TITLE: Check whether a point is within a 3D Triangle
QUESTION [13 upvotes]: I have a 3D plane defined by three points: $P_0$ , $P_1$ and $P_2$. How to check whether a point $P$ is located right on and inside the 3D triangle?
So, for example, if I have a plane defined by $({0,0,0})$, $({10,0,0})$ and $({0,10,0})$, then the point $({50,0,0})$ is considered not located on the plane, whereas the point  $({5,0,0})$ is.

REPLY [2 votes]: Vectors $V_{01}=P_1-P_0$ and $V_{02}=P_2-P_0$ lie in the plane of the triangle, and $V_{01}\times V_{02}$ is normal to this plane. Let $V_{0p}=P-P_0$, and if $V_{0p} \cdot (V_{01}\times V_{02}) = 0$ then $P$ lies in the plane.
Let $P=cP_0+aP_1+bP_2$ where $c=1-a-b$. Then $P=(1-a-b)P_0+aP_1+bP_2$ or $V_{0p}=aV_{01}+bV_{02}$. If $0 \le a,b,c \le 1$ then $P$ lies in the triangle or on its edge.
Vector $V_{01} \times (V_{01}\times V_{02})$ is orthogonal to $V_{01}$ so $V_{02} \cdot(V_{01} \times (V_{01}\times V_{02}))b=V_{0p} \cdot(V_{01} \times (V_{01}\times V_{02}))$ can be solved for b.
Likewise $V_{02} \times (V_{01}\times V_{02})$ is orthogonal to $V_{02}$ so $V_{01} \cdot(V_{02} \times (V_{01}\times V_{02}))a=V_{0p} \cdot(V_{02} \times (V_{01}\times V_{02}))$ can be solved for a.