The vector $2 i + 2 j + 3 k$ is rotated about origin through an angle $θ$ and becomes $i + 3 j + 2 k$ then $θ =$

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$\sin^{- 1} (\frac{\sqrt{43}}{7})$

$\tan^{- 1} (\frac{\sqrt{37}}{16})$

$\cos^{- 1} (\frac{16}{17})$

$\sin^{- 1} (\sqrt{\frac{33}{17}})$

answer is A.

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Detailed Solution

. $a = 2 i + 2 j + 3 k b = 2 i + 3 j + 2 k cosθ = \frac{a \cdot b}{| a | | b |} = \frac{4 + 6 + 6}{\sqrt{4 + 4 + 9} \sqrt{4 + 9 + 4}} = \frac{16}{17}$

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