Point in a triangle or parallelogram

You can also check whether a point is in the triangle or in the parallelogram. However, this only works with the parametric form.

The parametric equation is set up using the two vectors which make of the triangle or parallelogram.

With triangle ABC:

And with the ABCD parallelogram:

Example

Is the point $P(-0.5|1|1)$ in the triangle ABC with $A(0|1|0)$, $B(0|0|2)$ and $C(-2|2|2)$?

1. Set up parametric equation

   A parametric equation is set up with the 3 points of the triangle.

   $\text{E: } \vec{x} = \vec{OA} + r \cdot \vec{AB}$ $+ s \cdot \vec{AC}$

   $\vec{x}=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + r \cdot \begin{pmatrix} 0 \\ -1 \\ 2 \end{pmatrix}$ $+ s \cdot \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}$

2. Insert point

   The position vector of $P$ is used for $\vec{x}$ in $E$.

   $\begin{pmatrix} -0.5 \\ 1 \\ 1 \end{pmatrix}$ $=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + r \cdot \begin{pmatrix} 0 \\ -1 \\ 2 \end{pmatrix}$ $+ s \cdot \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}$

   Now we set up an equation system and solve it. Every line is an equation.
   1. $-0.5=-2s$
   2. $1=1-r+s$
   3. $1=2r+2s$

   From I. one obtains $s=\frac14$, which is used in II..

   $1=1-r+\frac14\quad|-1$
   $0=-r+\frac14\quad|+r$
   $r=\frac14$

   $r$ and $s$ are inserted in III..

   $1=2\cdot\frac14+2\cdot\frac14$
   $1=1$

   => Point lies on the plane

3. Check conditions

   To find out whether the point lies in a triangle, the conditions for $r$ and $s$ must be checked.
   
   
   1. $0\le \frac14\le1$
   2. $0\le \frac14\le1$
   3. $0\le \frac14+\frac14 \le1$

   All conditions apply.
   => P lies in the triangle ABC.