Describing point inside square using its corners

Question

I am faced with a problem of transforming point from 3d-coordinate system $\mathcal{S} \subset \mathbb{R}^3$ to 2d-coordinate system $\mathcal{T} \subset \mathbb{R}^2$.

I have 4 corners points of a square $S$, $s_i \in \mathcal{S}, \forall i \in \{1, 2, 3, 4\}$. The points $t_i \in \mathcal{T}, \forall i \in \{1, 2, 3, 4\}$ represent the corner points of the same square $S$. Here $t_1$ is the origin.

In order to map a point $s \in \mathcal{S}$ to a point $t \in \mathcal{T}$, I assume the following:

By calculating a weight vector $w \in \mathbb{R}^4$ such that $s = \sum_iw_is_i$ with $\forall i, w_i \geq 0, \sum_{i=1}^{4}w_i = 1$, I can calculate $t$ using $t = \sum_iw_it_i$.

I calculate the weight vector as follows: $\forall i, w_i = \frac{\exp\left(\frac{1}{\lVert s - s_i \rVert_2}\right)}{\sum_i\exp\left(\frac{1}{\lVert s - s_i \rVert_2}\right)}$.

However, the I am getting errors based on this calculation. Is my logic wrong ? Is there any other better way to perform this transformation ? Any help in this regard is highly appreciated.

Answer 1

For every point $s$ in the square there exist unique $a,b\in[0,1]$ such that $$s=s_1 + a(s_2-s_1)+b(s_3-s_1).$$ You can then map $s\in\mathcal{S}$ to $t\in\mathcal{T}$ by replacing the $s_i$ by the $t_i$.

Writing $s-s_1=(x,y)$ and $s_i=(x_i,y_i)$ for $i=2,3$ we can write this as $$\begin{pmatrix} x_2&x_3\\ y_2&y_3 \end{pmatrix} \cdot \begin{pmatrix} a\\ b \end{pmatrix} = \begin{pmatrix} x\\ y \end{pmatrix} ,$$ which shows that you can find $a$ and $b$ by inverting the matrix.