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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?

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Zaz
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  • The linked question answers in 2D, you could do the same for 3D. – StubbornAtom Aug 17 '16 at 16:49
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    Isn't $\left( {{P_1} - {P_0}} \right) \times \left( {{P_2} - {P_0}} \right)$ a vector? – ITA Aug 17 '16 at 16:51
  • @IvanAbraham: Whoops. Indeed it is. Edited. – Zaz Aug 17 '16 at 17:30
  • @StubbornAtom: That's what I thought, but I haven't yet been able to wrap my head around how you would do that. – Zaz Aug 17 '16 at 17:32
  • $$ \begin{bmatrix} 1 & 1 & 1 \ x_0 & x_1 & x_2 \ y_0 & y_1 & y_2 \ z_0 & z_1 & z_2 \end{bmatrix} $$ isn't square, so has no determinant. – Zaz Aug 17 '16 at 17:34

2 Answers2

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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

A. Thomas Yerger
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Even if we stick with cross products, we can make the result considerably more elegant in the sense of making the symmetry in indices manifest, viz.$$(P_1-P_0)\times (P_2-P_0)=P_1\times P_2-P_1\times P_0-P_0\times P_2+P_0\times P_0\\=P_0\times P_1+P_1\times P_2+P_2\times P_0.$$

J.G.
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