Question

The vector $\hat{i}+x\hat{j}+3\hat{k}$ is rotated through an angle $\theta$ and doubled in magnitude, then it becomes $4\hat{i}+\left(4x-2\right)\hat{j}+2\hat{k}$ . The value of $x$ is

• A. $\left\{-\frac{2}{3},2\right\}$
• B. $\left\{\frac{1}{3},2\right\}$
• C. $\left\{\frac{2}{3},0\right\}$
• D. $\{2,7\}$

Answer 1

Answer: (A)

We have, $2\left|\hat{i}+x\hat{j}+3\hat{k}\right|$
$=\left|4\hat{i}+\left(4x-2\right)\hat{j}+2\hat{k}\right|$
$\Rightarrow 2\sqrt{1+x^2+9}=\sqrt{4^2+{\left(4x-2\right)}^2+2^2}$
$\Rightarrow 4\left(x^2+10\right)=16+16x^2+4-16x+4$
$\Rightarrow 12x^2-16x-16=0$
$\Rightarrow 3x^2-4x-4=0$
$\Rightarrow \left(3x+2\right)\left(x-2\right)=0$
$\Rightarrow x=2$, $-\frac{2}{3}$