Question

The vector $\hat{i}+x\hat{j}+3\hat{k}$ is rotated through an angle $\theta$ and is doubled in magnitude. It now becomes $4\hat{i}+(4x-2)\hat{j}+2\hat{k}$. The values of $x$ are

• A. $1$
• B. $-\frac{2}{3}$
• C. $2$
• D. $\frac{4}{3}$

Answer 1

Answer: (C)

$2$

Given:

$\overrightarrow{v_1}=\hat{i}+x\hat{j}+3\hat{k}$

After rotating by$\angle \theta$, $\overrightarrow{v_1}$ becomes $\overrightarrow{v_2}$ such that

$\left|\overrightarrow{v_1}\right|=\left|\overrightarrow{v_2}\right|$

$\overrightarrow{v_3}=4\hat{i}+(4x-2)\hat{j}+2\hat{k}$

$\left|\overrightarrow{v_3}\right|=2\left|\overrightarrow{v_1}\right|$ $\left(1\right)$

$|\overrightarrow{v_1}|=\sqrt{1^2+x^2+3^2}$

$|\overrightarrow{v_3}|=\sqrt{4^2+(4x-2{)}^2+2^2}$

Substituting $\left|\overrightarrow{v_1}\right|$ and $\left|\overrightarrow{v_3}\right|$ in Eq. $\left(1\right)$:

$\sqrt{4^2+(4x-2{)}^2+2^2}=2\sqrt{1^2+x^2+3^2}$

$\sqrt{2^2+(2x-1{)}^2+1^2}=\sqrt{1^2+x^2+3^2}$

$\sqrt{5+(2x-1{)}^2}=\sqrt{10+x^2}$

$3x^2-4x-4=0$

$\Rightarrow x=2,-\frac{2}{3}$

Case 1: $x=2$

$\overrightarrow{v_1}=\hat{i}+2\hat{j}+3\hat{k}$

$\overrightarrow{v_3}=4\hat{i}+6\hat{j}+2\hat{k}$

$\left|\overrightarrow{v_1}\right|=\sqrt{1+4+9}=\sqrt{14}$

$\left|\overrightarrow{v_3}\right|=\sqrt{16+36+4}=\sqrt{56}=2\sqrt{14}$

$\therefore \left|\overrightarrow{v_3}\right|=2\left|\overrightarrow{v_1}\right|$

Case 2: $x=-\frac{2}{3}$

$\overrightarrow{v_1}=\hat{i}-\frac{2}{3}\hat{j}+3\hat{k}$

$\overrightarrow{v_3}=4\hat{i}-\frac{14}{3}\hat{j}+2\hat{k}$

$\left|\overrightarrow{v_1}\right|=\sqrt{\frac{94}{9}}$

$\left|\overrightarrow{v_3}\right|=2\sqrt{\frac{69}{9}}$

$\left|\overrightarrow{v_3}\right|\ne 2\left|\overrightarrow{v_1}\right|$

$\therefore x=2$

Option C