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PointVector

Problem with the equation of a plane Ax + By + Cz + D = 0

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Problem with the equation of a plane Ax + By + Cz + D = 0? calculate the normal vector of one triangle use it''s three points value, the result is two of normal vector values are zero, x = 0, y = 0, z = 1, for example. then use the equation of a plane: Ax + By + Cz + D = 0, to calculate one point whether into the triangle will wrong, any x y value will ok if z is ok , what can i do?

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quote:

what can I do?



You could rewrite your post so that it´s possible to understand what you want.

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Different username, but nearly identical wording as question posted in this thread a few posts down. You''re not using multiple user accounts, are you?

Anyway, as I explained in that other thread, for a plane defined by the normal vector (0,0,1), any point (x,y,z) will be on the plane so long as z==-D.

If you want to find out if that point lies within a specific triangle on the plane, you need to employ a test such as I described in the above-mentioned thread; merely plugging x and y into the equation and solving for z will not determine if it intersects a triangle, just if it lies on the plane.

As Atheist said, perhaps you could make yourself more understandable.

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