Question

Calculate the area of a triangle whose vertices are given by (3, -1, 2), (1, -1, 2)
and (4, -
2, 1).

Answer 1

Let A=(3, -1, 2), B=(1, -1, 2), C=(4, -2, 1).

Then find the distance between the vertices of triangle:

c = AB = $\sqrt{\smash[b]{(3 - 1)^2 + (-1 - (-1))^2 + (2 - 2)^2}} = 2$,

a = BC = $\sqrt{\smash[b]{(1 - 4)^2 + (-1 - (-2))^2 + (2 - 1)^2}} = \sqrt{\smash[b]{11}},$

b =AC = $\sqrt{\smash[b]{(3 - 4)^2 + (-1 - (-2)^2 + (2 - 1)^2)}} = \sqrt{\smash[b]{3}}$ .

The last step is to use the Heron's formula for the area of triangle:

$A=\sqrt{\smash[b]{\frac{(a+b+c)}{2}\frac{(b+c-a)}{2}\frac{(a-b+c)}{2}\frac{(a+b-c)}{2}}} =\\= \frac{1}{4}\sqrt{\smash[b]{(2+ \sqrt{\smash[b]{11}} + \sqrt{\smash[b]{3})}}}\cdot\sqrt{\smash[b]{( \sqrt{\smash[b]{11}} + \sqrt{\smash[b]{3}} - 2)}}\cdot \sqrt{\smash[b]{(2- \sqrt{\smash[b]{11}} + \sqrt{\smash[b]{3})}}}\cdot \sqrt{\smash[b]{(2+ \sqrt{\smash[b]{11}} - \sqrt{\smash[b]{3})}}} =\\= \frac{1}{4}\sqrt{\smash[b]{(7+ 4 \sqrt{\smash[b]{3}}-11})}\cdot\cdot\sqrt{\smash[b]{(2\sqrt{\smash[b]{11}} +11 - \sqrt{\smash[b]{33}} + 2\sqrt{\smash[b]{3}} + \sqrt{\smash[b]{33}} -3 -4 - 2\sqrt{\smash[b]{11}} + 2\sqrt{\smash[b]{3}})}} =\\= \frac{1}{4} \sqrt{\smash[b]{(4\sqrt{\smash[b]{3}} - 4})(4 + 4\sqrt{\smash[b]{3}})} = \frac{1}{4}\cdot 4 \sqrt{\smash[b]{(\sqrt{\smash[b]{3}}- 1)}(\sqrt{\smash[b]{3}} +1)} = \sqrt{\smash[b]{2}}.$