If the lines are parallel, u x v (Gibbs vector product) vanishes ... Also, from an algorithmic point of view, putting your lines into this (non-canonical) form in the first place is not entirely trivial --- several extra branches are required. I'd guess that this wasn't what Henry wanted either! WFL On 3/27/11, Dan Asimov <dasimov@earthlink.net> wrote:

If two lines are given parametrically by {u0 + su} and {v0 + tv} for (s,t) in R^2, vectors u0, v0, and unit vectors u, v, then the distance between the lines is just (setting w0 = u0-v0)

|w0 * u x v|,

or am I missing something?

--Dan