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

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$\{\frac{2}{3}, 0\}$

$\{\frac{1}{3}, 2\}$

$\{- \frac{2}{3}, 2\}$

${2, 7}$

answer is A.

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Detailed Solution

$\begin{array}{l} 2 | \bar{i} + x \bar{j} + 3 \bar{k} | = | 4 \bar{i} + (4 x - 2) \bar{j} + 2 \bar{k} | \\ \Rightarrow 4 (x^{2} + 10) = 20 + (4 x - 2)^{2} \\ \Rightarrow 3 x^{2} - 4 x - 4 = 0 x = 2, \frac{- 2}{3} \end{array}$

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