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If $|\vec{A} \times \vec{B}|=\sqrt{3}$ A. $B$, then the value of $|\vec{A}+\vec{B}|$ is:
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Verified Answer
The correct answer is:
$\left(A^2+B^2+A B\right)^{1 / 2}$
$|\vec{A} \times \vec{B}|=\sqrt{3}(\vec{A} \cdot \vec{B})$
$A B \sin \theta=\sqrt{3} A B \cos \theta \Rightarrow \tan \theta=\sqrt{3} \therefore \theta=60^{\circ}$
Now $|\vec{R}|=|\vec{A}+\vec{B}|=\sqrt{A^2+B^2+2 A B \cos \theta}$
$=\sqrt{A^2+B^2+2 A B\left(\frac{1}{2}\right)}=\left(A^2+B^2+A B\right)^{1 / 2}$
$A B \sin \theta=\sqrt{3} A B \cos \theta \Rightarrow \tan \theta=\sqrt{3} \therefore \theta=60^{\circ}$
Now $|\vec{R}|=|\vec{A}+\vec{B}|=\sqrt{A^2+B^2+2 A B \cos \theta}$
$=\sqrt{A^2+B^2+2 A B\left(\frac{1}{2}\right)}=\left(A^2+B^2+A B\right)^{1 / 2}$
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