How to calculate the area of a triangle ABC when given three position vectors $a, b$, and $ c$ in 3D?

Question

Where $a = ( 1, 2, 3), b = (2, 1, 3) $, and $c = (3,1,2). $

Answer 1

Heron works of course but it would be simpler to take half the length of the cross product $(b-a)\times(c-a)$.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.