There is a line and a face in $\Bbb R^3$, does the line inersect the face?
I have a plane (infinite area) in $\Bbb R^3$ defined by a point $(x_0,y_0,z_0)$ and its normal $n$. The plane contains a face (finite area) defined by $m$-coordinates $\{(x_1,y_1,z_1), ...,(x_m,y_m,z_m)\}$. The $m$-coordinates are not random spread over the plane, but form a triangle, square, pentagon, hexagon, ... . There's also a line $g$ in $\Bbb R^3$ with known begin- and end- coordinates $\{(x_b,y_b,z_b),(x_e,y_e,z_e)\}$.
I can already calculate where line $g$ will intersect the plane, so I've got a point $P$ $(x_p,y_p,z_p)$ on the plane and want to know if it is contained by the face.
Sounds like a problem from computer science, is that true?
Suppose the intersection point lies outside the face. If the face is convex, then there is a line $\ell$ contained in the plane and passing through the intersection point $(x_p,y_p,z_p)$ so that the face lies entirely on one side of $\ell$.
Let $v_k$ denote the vector from $(x_p,y_p,z_p)$ to $(x_k,y_k,z_k)$ on the face. Using dot product, you could calculate the angles between $v_1$ and each $v_k$, $k=2,\dots,m$. Every angle would be between $-\pi/2$ and $\pi/2$. Equivalently, the dot products would be positive.
On the other hand, if it lies inside the face there will exist some angle between $v_1$ and another vector which will be greater than $\pi/2$. In this case the dot product will be negative.
So the algorithm would be to calculate the $\{ v_k \}_{k=1}^m$, then the numbers $v_1 \cdot v_k$ for $k = 2, \dots, m$. If all these results are positive then the point lies outside the face.
I'll reiterate that this relies on convexity of the face.