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For a triangle $ABC$ in 3D (each point has x, y, z coordinates) is it possible to generate uniform random points on the triangle from only the following data:

Normal of the triangle plane $N = \vec{\text{AB}}\times \vec{\text{AC}}$

Normal of the $AC$ edge plane $U = \cfrac{\vec{AC} \times N}{| \text{N} |^2} $ and $d_u = -U.A$

Normal of the $AB$ edge plane $V = \cfrac{ N \times \vec{AB} }{| \text{N} |^2} $ and $d_v = -V.A$

The two edge normals U and V are calculated such that $\vec{U}.B + d_u = 1$ and $\vec{V}.C + d_v = 1$ (As of yet I don't have the reputation to post an image so of the arrangement but will as soon as I can).

If this cannot be done with the available data what is the minimum amount of extra data required to generate the uniform random points (e.g. A, B, C)?

I am aware that the you can generate uniform random points on a triangle using the vertex $A$ and edges $\vec{AB} = B - A$ and $\vec{BC} = C-B$ as follows:

$S = A +\sqrt{R_1}\vec{AB} + \sqrt{R_1}R_2\vec{BC}$

Where $R_1 \text{and} R_2$ are random numbers between 0 and 1.

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There is no such thing as an "$AC$ edge plane". –  Christian Blatter Jan 8 '12 at 12:09
1  
It may have some other name however the peer review journal I am reading refers to it as such. It is the plane which the edge AC lies in which is perpendicular to the triangle plane –  cubiclewar Jan 8 '12 at 20:42

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