If you are interested in the point with shortest distance on the triangle, additionally: Checkout http://www.mathworks.com/matlabcentral/fileexchange/22857-distance-between-a-point-and-a-triangle-in-3d

I think my file is slower (not checked), but returns the said point as well.
Cheers Gwendolyn

Hi John,
I like your submission, and it works fine for my puposes even in R14. Using your old factor routines, as the new factor command uses a mat file (_primeslist_), which the old matlab version can not read. Is it possible to provide a compatible mat file for MATLAB R14?

But tridist should be the same as the norm from the initial point to the closest point (output from the function as PP0). Plotting it, it looks like PP0 is getting to the right spot, so there must be an issue with that particular path through the code. Notice that if you truncate the values I gave above, the correct answer is reached (i.e. if you only use 4 places past the decimal).

Ilya and Jan, because of floating point arithmetic, it's impossible to find intersections perfectly in all cases. Jan, your example has two curves that touch at a single point; some people might define this as an intersection. Your assertion that (0,0) isn't an intersection is debatable.

It also erroneously finds contact points (no real intersections).
Example: [x0, y0] = intersections([-1,0,-1], [-1,0,1], [1, 0, 1], [-1,0,1], 1);
returns point (0, 0) as intersection point although it isn't.

An excellent function! However, if both intersecting curves already include the intersection point, weird results are possible (see the provided example). This problem was already touched by John Mahoney (the results were different).
My test case: several lines pass through one point and every line must eventually include this point. However, it's not known in advance that we have this situation, it simply may appear.
x1=[-0.49313932739246, -0.02127781500161, 0.450583697389237];
y1=[-10.01, 0, 10.01];
x2=[-1.22073877122679, -0.02127781500161, 1.17818314122357];
y2=[10.01, 0, -10.01];
[x,y,seg1,seg2] = intersections(x1, y1, x2, y2)
The case is published because it may be useful to know about. Otherwise I agree that the function works fine in 99,99999999% of cases.

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