Question

Let $\overrightarrow{x}$ be a vector in the plane containing vectors $\overrightarrow{a}=2\hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{b}=\hat{i}+2\hat{j}-\hat{k}$. If the vector $\overrightarrow{x}$ is perpendicular to $(3\hat{i}+2\hat{j}-\hat{k})$ and its projection on $\vec{a}$ is $\frac{17\sqrt{6}}{2}$, then the value of $|\overrightarrow{x}{|}^2$ is equal to ________.

Answer 1

Answer: 486

Let $\overrightarrow{x}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}(\lambda$ and $\mu$ are scalars $)$
$\overrightarrow{x}=\hat{i}(2\lambda +\mu )+\hat{j}(2\mu -\lambda )+\hat{k}(\lambda -\mu )$
Since $\overrightarrow{x}\cdot (3\hat{i}+2\hat{j}-\hat{k})=0$
$3\lambda +8\mu =0\dots \dots (1)$
Also Projection of $\overrightarrow{x}$ on $\overrightarrow{a}$ is $\frac{17\sqrt{6}}{2}$
$\frac{\overrightarrow{x}\cdot \overrightarrow{a}}{|\overrightarrow{a}|}=\frac{17\sqrt{6}}{2}$
$6\lambda -\mu =51\dots$(2)
From (1) and (2)
$\lambda =8,\mu =-3$
$\overrightarrow{x}=13\hat{i}-14\hat{j}+11\hat{k}$
$|\overrightarrow{x}{|}^2=486$