[Math] Find the parametric form $S(u, v)$ where $a \le u \le b$ and $c \le v \le d$ for the triangle with vertices $(1, 1, 1), (4, 2, 1),$ and $(1, 2, 2)$.

geometryparametricparametrizationsurfacestriangles

Find the parametric form $S(u, v)$ where $a \le u \le b$ and $c \le v \le d$ for the triangle with vertices $(1, 1, 1), (4, 2, 1),$ and $(1, 2, 2)$.

I am told that one parameterisation is $S(u, v) = (1 + 3u, 1 + u + v, 1 + v)$ for $0 \le u \le 1, 0 \le v \le 1 – u$. However, I do not understand, nor am I shown, the reasoning behind such a solution.

In general, I'm not sure how to parameterise a triangle in such a way. At first, I tried to use the equation of a point on a plane $P(x, y, z) = A + Bu + Cv$ to find a parameterisation, but I could not find a way to bound $u$ and $v$.

I would greatly appreciate it if people could please take the time to show me the reasoning behind the above solution.

EDIT: I have a suspicion that this parameterisation has something to do with a parallelogram. Although, I'm not sure how to parameterise a parallelogram either…

Best Answer

I believe that the general formula you seek is

$$ S(u,v)=A+(B-A)u+(C-A)v$$

for $0\le u\le1$ and $0\le u+v\le1$ so that $S$ is defined on the triangle with coordinates $(0,0),\,(1,0),\,(0,1)$. The transformation maps this triangle onto any $\triangle ABC$. It accomplishes this as follows:

  1. $S$ maps $\mathbf{0}=(0,0)$ to the position vector $\vec{OA}$.
  2. $S$ maps the vector $\mathbf{i}=(1,0)$ to the position vector $\vec{OB}$
  3. $S$ maps the vector $\mathbf{j}=(0,1)$ to the position vector $\vec{OC}$

where $O$ has coordinates $(0,0,0)$.

In this case

\begin{eqnarray} S(u,v)&=&(1,1,1)+(3,1,0)u+(0,1,1)v\\ &=& (1 + 3u, 1 + u + v, 1 + v) \end{eqnarray}

Additional note: If you extended $S$ to the unit square it would map onto the parallelogram $ABA^\prime C$ where $A^\prime=(4,3,2)$ is the reflection of $A$ in the line $BC$.

Unit triangle

If one wishes to have $a\le u\le b$ and $c\le v\le d$ then one must first map the following rectangle onto the unit square and the shaded triangle onto the "unit triangle." $u$ and $v$ must be constrained below the line connecting the points $(a,d)$ and $(b,c)$.

basic rectangle

The resulting composition and corresponding restrictions on $u$ and $v$ are:

\begin{equation} S(u,v)=A+(B-A)\left(\frac{u-a}{b-a}\right)+(C-A)\left(\frac{v-c}{d-c}\right)\tag{1} \end{equation}

with the restriction

$$ \frac{u-a}{b-a}+\frac{v-c}{d-c}\le1\tag{2}$$

For example, suppose we want $2\le u\le 5$ and $3\le v\le 4$ subject to constraints on $u$ and $v$ from equation $(2)$ which resolve to the requirement that

$$ u+3v\le 14$$

Then equation $(1)$ gives the transformation

\begin{eqnarray} S(u,v)&=&(1,1,1)+(3,1,0)\left(\frac{u-2}{3}\right)+(0,1,1)(v-3)\\ &=&\left(u-1,\frac{u-8}{3}+v,v-2\right) \end{eqnarray}

It is then easily verified that

  1. $S(2,3)=(1,1,1)$
  2. $S(5,3)=(4,2,1)$
  3. $S(2,4)=(1,2,2)$
Related Question