Volume of tetrahedron

A pyramid has vertices the points $A(2, 0, 0), \; B(0, 3, 0) , \;C (0, 0, 6), \; D(2, 3, 8)$. Evaluate its volume and the lenght of the  height from the vertice $D$.

Solution



The volume of a tetrahedron (and in our case the vol. of the pyramid) is given by the $4 \times 4$ determinant:

$$V= \begin{vmatrix}
2 &0  &0  &2 \\
 0& 3 & 0 &3 \\
 0& 0 &6  &8 \\
 1& 1 &1  &1
\end{vmatrix}=14$$

The area of the base's triangle is given by:

$$ A=\frac{1}{2}\left | \overrightarrow{AB} \times \overrightarrow{AC} \right | = \frac{1}{2}\left | (-2, 3, 0) \times (-2, 0, 6) \right |= \frac{1}{2}\left | 18, 12, 6 \right |= 3\sqrt{14}$$

since the outer product geometrically means the area of a parallelogram.

Hence the lenght of the height can be evaluated by invoking the formula $V= \frac{1}{3}A h $. Thus:

$$V= \frac{1}{3}A h \Leftrightarrow 14 = \frac{1}{3}\cdot 3 \sqrt{14} \cdot h \Leftrightarrow h =\sqrt{14}$$