I want to calculate the volume of the tetrahedron defined by those 4 points:
$$ P_1 = (-0.0865403, -0.122347, 0.898904)\\ P_2 = (-0.436523, -0.30131, 1.92251)\\ P_3 = (-0.459102, -0.0670386, 1.68168)\\ P_4 = (0,0,0) $$
How would you calculate this volume?
I'm doing
$$B_1 = P_1-P_4 = P_1\\B_2 = P_2-P_4 = P_2\\B_3 = P_3-P_4 = P_3$$
which is an equation that I found (is it correct?)
then
$$V = \frac{|B_1\cdot(B_2\times B_3)|}{6} = \frac{|(-0.0865403, -0.122347, 0.898904)\cdot((-0.436523, -0.30131, 1.92251)\times (-0.459102, -0.0670386, 1.68168))|}{6}$$
Which is 0.00786195 according to wolfram alpha (see here and divide by 6).
I have 2 questions: am I calculating the tetrahedron volume correctly? Does the order of points matter in the equation?
$\endgroup$ 11 Answer
$\begingroup$A more convenient formula for the volume is $$V = \frac{1}{3!} \left|\begin{matrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1 \\ \end{matrix}\right|$$where the vertices are $(x_i, y_i, z_i)$ for $i = 1, 2, 3, 4$ in any order. Take the absolute value of the result.
In WolframAlpha, you can compute this with the input
Abs[Det[{{-0.0865403, -0.122347, 0.898904, 1}, {-0.436523, -0.30131, 1.92251, 1}, {-0.459102, -0.0670386, 1.68168, 1}, {0, 0, 0, 1}}]/3!]The advantage of this approach is that since one of your vertices is the origin, you could even compute the determinant fairly easily by hand as it reduces to a $3 \times 3$ determinant.
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