Question

Find the area of triangle determinied by the point P(-3,1-1) Q(-2,0,5) and R(-4,-1,2).

Answer 1

Firstly we should find vectors $\overline{PQ}$ and $\overline{QR}$ :

$\overline{PQ}=(1,-1,6)$

$\overline{QR}$ $=(-2,-1,-3)$

The area of ​​a triangle PQR is half the area of ​​a parallelogram built on vectors $\overline{PQ}$ and $\overline{QR}$. The area of ​​a parallelogram built on vectors $\overline{PQ}$ and $\overline{QR}$ is the modulus of the vector product $\overline{PQ}$ and $\overline{QR}$, and therefore the area of ​​triangle PQR is:

$S_{PQR}=\frac{1}{2}|\overline{PQ}\times\overline{QR}|$

$\overline{PQ}\times\overline{QR}=$ 9*$\overline{i}$ -9*$\overline{j}$ -3*$\overline{k}$

$S_{PQR}=\frac{1}{2}\sqrt{81+81+9}=\frac{3\sqrt{19}}{2}$