JEE Main 2021 — Vector Algebra Question with Solution

A plane containing two vectors can be expressed as a linear combination of the vectors.

Hence, let $\overrightarrow{x}=\lambda \overrightarrow{a}+\mu \overrightarrow{b} (\lambda$ and $\mu$ are scalars$).$

$\Rightarrow \overrightarrow{x}=\hat{i}(2\lambda +\mu )+\hat{j}(2\mu -\lambda )+\hat{k}(\lambda -\mu )$

Since $\overrightarrow{x}$ is perpendicular to $3\hat{i}+2\hat{j}-\hat{k},$

$\Rightarrow \overrightarrow{x}·\left(3\hat{i}+2\hat{j}-\hat{k}\right)=0$

$\Rightarrow \left(\hat{i}(2\lambda +\mu )+\hat{j}(2\mu -\lambda )+\hat{k}(\lambda -\mu )\right)·\left(3\hat{i}+2\hat{j}-\hat{k}\right)=0$

$\Rightarrow 3(2\lambda +\mu )+2(2\mu -\lambda )-(\lambda -\mu )=0$

$\Rightarrow 3\lambda +8\mu =0 ...\left(1\right)$

Also, the projection of $\overrightarrow{x}$ on $\overrightarrow{a}$ is $\frac{17\sqrt{6}}{2}$

$\Rightarrow \frac{\overrightarrow{x}·\overrightarrow{a}}{\left|\overrightarrow{a}\right|}=\frac{17\sqrt{6}}{2}$

$\Rightarrow \frac{\left(\hat{i}(2\lambda +\mu )+\hat{j}(2\mu -\lambda )+\hat{k}(\lambda -\mu )\right)·\left(2\hat{i}-\hat{j}+\hat{k}\right)}{\sqrt{2^2+1^2+1^2}}=\frac{17\sqrt{6}}{2}$

$\Rightarrow \frac{2(2\lambda +\mu )-(2\mu -\lambda )+(\lambda -\mu )}{\sqrt{6}}=\frac{17\sqrt{6}}{2}$

$\Rightarrow 6\lambda -\mu =51 ...\left(2\right)$

On solving the equations $\left(1\right)$ and $\left(2\right),$ we get $\lambda =8, \mu =-3.$

Thus, $\overrightarrow{x}=13\hat{i}-14\hat{j}+11\hat{k}$

$\Rightarrow \left|\overrightarrow{x}\right|=\sqrt{{13}^2+{\left(-14\right)}^2+{11}^2}$

$\Rightarrow |\overrightarrow{x}{|}^2=486.$