I did the following:
Taking the tetrahedron $OABC$, one can decompose it in: $OA,OB,OC, AB,BC$. And then, writing:
$$x(BC-AB)+AB\quad x\in[0,1]$$
We obtain all the points in the line segment from $AB$ to $BC$. Now if we multiply it by another scalar $y$, we have:
$$y(x(BC-AB)+AB)\quad x,y\in[0,1]$$
This would give me all the points inside $ABC$. The condition would be that:
$$y(x(BC-AB)+AB)= \lambda OA + \zeta OB + \psi OC$$
Writing:
$$BC=(a_1,a_2,a_3)\quad AB=(a_4,a_5,a_6) \quad OA=(a_7,a_8,a_9)\\ OB=(a_{10},a_{11},a_{12}) \quad OC=(a_{13},a_{14},a_{15})$$
We have:
\begin{bmatrix} -a_{10} \zeta -a_7 \lambda +a_1 x y+a_4 (y-x y)-a_{13} \psi =0 \\ -a_{11} \zeta -a_8 \lambda +a_2 x y+a_5 (y-x y)-a_{14} \psi =0 \\ -a_{12} \zeta -a_9 \lambda +a_3 x y+a_6 (y-x y)-a_{15} \psi =0 \\ \end{bmatrix}
With $x,y \in [0,1]$.
I have actually two questions:
- Is there a simpler way?
- Is it possible to answer it the way I'm doing? I guess that there is some kind of matrix property that could be used to show the conditions on $\zeta, \lambda, \psi$. Perhaps determinants?