Let
$A(2,1,3), B(3,2,5), C(3,3,6), D(4,4,2)$
Find the volume $V$ of tetrahedron $ABCD$.
My Solution:
$\vec{AB} = (1, 1, 2), \vec{AC} = (1, 2, 3), \vec{AD} = (2, 3, -1)$
$$V=\begin{vmatrix}1&1&2\\1&2&3\\2&3&-1 \end{vmatrix}=-6$$
My question is: Is my solution right and can the volume be a negative number ?
Note that the volume of a Tetrahedron = volume of a Pyramid with triangular base = $$= 1/3\,\text{ * }\,\text{base}\,\text{area}\,\text{ * }\,\text{height}$$ Then, if $\vec a,\,\vec b,\,\vec c$ are the vectors corresponding to three concurrent sides, the above is obtained as $$ V = 1/3\left| {\left( {1/2\,\vec b \times \,\vec a} \right) \cdot \vec c} \right| = 1/6\left| {\;\left| {\begin{array}{*{20}c} {c_x } & {c_y } & {c_z } \\ {b_x } & {b_y } & {b_z } \\ {a_x } & {a_y } & {a_z } \\ \end{array} } \right|\;} \right| $$ You shall take the absolute value, because $\vec b \times \,\vec a$ is a vector, whose orientation respect to $\vec c$ will determine a positive or negative scalar result of the double product.
The volume of a 3D parallelepiped with sides AB, AC and AD is given by
$$\left|\det(AB, AC, AD)\right|.$$
The parallelepiped is always $6$ times as voluminous as the corresponding tetrahedron $ABCD$, so the volume of the tetrahedron $ABCD$ is given by
$$\frac{1}{6}\left|\det(AB, AC, AD)\right|.$$
In this case, this means that the tetrahedron has a volume of exactly $1$.
If you did this using $B$ as the origin point (using $BA, BC, BD$ in that order) you would find $$ V=\frac16 \left| \pmatrix{-1&-1&-2\\0&1&1\\1&2&-3} \right| = \frac16 \cdot 6 $$ The determinant is equal to the volume only up to a factor of $\pm 1$; you can see this because if you change the order of any two of the sides, the determinant changes sign.
BTW, the $\frac16$ comes from $V = \frac13 Bh$ where $B$ is the area of the base, which is $\frac12 \ell w$.
