Question

Find the volume of a tetrahedron whose vertices are $A(-1,2,3),B(3,-2,1),C(2,1,3)$ and $D(-1,-2,4)$

Answer 1

Let $\overline{a},\overline{b},\overline{c},\overline{d}$ be the position vectors of points A,B,C,D respectively of a tetrahedron.

$\begin{array}{c}\overline{a}=-\hat{i}+2\hat{j}+3\hat{k},\overline{b}=3\hat{i}+-2\hat{j}+\hat{k},\\ \overline{c}=2\hat{i}+\hat{j}+3\hat{k},\overline{d}=-\hat{i}-2\hat{j}+4\hat{k}\end{array}$

Now,

$\begin{array}{c}\overline{A}\overline{B}=\overline{b}-\overline{a}=\left(3\hat{i}+-2\hat{j}+\hat{k}\right)-\left(-\hat{i}+2\hat{j}+3\hat{k}\right)\\ =4\hat{i}+-4\hat{j}-2\hat{k}\\ \overline{A}\overline{C}=\overline{c}-\overline{a}=\left(2\hat{i}+\hat{j}+3\hat{k}\right)-\left(2\hat{i}+\hat{j}+3\hat{k}\right)\\ =3\hat{i}-\hat{j}\\ \overline{A}\overline{D}=\overline{d}-\overline{a}=\left(-\hat{i}-2\hat{j}+4\hat{k}\right)-\left(2\hat{i}+\hat{j}+3\hat{k}\right)\\ =-4\hat{j}+\hat{k}\end{array}$

Volume of a tetrahedron whose coterminus edges are $\overline{A}\overline{B},\overline{A}\overline{C},\overline{A}\overline{D}is\frac{1}{6}\left[\overline{A}\overline{B}\overline{A}\overline{C}\overline{A}\overline{D}\right]$

Volume of the tetrahedron

$\begin{array}{c}=\frac{1}{6}\left(\begin{array}{ccc}& -4& -2\\ 3& -1& 0\\ 0& -4& 1\end{array}\right)\\ =\frac{1}{6}\left[4\left(-1-0\right)+4\left(3-0\right)-2\left(-12-0\right)\right]\\ =\frac{1}{6}\left(4+12+24\right)\\ =\frac{1}{6}\left(32\right)\\ =\frac{16}{3}\end{array}$

volume of the tetrahedron is $=\frac{16}{3}$ cubic units .