Question

Using vectors, find the area of the triangle ABC with vertices $A(1,2,3),B(2,-1,4)$ and $C(4,5,-1)$.

Answer 1

We form the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$
$\overrightarrow{AB}=2-1,-1-2,4-3\Rightarrow 1,-3,1$
$\overrightarrow{AC}=4-1,5-2,-1-3\Rightarrow 3,3,-4$
We find their cross product $\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 1& -3& 1\\ 3& 3& -4\end{array}\right|$
= $\overrightarrow{i(-4(-3)-3)-\overrightarrow{j}(-4-3)+\overrightarrow{k}(3+9)}$
= $9\overrightarrow{i}+7\overrightarrow{j}+12\overrightarrow{k}$
So, the magnitude is {9, 7, 12}
$=\sqrt{9^2+7^2+{12}^2}=\sqrt{81+49+144}=\sqrt{274}\approx 16.55$
The triangle's area is $\frac{1}{2}$ the area of that parallelogram.
So, Area of traingle = $\frac{1}{2}\times 16.55\Rightarrow \frac{16.55}{2}=8.275$ Sq.units