Timeline for Check whether a point is within a 3D Triangle
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 9, 2010 at 9:35 | comment | added | Andrea Mori | Moreover the conditions on the coefficients are those that characterize the points in the simplex spanned by the vectors (just two vectors, in this case), which is the convex hull of the points P_0. P_1 and P_2, i.e. the triangle that they define. Everything generalizes straightforwardly in arbitrary dimension. | |
| Sep 9, 2010 at 9:33 | comment | added | Andrea Mori | Michael's solution is correct. If P is your test point, first of all you want the three vectors P_1-P_0, P_2-P_0 and P-P_0 to be coplanar, i.e. linerly dependent. That's the linear equation in x and y whose solvability is equivalent to the vanishing of some determinant (as noted also by others). | |
| Sep 9, 2010 at 8:49 | comment | added | J. M. ain't a mathematician | Hmm... so it's an overdetermined system. Would least squares apply here, or is there a specialized method for this problem? | |
| Sep 9, 2010 at 8:32 | history | edited | Michael Ulm | CC BY-SA 2.5 | added 5 characters in body; added 5 characters in body |
| Sep 9, 2010 at 8:31 | comment | added | Michael Ulm | @Jonas Kibelbek: Good Catch. Corrected. | |
| Sep 9, 2010 at 8:31 | comment | added | Michael Ulm | @J.M. Yes, these are three equations (one for each dimension) in the two skalar unknowns x and y. | |
| Sep 9, 2010 at 8:19 | comment | added | J. M. ain't a mathematician | Michael, I'm trying to understand your solution... is this three equations in two unknowns? | |
| Sep 9, 2010 at 8:16 | comment | added | Jonas Kibelbek | I believe this should say x*(P_1-P_0)+y*(P_2-P_0)=P-P_0. | |
| Sep 9, 2010 at 7:54 | vote | accept | Graviton | ||
| Sep 9, 2010 at 7:18 | history | answered | Michael Ulm | CC BY-SA 2.5 |