This is related to this answer: https://math.stackexchange.com/a/214194/146115
We have 4 points $a,b,c,x \in \mathbb R^3$.
From them we create the following vectors:
$B'= b-a \\ C'= c-a \\ X'= x-a$
Then compute the determinant of the matrix $(B' | C' | X')$
What I don't understand is:
The sign of the resulting determinant will be positive for $X$ on one side of the plane and negative on the other side.
Why does the position of $X$ with relation to the $BC$ plane affect the sign of the volume that is being calculated by $det(B' | C' | X')$?
Is it related to the normal of the plane pointing to the side that $X$ is on?
With your notation, the determinant in question can be computed as $X'\cdot(B'\times C')$, where $\times$ denotes the ordinary cross product, and the dot is scalar product. Recall that $B'\times C'$ is orthogonal to both $B'$ and $C'$, and nonzero if $B'$ and $C'$ are not parallel. So $B'\times C'$ is a normal vector to the plane we're looking at, and then the sign of the dot product with $X'$ determines which side of the plane $X'$ points towards.
A very different approach is to note that the determinant is nonzero if, and only if, the three vectors are linearly independent. That is, iff $X$ does not lie in the subspace spanned by $B'$ and $C'$. By continuity, the sign of the determinant must then be constant when $X'$ points to one side. Since replacing $X'$ by $-X'$ changes the sign of the determinant, the sign on the opposite side must then be opposite.