In computational geometry if three points $P_0, P_1, P_2$ are on a plane then we can tell if $P_2$ is to the left or to the right of $\overrightarrow{P_0 P_1}$ by examining the cross product $\overrightarrow{P_0P_1} \times \overrightarrow{P_0P_2}$ and particularly by examining the sign of: $$P = (x_2-x_1)(y_3-y_1) - (y_2-y_1)(x_3-x_1).$$
Question: If our plane is in 3D space and points $P_0, P_1, P_2$ have a $z$ component (they did before too, but it was zero) then how does the above expression changes? $P = $???
In general we have
$\vec {P_0P_1}=(x_1-x_0,y_1-y_0,z_1-z_0)$
$\vec {P_0P_2}=(x_2-x_0,y_2-y_0,z_2-z_0)$
and thus
$$\vec {P_0P_1}\times\vec {P_0P_2} =\begin{vmatrix} \vec i&\vec j&\vec k\\ x_1-x_0&y_1-y_0&z_1-z_0\\ x_2-x_0&y_2-y_0&z_2-z_0 \end{vmatrix}=$$ $$=\begin{vmatrix} y_1-y_0&z_1-z_0\\ y_2-y_0&z_2-z_0 \end{vmatrix}\vec i =-\begin{vmatrix} x_1-x_0&z_1-z_0\\ x_2-x_0&z_2-z_0 \end{vmatrix}\vec j =\begin{vmatrix} x_1-x_0&y_1-y_0\\ x_2-x_0&y_2-y_0 \end{vmatrix}\vec k$$