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I have read that the volume of a tetrahedron can be described as $$V=\pm \frac{1}{6} det \begin{bmatrix} x_{1} & y_{1} & z_{1} & 1\\ x_{2} & y_{2} & z_{2} & 1\\ x_{3} & y_{3} & z_{3} & 1\\ x_{4} & y_{4} & z_{4} & 1 \end{bmatrix}$$ Where $(x_{1},y_{1}, z_{1})$, $(x_{2},y_{2}, z_{2})$, $(x_{3},y_{3}, z_{3})$, $(x_{4},y_{4}, z_{4})$ are the vertices of the tetrahedron, but I can't find a proof. I know it is somewhat related to the formula for the area of a triangle. $$A=\pm \frac{1}{2} \begin{vmatrix} x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{vmatrix}$$

Shambhala
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  • Cf. [this question](https://math.stackexchange.com/questions/3626329/proof-that-the-volume-of-a-tetrahedron-is-given-by-a-4-times-4-determinant) – J. W. Tanner May 06 '21 at 20:32
  • I don't understand why it is being asked wether $\vec{BC} \cdot (\vec{BD} \times \vec{BA})=det(M)$. It seems to me the proof is incomplete – Shambhala May 06 '21 at 20:35

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This post linked in J. W. Tanner's comment describes one approach. Here is an alternative. First, using row-reduction, we note that $$ \det \pmatrix{x_{1} & y_{1} & z_{1} & 1\\ x_{2} & y_{2} & z_{2} & 1\\ x_{3} & y_{3} & z_{3} & 1\\ x_{4} & y_{4} & z_{4} & 1 } = \det \pmatrix{ x_{1}-x_4 & y_{1}-y_4 & z_{1}-z_4 & 0\\ x_{2} - x_4 & y_{2}-y_4 & z_{2}-z_4 & 0\\ x_{3}-x_4 & y_{3}-y_4 & z_{3}-z_4 & 0\\ x_{4} & y_{4} & z_4 & 1 } \\=\det \pmatrix{ x_{1}-x_4 & y_{1}-y_4 & z_{1}-z_4 \\ x_{2} - x_4 & y_{2}-y_4 & z_{2}-z_4 \\ x_{3}-x_4 & y_{3}-y_4 & z_{3}-z_4 }. $$ This is essentially equivalent to the first step of the linked proof. From there, we consider the volume of the tetrahedron formed by connecting the origin and the standard basis vectors, which I will call the "standard tetrahedron". As is explained here for instance, this volume is equal to $1/3! = 1/6$.

Now, the volume of the tetrahedron formed by connecting the four given points is equal to the volume of the tetrahedron formed by connecting the origin to the points $(x_i - x_4,y_i - y_4,z_i - z_4)$ for $i = 1,2,3$. This tetrahedron is the image of the standard tetrahedron under multiplication by the matrix $$ \pmatrix{ x_{1}-x_4 & y_{1}-y_4 & z_{1}-z_4 \\ x_{2} - x_4 & y_{2}-y_4 & z_{2}-z_4 \\ x_{3}-x_4 & y_{3}-y_4 & z_{3}-z_4 }. $$ Because the determinant of $A$ can be interpreted as the volume expansion resulting from multiplication by $A$, we conclude that the desired volume is $$ \frac 16 \cdot \det \pmatrix{ x_{1}-x_4 & y_{1}-y_4 & z_{1}-z_4 \\ x_{2} - x_4 & y_{2}-y_4 & z_{2}-z_4 \\ x_{3}-x_4 & y_{3}-y_4 & z_{3}-z_4 } = \frac 16 \cdot \det \pmatrix{x_{1} & y_{1} & z_{1} & 1\\ x_{2} & y_{2} & z_{2} & 1\\ x_{3} & y_{3} & z_{3} & 1\\ x_{4} & y_{4} & z_{4} & 1}, $$ which was what we wanted.

Ben Grossmann
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