Hi,
You are overthinking. As the others have said the trick is to use the
dot product with the normal.
Now in terms of CS code it is even simpler:
csPlane3 myPlane(point1,point2,point3);
if (myPlane.Classify(anotherPoint) < 0)
{ //then }
else
{ //otherwise }
-Marten
Quoting Scott Johnson <[EMAIL PROTECTED]>:
> 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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