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 value of x are

• A. $-\frac{2}{3},2$
• B. $\frac{1}{3},2$
• C. $\frac{2}{3},2$
• D. $zero,2$

Answer 1

The correct option is A $-\frac{2}{3},2$

The vector's magnitude after rotation through an angle $\theta$ does not change.Therefore we have,

$\Rightarrow 2\sqrt{1^2+x^2+3^2}=\sqrt{4^2+{(4x-2)}^2+2^2}\Rightarrow 2\sqrt{1^2+x^2+9}=\sqrt{16+16x^2+4-16x+4}\Rightarrow 4(1^2+x^2+9)=16+16x^2+8-16x\Rightarrow 4+4x^2+36=16+16x^2+8-16x\Rightarrow 12x^2-16x-16=0\Rightarrow 3x^2-4x-4=0\Rightarrow x=\frac{4\pm \sqrt{16+4\times 3\times 4}}{2+3}=\frac{4\pm \sqrt{64}}{6}=\frac{4\pm 8}{6}=2,\frac{-2}{3}$