# Area of a triangle

#### Petrus

##### Well-known member
Hello MHB,
I am working with an old exam that I don't get same answer.
Line $$\displaystyle l_1(x,y,z)=(1,0,1)+t(2,-1,-2)$$ and $$\displaystyle l_2(x,y,z)=(2,-5,0)+s(-1,2,1)$$ intercept on point A also interpect on plane $$\displaystyle \pi:-x+2y-z+4=0$$ in point B and C, decide area of triangle ABC

Progress:
Point A:
If we equal them we get $$\displaystyle t=\frac{7}{3}$$ and $$\displaystyle s=-\frac{11}{3}$$ that means $$\displaystyle A=(\frac{17}{3},-\frac{7}{3},-\frac{8}{3})$$
Point B:
We replace $$\displaystyle x,y,z$$ in the plane with $$\displaystyle l_1$$ and get $$\displaystyle t=1$$
so $$\displaystyle B=(3,-1,-1)$$
Point C:
We replace $$\displaystyle x,y,z$$ in the plane with $$\displaystyle l_2$$ and get $$\displaystyle s=2$$ that means $$\displaystyle C=(0,-1,2)$$
Remember area of a triangle is half of the area of paralellogram that means
$$\displaystyle \text{Area}=\frac{1}{2}\left|\textbf{AB}\times \textbf{AC} \right|$$
and $$\displaystyle AB=(-\frac{8}{3},\frac{4}{3},-\frac{5}{3})$$
$$\displaystyle AC=(-\frac{17}{3},\frac{4}{3},\frac{14}{3})$$
so the area is $$\displaystyle \frac{\sqrt{45881}}{19}$$ which does not agree with the facit, what I am doing wrong?
edit: I forgot to say that cross product of AB x AC is $$\displaystyle (\frac{76}{9}, \frac{197}{9}, 4)$$

Regards,
$$\displaystyle |\pi\rangle$$

Last edited:

#### MarkFL

To denote the area with vectors and their cross, I would use the $\LaTeX$ code:
$$\displaystyle \text{Area}=\frac{1}{2}\left|\textbf{AB}\times \textbf{AC} \right|$$