I'm trying to understand what does the algorithm does and I'm stuck in both 2.1.1 and 3 sections where the algorithm is explained.
Author says:
The triangles $ \triangle ABC $ and $ \triangle PQR $ are
$ X = A + u U + v V $ with $ U = B - A $, $ V = C - A $, $ 0 \le u , v $, $ u + v \le 1 $
$ X = P + s S + t T $ with $ S = Q - P $, $ T = R - P $, $ 0 \le s, t $, $ s + t \le 1 $
Why triangles are defined as sum point + vector * scalar + vector * scalar and where these scalars (small letters) $u$, $v$, $s$, $t$ come from?
Is $X$ just a form of writing common operator to equate the equalities to find common points (intersection(s))?
Starting from the concept of affine space, you can use the operation \begin{align} & \text{point}\ +\ \text{vector}\ =\ \text{point}\\ & A + \mathbf{v} = B \end{align} to define the point $B$ that you can reach starting from $A$ and moving from there following the vector $\mathbf{v},$ and vice versa, the vector needed to go from $A$ to $B$ as $$ B-A=\mathbf{v}. $$
So $$ A+t\mathbf{v},\qquad 0\leq t\leq 1 $$ or equivalently $$ A+t(B-A),\qquad 0\leq t\leq 1 $$ represents the set of point from $A$ to $B$, with $A$ corresponding to $t=0$ and $B$ corresponding to $t=1$.
In the same manner, given another point $C$, you can represent all point from $A$ to $C$ with $$ A+s(C-A),\qquad 0\leq s\leq 1. $$
Now, each point $P$ of the segment from $B$ to $C$ can be expressed as a linear combination of $B-A$ and $C-A$ $$ P-A=t(B-A)+s(C-A) $$ with positive coefficients $%t, s$ such that $t+s=1,$ so that $$ P=A+t(B-A)+s(C-A) $$
Finally, to represent all points of segment from $A$ to $P$, for all $P$ on the segment from $B$ to $C$, i.e to represent all the points of the triangle $ABC$, we can write $$ P=A+t(B-A)+s(C-A),\quad 0\leq t\leq 1,\ 0\leq s\leq1,\ t+s\leq 1. $$