Q. The vector $\widehat{\dot{\mathbf{i}}} + x\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}}$ is rotated through an angle $\theta$ and doubled in magnitude, then it becomes $4\widehat{\dot{\mathbf{i}}} + (4x - 2)\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}}.$ The values 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) $\left\{ 2,7 \right\}$
Explanation
(a)
Since, $2\left| \widehat{\dot{\mathbf{i}}} + x\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}} \right| = \left| 4\widehat{\dot{\mathbf{i}}} + (4x - 2)\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}} \right|$
$\Longrightarrow 2\sqrt{1 + x^{2} + 9} = \sqrt{4^{2} + (4x - 2)^{2} + 2^{2}}$
$\Longrightarrow \ {12x}^{2} - 16x - 16 = 0$
$\Longrightarrow (3x + 2)(x - 2) = 0$
$\Longrightarrow x = 2, - \frac{2}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.