Problem 5. a) Find the area of the space triangle with vertices $P_0, P_1, P_2$:
$$ P_0 = (2,1,0),\ \ P_1=(1,0,1),\ \ P_2=(2,-1,1) $$
My current solution is to use $\frac{1}{2}\Big|(P_1-P_0)×(P_2-P_0)\Big|$, which is great but leaves me wanting for a more elegant solution similar to taking the determinant for triangles in the plane.
Is there a better solution, either in terms of elegance or ease of mental calculation?
Best Answer
I like your solution, but another way to do it would be to compute the lengths of the sides using the distance formula, and then use Heron's formula.
http://www.mathopenref.com/heronsformula.html