Question

If the volume of a parallelepiped, whose coterminous edges are given by the vectors $\overrightarrow{a}=\hat{i}+\hat{j}+n\hat{k},\overrightarrow{b}=2\hat{i}+4\hat{j}-n\hat{k}$ and $\overrightarrow{c}=\hat{i}+n\hat{j}+3\hat{k}(n\ge 0),$ is $158$ cu.units, then

• A. $n=9$
• B. $\overrightarrow{b}.\overrightarrow{c}=10$
• C. $\overrightarrow{a}.\overrightarrow{c}=17$
• D. $n=7$

Answer 1

The correct option is B $\overrightarrow{b}.\overrightarrow{c}=10$

Volume of parallelepiped $=\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$

$\left|\begin{array}{ccc}1& 1& n\\ 2& 4& -n\\ 1& n& 3\end{array}\right|=158$

$\Rightarrow (12+n^2)-1(6+n)+n(2n-4)=158$
$\Rightarrow 3n^2-5n-152=0$
$\Rightarrow 3n^2-24n+19n-152=0$
$\Rightarrow 3n(n-8)+19(n-8)=0$
$\Rightarrow n=8$
$\therefore \overrightarrow{a}=\hat{i}+\hat{j}+8\hat{k},\overrightarrow{b}=2\hat{i}+4\hat{j}-8\hat{k}$ and
$\overrightarrow{c}=\hat{i}+8\hat{j}+3\hat{k}$
$\overrightarrow{a}.\overrightarrow{c}=1+8+24=33$
$\overrightarrow{b}.\overrightarrow{c}=2+32-24=10$