If two vectors vec{A} and vec{B} having equal | vedclass

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(C) The magnitude of the sum of two vectors is given by $|\vec{A}+\vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.Given $A = B = R$,we have:$|\vec{A}+\

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$|\vec{A}+\vec{B}|=2 R \cos \left(\frac{	heta}{2}ight)$

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(C) The magnitude of the sum of two vectors is given by $|\vec{A}+\vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.Given $A = B = R$,we have:$|\vec{A}+\vec{B}| = \sqrt{R^2 + R^2 + 2R^2 \cos 	heta} = \sqrt{2R^2(1 + \cos 	heta)}$.Using the identity $1 + \cos 	heta = 2 \cos^2(	heta/2)$:$|\vec{A}+\vec{B}| = \sqrt{2R^2 \cdot 2 \cos^2(	heta/2)} = \sqrt{4R^2 \cos^2(	heta/2)} = 2R \cos(	heta/2)$.Similarly,for the difference of two vectors:$|\vec{A}-\vec{B}| = \sqrt{A^2 + B^2 - 2AB \cos 	heta} = \sqrt{2R^2(1 - \cos 	heta)}$.Using the identity $1 - \cos 	heta = 2 \sin^2(	heta/2)$:$|\vec{A}-\vec{B}| = \sqrt{2R^2 \cdot 2 \sin^2(	heta/2)} = 2R \sin(	heta/2)$.Comparing with the given options,option $C$ is correct.

If two vectors $\vec{A}$ and $\vec{B}$ having equal magnitude $R$ are inclined at an angle $\theta$,then

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