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I'm reading the chapter 8 of the Cg tutorial and I could not understand 8.4.1 :

Because all these coordinates lie in the plane of the same triangle, it is possible to derive plane equations for x, y, and z in terms of s and t:

A_0 x + B_0 s + C_0 t + D_0 = 0

A_1 y + B_1 s + C_1 t + D_1 = 0

A_2 z + B_2 s + C_2 t + D_2 = 0

How are these equations derived?

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In this context, deriving the plane equations does not refer to the equations themselves, but to the coefficients $A$, $B$, $C$ and $D$.

The equations are simply plane equations for a plane defined by three points, where the first plane is defined by the three points $(x_i,s_i,t_i)_{i=1..3}$, the second plane is defined by $(y_i,s_i,t_i)_{i=1..3}$ and the third plane by $(z_i,s_i,t_i)_{i=1..3}$. They can be seen as auxiliary objects in deriving equations for $x$, $y$ and $z$ in terms of $s$ and $t$.

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  • $\begingroup$ I still could not figure out why choosing plane equations, not other equations? $\endgroup$ – Cu2S Oct 31 '18 at 5:26
  • $\begingroup$ Because you want to express $x$ (or $y$ or $z$) in terms of $s$ and $t$ in the simplest possible way. Intuitively you would expect this to be a linear combination of $s$ and $t$, which is precisely what you get when you rearrange the plane equation. This works if the triangle is non-degenerate in $(s,t)$-space, because then the points $(x_i,s_i,t_i)_{i=1..3}$ that determine the plane are non-collinear. $\endgroup$ – Chris Oct 31 '18 at 11:15

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