Question

The volume of the tetrahedron whose co-terminous edges are $(-12\hat{i}+\lambda \hat{k}),(3\hat{j}-\hat{k})$ and $(2\hat{i}+\hat{j}-15\hat{k})$ is $546$ cubic units.Then the value of $\lambda$ is

• A. $458$
• B. $-458$
• C. $300$
• D. $-3$

Answer 1

The correct option is B $-458$

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three co-terminous edges of tetrahedron. And, $\overrightarrow{a}=(-12\hat{i}+\lambda \hat{k}),\overrightarrow{b}=(3\hat{j}-\hat{k}),\overrightarrow{c}=(2\hat{i}+\hat{j}-15\hat{k})$
Where, Volume of tetrahedron $=\left|\frac{1}{6}\right[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\left]\right|$
$=\left|\frac{1}{6}\left|\begin{array}{ccc}-12& 0& \lambda \\ 0& 3& -1\\ 2& 1& -15\end{array}\right|\right|=\left|\frac{1}{6}\right(-12(-45+1)+\lambda (0-6)\left)\right|$
$\Rightarrow |88-\lambda |=546\Rightarrow 88-\lambda =\pm 546$
$\Rightarrow \lambda =-458,634$