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If \( |\vec{A} \times \vec{B}|=\sqrt{3} \vec{A} \cdot \vec{B} \), then the value of \( |\vec{A}+\vec{B}| \) is 

\( (a)\left(A^{2}+B^{2}+A B\right)^{1 / 2} \) 

(b) \( \left(A^{2}+B^{2}+\frac{A B}{\sqrt{3}}\right)^{1 / 2} \) 

(c) \( A+B \) 

\( (d)\left(A^{2}+B^{2}+\sqrt{3} A B\right)^{1 / 2} \)

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Correct option is (a) \((A^2 + B^2 + AB)^\frac12\)

\(|\vec A \times \vec B| = \sqrt 3\vec A .\vec B\)

⇒ \(|\vec A| |\vec B| sin \theta = \sqrt 3 |\vec A| |\vec B| cos \theta\) 

⇒ \(sin \theta = \sqrt 3 \,cos\theta\)

⇒ \(tan \theta = \sqrt 3 = tan 60°\)

⇒ \(\theta = 60° \) 

\(|\vec A + \vec B|^2 = (\vec A + \vec B). (\vec A + \vec B)\)

\(= \vec A . \vec A + 2\vec A. \vec B + \vec B.\vec B\)

\(= |\vec A| ^2 + |\vec B|^2 + 2|\vec A| |\vec B| cos 60° \)  \((\because \theta = 60° )\) 

\( = |\vec A|^2 + |\vec B|^2 + 2|\vec A| |\vec B| \times \frac 12\)

\( = |\vec A|^2 + |\vec B|^2 + |\vec A| |\vec B| \) 

\(\therefore |\vec A + \vec B| = \left( |\vec A|^2 + |\vec B|^2 + |\vec A||\vec B|\right)^\frac12\)

\(= (A^2 + B^2 + AB)^\frac12\)

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