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?
Best Answer
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
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.