Question

The vector $\hat{i}+x\hat{j}+3\hat{k}$ is rotated through an angle $\theta$ and is doubled in magnitude. It now becomes $4\hat{i}+(4x-2)\hat{j}+2\hat{k}$. The value(s) of $x$ are

• A. $1$
• B. $\frac{-2}{3}$
• C. $2$
• D. $\frac{4}{3}$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{-2}{3}$
C $2$

Let $\overrightarrow{\alpha }=\hat{i}+x\hat{j}+3\hat{k},\overrightarrow{\beta }=4\hat{i}+(4x-2)\hat{j}+2\hat{k}$
Given, $2|\overrightarrow{\alpha }|=|\overrightarrow{\beta }|$
or $2\sqrt{10+x^2}=\sqrt{20+4(2x-1{)}^2}$
or $10+x^2=5+(4x^2-4x+1)$
or $3x^2-4x-4=0$
or $x=2,-\frac{2}{3}$