We will look at the 3 dimensional space $\mathbb{R}^3$. Say we have a tetrahedron with the vertices $p_1$, $p_2$, $p_3$, and $p_4$, each with corresponding value $v_1$, $v_2$, $v_3$, and $v_4$. Given a point P inside the tetrahedron (could also be on a face), how would one calculate the corresponding value of P?
I tried interpolating in some way but I'm a little too inexperienced on that side. Also tried to use a hyperplane running through the 4 points, but this also leads nowhere for me.
You probably want linear interpolation. First you express $P$ as a linear combination of the vertices in such a way that the coefficients add up to $1$.
$$ a_1p_1+a_2p_2+a_3p_3+a_4p_4=P\\ a_1+a_2+a_3+a_4=1 $$
You can combine these into a single matrix equation:
$$ \begin{pmatrix} x_1&x_2&x_3&x_4\\ y_1&y_2&y_3&y_4\\ z_1&z_2&z_3&z_4\\ 1&1&1&1 \end{pmatrix} \cdot \begin{pmatrix} a_1\\a_2\\a_3\\a_4 \end{pmatrix} = \begin{pmatrix} x\\y\\z\\1 \end{pmatrix} $$
Now you can employ the standard tools for solving systems of linear equations to get the $a_i$.
Then you use those coefficients to combine values, too.
$$v=a_1v_1+a_2v_2+a_3v_3+a_4v_4$$
Note that as long as your point $P$ lies inside the tetrahedron, you get $0\lt a_i\lt 1$. On the boundary you get some equalities. And outside you get some negative numbers and may exceed $1$ to compensate.