Question

The vector $\hat{ i }+x \hat{ j }+3 \hat{ k }$ is rotated through an angle $\theta$ and doubled in magnitude, then it becomes $4 \hat{ i }+(4 x-2) \hat{ j }+2 \hat{ k }$. The value 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. $\{2,7\}$

Answer 1

Answer: (A)

The vector $\hat{ i }+x \hat{ j }+z \hat{ k }$ doubled in magnitude, then it becomes $4 \hat{ i }+(4 x-2) \hat{ j }+2 \hat{ k }$
$ \therefore 2|\hat{ i }+x \hat{ j }+3 \hat{ k }|=\mid 4 \hat{ i }+(4 x-2) \hat{ j }+2 \hat{ k }$
$\Rightarrow 2 \sqrt{1+x^{2}+9}=\sqrt{16+(4 x-2)^{2}+4}$
$\Rightarrow 40+4 x^{2}=20+(4 x-2)^{2}$
$\Rightarrow 3 x^{2}-4 x-4=0$
$\Rightarrow 3 x^{2}-6 x+2 x-4=0$
$\Rightarrow 3 x(x-2)+2(x-2)=0$
$\Rightarrow (x-2)(3 x+2)=0$
$\Rightarrow x=2,-\frac{2}{3}$