Calculate $\bigtriangleup$ ABC where $A(-2,-3,0)$,$B(-1,0,5)$,$C(4,2,2)$

Question

I want to calculate $\bigtriangleup$ ABC where $A(-2,-3,0)$,$B(-1,0,5)$,$C(4,2,2)$
What I did was to mark the triangle vertices randomly
1) calculate the middle of AB ( I call it G ) to find the vertical vector CG then what I do is to calculate $\frac{CG*AB}{2}$ but I dont get the right answer, this is the right way to do that? or I need to do something else?
thanks!

Answer 1

Use the cross product! $\|\vec u \times\vec v\|$ gives the area of the parallelogram spanned by $\vec u$ and $\vec v$. So $1/2$ times this gives the area of the triangle with vertices $\vec 0$, $\vec u$, and $\vec v$.

Answer 2

If you want find $G$ so that $\vec{CG}\perp \vec{AB}$.

Set $\vec{OG}=t\vec{OA}+(1-t)\vec{OB}$,

then let $\vec{AG}\cdot{}\vec{CG}=0$.

You can solve $t$.

The area is $\frac{CG\cdot{}AB}{2}$.