Question

If the three co-terminous sides of a tetrahedron is given as $(\hat{i}+2\hat{j}+4\hat{k}),(2\hat{i}+3\hat{j}-\hat{k})$ and $(-3\hat{i}+\hat{j}+\hat{k}).$ The volume (in cubic units) of above tetrahedron is $V,$ then value of $3V$ is

Answer 1

Let sides of tetrahedron are $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$
And,
Given: $\overrightarrow{a}=(\hat{i}+2\hat{j}+4\hat{k}),\overrightarrow{b}=(2\hat{i}+3\hat{j}-\hat{k}),\overrightarrow{c}=(-3\hat{i}+\hat{j}+\hat{k})$
$\therefore$ Volume of tetrahedron $=\frac{1}{6}|\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]|=\frac{1}{6}\left|\left|\begin{array}{ccc}1& 2& 4\\ 2& 3& -1\\ -3& 1& 1\end{array}\right|\right|=\frac{1}{6}\left|\right(1(3+1)-2(2-3)+4(2+9)\left)\right|=V=\frac{50}{6}=\frac{25}{3}\text{cubic units}$.