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.

Solution 1:

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.

Solution 2:

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.