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Due to perspective foreshortening, the projections of parallel lines (1D geometric objects) meet in their vanishing point (a 0D geometric object). Higher dimensional geometric objects also has higher dimensional vanishing objects. I am interested in vanishing points of the edges of the triangle

∆(p0, p1, p2)

if p0T,p1T and p2T are given ?

Any idea ?

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  • $\begingroup$ I don't entirely follow. The edges of a triangle are not parallel, so how would they have a vanishing point? $\endgroup$ – Dan Hulme Dec 11 '17 at 10:13
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Vanishing points are "points at infinity" in projective geometry, which are represented by $w = 0$ in homogeneous coordinates. You can construct the vanishing point of a ray or line by taking its $xyz$ direction vector and adjoining $w = 0$. (For a line, use the direction vector and its negation to get vanishing points in both directions.)

Then you can transform this homogeneous vector by your view/projection matrices as usual, and divide out $w$ from the result to find where the point lies in screen space. (If the result has $w \leq 0$ then it's behind the camera or off to the side.)

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