Question

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

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

Answer 1

The correct option is A

$\left\{-\frac{2}{3},2\right\}$

Explanation for correct option

The vector $\hat{i}+x\hat{j}+3\hat{k}$ is double in magnitude .

$$\begin{gathered} \therefore 2\left|\hat{i}+x\hat{j}+3\hat{k}\right|=\left|4\hat{i}+(4x-2)\hat{j}+2\hat{k}\right| \\ \Rightarrow 2\sqrt{1^2+x^2+3^2}=\sqrt{4^2+{\left(4x-2\right)}^2+2^2} \\ \Rightarrow 4\left(10+x^2\right)=\left(20+{\left(4x-2\right)}^2\right) \\ \Rightarrow 40+4x^2=20+16x^2-16x+4 \\ \Rightarrow 12x^2-16x-16=0 \\ \Rightarrow \left(x-2\right)\left(3x+2\right)=0 \\ \therefore x=2,x=-\frac{2}{3} \end{gathered}$$

Hence option (A) is correct