Write a triangle in the space in parametric form I want to write the triangle of vertices $A(0,0,0)$, $B(1,1,1)$ and $C(0,0,2)$ in parametric form.
I did the following:
I know that the plane that contains these 3 points is
$x=y$
so a parametric equation of the plane is $(x,y,z)=(u,u,v)$ with $u,v\in\mathbb{R}$.
Following my intuition, the triangle can be wrote parametrically as follows:
$(x,y,z)=(u,u,v)$ with $0\leq u\leq 1$ and $u\leq v\leq 2-u$.
I am right? What I finally want is to calculate an integral over the triangle, and for that I need this, to have the right limit for the integral.
 A: Any point $P$ on the perimeter or the interior of $\triangle ABC$  can be written as a linear combination of the coordinates of $A,B,C$ (Barycentric coordinates)
$P = t A + s B + (1 - t - s) C $
where $0 \le t \le 1$ , and $ 0 \le s \le  1 - t $.
Using the coordinates for $A, B, C$ in the problem, we have
$P = t (0, 0, 0) + s (1, 1, 1) + (1 - t - s) (0, 0, 2) = (s , s, 2 - 2 t - s ) $
Comparing this to the form you got which is $(u, u, v)$ we get $u = s$ and $v = 2 - 2 t - s $
Imposing $ 0 \le s \le 1 $ and $ 0 \le t \le 1 - s $ leads to
$ 0 \le u \le 1 $ and $ 0 \le  \frac{1}{2} (2 - u - v) \le 1 - u $, which simplifies to
$ 0 \le 2 - u - v \le 2 - 2 u $
Multiplying by $(-1)$
$ 0 \ge u + v - 2 \ge 2 u - 2 $
Adding $2 - u$ throughout
$ 2 - u \ge v \ge u $
which is the same as $ u \le v \le 2 - u $  , the inequality that you derived in your question.
A: I think the following will do the job:
$r:[0, 1]\times [0, 1] \to \mathbb{R}^3$ defined by
$$
\begin{aligned}
r(u,v)
&= (Au + B(1-u))v + C(1-v)\\
&= Auv + B(1-u)v + C(1-v)
\end{aligned}
$$
where $A$, $B$, and $C$ are the position vectors of the points.
The underlying idea: First, find the point on the segment $AB$, $P = Au + B(1-u)$, and find a point on the segment $P$ and $C$, $Pv + C(1-v)$. Thank you.
