wedging by each vector in turn to cancel out the other ones. This gives:

Both of these are equalities of bi-vectors, but if \((b-a) \^ (d-c) \neq 0\) then all three are parallel, so we can divide through by magnitudes, or just divide through on any single non-zero component since they all have to be the same.

todo: it would be nice to have a 'closed form' for $$t,s$$ here instead of this 'algorithm'.

\(p\) is actually inside \((a,b)\) and \((c,d)\) if both \(s, t \in (0, 1)\).