The vector $\mathbf{x}$ is perpendicular to the vectors $\mathbf{a}=3 \hat{i}+2 \hat{j}+2 \hat{k},…
The vector $\mathbf{x}$ is perpendicular to the vectors $\mathbf{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}, \mathbf{b}=18 \hat{i}-22 \hat{j}-5 \hat{k}$ and make an obtuse angle with $\hat{j}$. If $|\mathbf{x}|=14$, then $\mathbf{x}=$
$8 \hat{i}+12 \hat{j}+24 \hat{k}$
$-8 \hat{i}+6 \hat{j}+24 \hat{k}$
$8 \hat{i}-12 \hat{j}-24 \hat{k}$
$-8 \hat{i}-12 \hat{j}+24 \hat{k}$
Solution
$
\mathbf{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}, \mathbf{b}=18 \hat{i}-22 \hat{j}-5 \hat{k}
$
$\mathbf{x}$ is perpendicular to $\mathbf{a}$ and $\mathbf{b}$.
$\therefore \mathbf{x}$ is parallel to $(\mathbf{a} \times \mathbf{b})$.
Now, $\mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 18 & -22 & -5\end{array}\right|$
$
\Rightarrow \mathbf{a} \times \mathbf{b}=34 \hat{i}+51 \hat{j}-102 \hat{k}
$
As $\mathbf{x}$ makes obtuse angle with $\hat{j}$, coefficient of $\hat{j}$ should be negative.
$
\begin{aligned}
\therefore \quad \mathbf{x} & =k(-2 \hat{i}-3 \hat{j}+6 \hat{k}), k>0 \\
|\mathbf{x}| & =k \sqrt{4+9+36} \\
& =7 k=14 \Rightarrow k=2 \\
\therefore \quad \mathbf{x} & =-4 \hat{i}-6 \hat{j}+12 \hat{k} \\
& \mathbf{x}=-8 \hat{i}-12 \hat{j}+24 \hat{k}
\end{aligned}
$