Question

If the volume of the tetrahedron with edges $\hat{i}+\hat{j}+\hat{k},\hat{i}+a\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}-\hat{k}$ is $6$ cubic units, then $a$ is

• A. $1$
• B. $-1$
• C. $2$
• D. $-17$

Answer 1

The correct option is D $-17$

Volume of the tetrahedron $=\frac{1}{6}\left|\left[\overrightarrow{A}\overrightarrow{B}\overrightarrow{C}\right]\right|$
So, $\frac{1}{6}\left|\begin{array}{ccc}1& 1& 1\\ 1& a& 1\\ 1& 2& -1\end{array}\right|=\pm 6$

$\Rightarrow \left|\begin{array}{ccc}1& 1& 1\\ 1& a& 1\\ 1& 2& -1\end{array}\right|=\pm 36$

Applying $C_3\to C_3-C_1,$ we get
$\left|\begin{array}{ccc}1& 1& 0\\ 1& a& 0\\ 1& -2& -2\end{array}\right|=\pm 36$

$\Rightarrow -2(a-1)=\pm 36$
$\Rightarrow a-1=\pm 18$
$\Rightarrow a=-17,19$