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

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

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

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

$2, 7$

answer is A.

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Detailed Solution

$\sqrt{1 b + {(4 x - 2)}^{2} + 4} = 2 \sqrt{1 + x^{2} + 9} S . B . S \Rightarrow 20 + 16 x^{2} + 4 - 16 x = (x^{2} + 10) 4 12 x^{2} - 16 x - 16 = 0 \Rightarrow 3 x^{2} - 4 x - 4 = 0 \Rightarrow x = 2, \frac{- 2}{3}$

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