Question

Find the value of $\lambda$ if the volume of a tetrahedron whose vertices are with position vectors $\hat{i}-6\hat{j}+10\hat{k},-\hat{i}-3\hat{j}+7\hat{k},5\hat{i}-\hat{j}+\lambda \hat{k}$ and $7\hat{i}-4\hat{j}+7\hat{k}$ is $11$ cubic unit

• A. $7$
• B. $9$
• C. $11$
• D. $13$

Answer 1

The correct option is A $7$

Let the vertices be $A,B,C,D$ and $O$ be the origin.
$\therefore \overrightarrow{OA}=\hat{i}-6\hat{j}+10\hat{k},\overrightarrow{OB}=\hat{i}-3\hat{j}+7\hat{k}$,
$\overrightarrow{OC}=-5\hat{i}-\hat{j}+\lambda \hat{k},\overrightarrow{OD}=7\hat{i}-4\hat{j}+7\hat{k}$
$\therefore \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=-2\hat{i}+3\hat{j}-3\hat{k}$
$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=4\hat{i}+5\hat{j}+(\lambda -10)k$
$\overrightarrow{AD}=\overrightarrow{OD}-\overrightarrow{OA}=6\hat{i}+2\hat{j}-3\hat{k}$

Volume of tetrahedron
$V=\frac{1}{6}\left[\overrightarrow{AB}\overrightarrow{AC}\overrightarrow{AD}\right]=\frac{1}{6}\left|\begin{array}{ccc}-2& 3& -3\\ 4& 5& \lambda -10\\ 6& 2& -3\end{array}\right|$
$V=\frac{1}{6}\{-2(-15-2\lambda +20)-3(-12-6\lambda +60)-3(8-10\left)\right\}$

$V=\frac{1}{6}\{4\lambda -10-144+18\lambda +66\}$

$V=\frac{1}{6}(22\lambda -88)=11$ $($given $)$

or $2\lambda -8=6$

or $\lambda =7$