Two vectors $\vec{A}$ and $\vec{B}$ have equal magnitudes. If magnitude of $\vec{A} + \vec{B}$ is equal to two times the magnitude of $\vec{A} - \vec{B}$ , then the angle between $\vec{A}$ and $\vec{B}$ will be :

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$\sin^{- 1} (\frac{1}{3})$

$\cos^{- 1} (\frac{3}{5})$

$\cos^{- 1} (\frac{1}{3})$

$\sin^{- 1} (\frac{3}{5})$

answer is C.

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Detailed Solution

$|\vec{A} + \vec{B}| = \sqrt{2 A^{2} (1 + \cos θ)}$

$|\vec{A} - \vec{B}| = \sqrt{2 A^{2} (1 - \cos θ)}$

$g i v e n |\vec{A} + \vec{B}| = 2 |\vec{A} - \vec{B}|$

$\frac{\sqrt{2 A^{2} (1 + \cos θ)}}{\sqrt{2 A^{2} (1 - \cos θ)}} = 2$

$\frac{(1 + \cos θ)}{(1 - \cos θ)} = 4; \cos θ = \frac{3}{5}$

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