Calculating the area of a triangle defined in $\mathbb{R}^3$

Question

In $\mathbb{R}^3:$

$$A = (2,0,-1)\\ B = (3,1,-1)\\ C = (4,-2,0)$$

Calculate the area of the triangle defined by $ABC$.

In $\mathbb{R}^2$, I usually perform an orthogonal projection to calculate the height of the triangle, and then

$$\frac{b\cdot h}{2}$$

It is not entirely clear to me how to do the same for $\mathbb{R}^3$. Well, let's do it either way:

Cast an orthogonal projection of $\vec{AC}$ over $\vec{AB}$:

$$Proj_{\vec{AB}}\vec{AC} = \frac{\vec{AC} \cdot \vec{AB}}{\vec{AB} \cdot \vec{AB}}\cdot \vec{AB}$$

$$\frac{(2,-2,1)\cdot (1,1,0)}{(1,1,0) \cdot (1,1,0)}\cdot(1,1,0)$$

$$\frac{0}{2}\cdot(1,1,0)$$

$$(0,0,0)$$

Huh, a null projection... Maybe I can't do it this way after all?

Answer 1

Your work makes no sense. The area is zero if $\vec{AC}\perp \vec{AB}$?? Also, dot returns a scalar and you are taking dot of a scalar with a vector.

The best way is $\frac 1 2|\vec {AB} \times \vec {AC}| $

Answer 2

Awesome's answer is pretty economical. If you're in the mood, a much more masochistic way would be to use the Pythagorean theorem to find the side lengths of the triangle, then apply Heron's formula.