Hi!

you could get the normal vector of AB and AC for the 3 coplanar points then 
given that the halfway vector from A to between BC is 0 degrees, see which 
direction the point you're testing lies on.

If > pi then its on left, otherwise on right. Don't know if i make any 
assumptions here though, and I think it should work in most cases (best to 
test though).

On Sunday 08 July 2007 22:12, Scott Johnson wrote:
> Hi Everyone!
>
> This may seem like a basic question, but in the interest of time, I am
> going to ask it anyway rather than struggle with it.
>
> All of the sideness tests I have seen in 2D (i.e. the Area() function in
> triangulate.cpp in libcrystalspace, which is essentially a copy of
> AreaSide() in
> http://orion.math.iastate.edu/burkardt/c_src/orourke/tri.c) uses a
> matrix determinant to find whether a point is on one side or another of
> a line, (essentially uses cross products).   The problem I have is that
> these are all assuming that the planes in which the vectors lie are
> parallel to the XY plane.  That is, the Z component can essentially be
> disregarded.  Of course, when I take cross products of 3D vectors, I
> obtain another 3D vector.  This happens in 2D of course, but there is a
> matrix which can be used to obtain a determinant telling the side on
> which a point lies, given a vector.
>
> Now, my question is this: Given two vectors (three points), we know they
> are all coplanar, so is there a way to utilize a matrix determinant
> trick similar to in 2D which will give us the side of the line the point
> lies on (i.e. negative or positive - it doesn't really mean 'right' or
> 'left', as these don't have meanings in arbitrary 3-dimesions).
>
> My initial intuition was to determine the angle the vector makes with
> the XY plane, in X and Y dimensions, then rotate all points by this, to
> ensure that we are parallel to the XY plane, then utilize the same
> method.  Would this be too much work?  I am wondering if I am
> overthinking the problem.
>
> Any suggestions would be helpful.  Thanks!
>
> ~Scott
>
>
>
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