Question

The vector $ \hat{i}+x\hat{j}+3\hat{k} $ is rotated through an angle $ θ $ 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, $2\left|\hat{i}+x\hat{j}+3\hat{k}\right|$
$=\left|4\hat{i}+\left(4x-2\right)\hat{j}+2\hat{k}\right|$
$\Rightarrow 2\sqrt{1+x^{2}+9}=\sqrt{4^{2}+\left(4x-2\right)^{2}+2^{2}}$
$\Rightarrow 4\left(x^{2}+10\right)=16+16x^{2}+4-16x+4$
$\Rightarrow 12x^{2}-16x-16=0$
$\Rightarrow 3x^{2}-4x-4=0$
$\Rightarrow \left(3x+2\right)\left(x-2\right)=0$
$\Rightarrow x=2$, $-\frac{2}{3}$