Question

The vector $\overline{i}+2x\overline{j}+2\overline{k}$ rotated through an angle $\theta$ and doubled in magnitude then it becomes $2\overline{i}+(2x+2)\overline{j}+(6x-2)\overline{k}$, then the values of $x$ are

• A. $1,\frac{1}{3}$
• B. $-1,\frac{1}{3}$
• C. $1,\frac{-1}{3}$
• D. In between $(0,3)$

Answer 1

The correct option is C $1,\frac{-1}{3}$

Let $\overrightarrow{V}=\hat{i}+2x\hat{j}+2\hat{k}$
$\left|\overrightarrow{V}\right|=\sqrt{1+4x^2+4}$
Let $\overrightarrow{U}=2\hat{i}+(2x+2)\hat{j}+(6x-2)\hat{k}$
$\left|\overrightarrow{U}\right|=\sqrt{4+4(x+1{)}^2+4(3x-1{)}^2}$
According to the condition given $2\left|\overrightarrow{V}\right|=\left|\overrightarrow{U}\right|$
$4(5+4x^2)=4(1+(x+1{)}^2+(3x-1{)}^2)$
$5+4x^2=1+x^2+1+2x+9x^2-6x+1$
$5+4x^2=3+10x^2-4x$
=$6x^2-4x-2$
$0=(x-1)(6x+2)$
$x=1,\frac{-1}{3}$