Where $a = ( 1, 2, 3), b = (2, 1, 3) $, and $c = (3,1,2). $
$\endgroup$ 16 Answers
$\begingroup$Heron works of course but it would be simpler to take half the length of the cross product $(b-a)\times(c-a)$.
$\endgroup$ $\begingroup$Solution: Construct the vectors $\hat{ab}$, $\hat{ac}$ and take $\frac{1}{2} |\hat{ab} \times \hat{ac}|$. We take half of the resulting since the original gives the area of the parallelogram decsribed by the vectors.
$\endgroup$ $\begingroup$use this formula: $$S=\sqrt{p(p-s_1)(p-s_2)(p-s_3)}$$ where $p=\frac{s_1+s_2+s_3}{2}$, and $s_1,s_2,s_3$ are the lengths of the three sides. You can get $s_1,s_2,s_3$ from the three positions.
$\endgroup$ $\begingroup$Compute the differences $a-b$ and $a-c$. then take the cross product of them $(a-c)\times(a-c)$. The norm of this, divided by two is the area of the triangle.
$\endgroup$ $\begingroup$One way to do it would be to first find out the length of each of the 3 sides, by simply applying Pythagorus. e.g.
AB = $\sqrt{(2-0)^2 + (1-1)^2 + (4-3)^2}=\sqrt5$
You can do this for all of the sides, and it is possible to find the area of a triangle given the length of each of its sides.
$\endgroup$ 0 $\begingroup$Cross product works great as a black box, but it lacks geometric intuition.
For triangle ABC, we construct a line segment intersecting C, and normal to side AB. The triangle's area is $0.5\cdot \left \| \overrightarrow{AB} \right \|\cdot \left \| \overrightarrow{CH} \right \|$$$\left \| \overrightarrow{CH} \right \|=\left \| \overrightarrow{AC} \right \| \cdot sin(θ)$$where$$θ = arccos(\frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{\left \| \overrightarrow{AB} \right \| \cdot \left \| \overrightarrow{AC} \right \|})$$
Thus:$$Area=0.5\cdot \left \| \overrightarrow{AB} \right \|\cdot \left \| \overrightarrow{AC} \right \|\cdot sin(θ)$$This is equal to:$$0.5\cdot \left \| \overrightarrow{AB} \times \overrightarrow{AC} \right \|$$
This is why $$0.5\cdot \left \| \overrightarrow{AB} \times \overrightarrow{AC} \right \|=0.5\cdot \left \| \overrightarrow{AB} \right \|\cdot \left \| \overrightarrow{AC} \right \|\cdot sin(θ)$$
$\endgroup$
