Recently, Jeff Massung wrote: > Given a line segment AB, and a point C, take the dot product of AC * AB: > > dot = [(B-A).x * (C-A).x] + [(B-A).y * (C-A).y] > > The dot product (of two, normalized vectors) will give you the cosine of the > angle between them. You actually don't care about the cosine, but the cosine > of [0,90] will be >= 0 and the cosine of (90,180] will be < 0. That's all > you care about. > > So, if you have a triangle made up of points A,B,C and a random point D, you > can do: > > dot[0] = AD * AB > dot[1] = BD * BC > dot[2] = CD * CA > > If any of those dot products are negative, then the point is outside the > triangle. If all of them are positive, then the point is inside the > triangle.
Dot products... normalized vectors... head starting to spin... vision getting blurry... OK, Jeff, so does (B-A).x mean the x coordinate of point B minus the x coordinate of point A? Thanks & Regards, Scott Rossi Creative Director Tactile Media, UX Design _______________________________________________ use-revolution mailing list [email protected] Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
