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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 :

Option: 1
$\sin ^{-1}\left(\frac{3}{5}\right)$

Option: 2
$\sin ^{-1}\left(\frac{1}{3}\right)$

Option: 3
$\cos ^{-1}\left(\frac{3}{5}\right)$

Option: 4
$\cos ^{-1}\left(\frac{1}{3}\right)$

Answers (1)

best_answer

If $\mathrm{\overrightarrow{\text{A}}\; and\; \overrightarrow{\text{B}}}$ angle is $\theta$ then angle between $\mathrm{\overrightarrow{\text{A}}\; and\; \overrightarrow{\text{-B}}}$ is $180^{\circ }-\theta$

$|\vec{A}+\vec{B}|=2|\vec{A}-\vec{B}|$

$\mathrm{\sqrt{A^{2}+A^{2}+2 A^{2} \cos \theta}=2 \sqrt{A^{2}+A^{2}-2 A^{2} \cos \theta}}$

$\mathrm{5\cos \theta =3}$

$\mathrm{\theta=\cos ^{-1}\left(\frac{3}{5}\right)}$

Hence, correct answer is Option (3)

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