If P is any point on your plane, n is the plane normal,
and V_i is the ith vertex of your candidate triangle,
then if (V_i - P) . n = 0 for all i, then your triangle lies
on that plane. The point P fixes your reference plane from
among the infinite parallel planes that all have the same
normal n.

Vaidya



>On Thu, 11 Sep 2003 16:08:35 +0530, Shyam Prakash
<[EMAIL PROTECTED]> wrote:

>George,
>            I believe this alone is not sufficient, there is one more
>catch. Consider two triangle parallel to each other but lying on two
>different planes(say with different z- depth). The below said algorithm
>will fail in this case since the angle between these triangles will be
>zero degrees but actually they are not on a single plane.
>
>Thanks
>Shyam
>
>-----Original Message-----
>From: Discussion list for Java 3D API
>[mailto:[EMAIL PROTECTED] On Behalf Of George Perera
>Sent: Thursday, September 11, 2003 3:58 PM
>To: [EMAIL PROTECTED]
>Subject: [JAVA3D] Antwort: Re: [JAVA3D] AW: [JAVA3D] Determining the
>triangles lying on a plane
>
>
>Hello Shyam !
>
>you could calculate the angle between the normal to the triangles in
>question ( one at a time of course) and the normal to your reference
>plain. If this angle is within the tolerance you set then the triangles
>are on the "same" plane.
>
>regards,
>George
>
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