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(C) Let the magnitudes of vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ be $A = B = a$.The magnitude of the sum is given by $|\overrightarrow{

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the magnitudes of vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ be $A = B = a$.The magnitude of the sum is given by $|\overrightarrow{A} + \overrightarrow{B}| = \sqrt{a^2 + a^2 + 2a^2 \cos 	heta} = \sqrt{2a^2(1 + \cos 	heta)} = \sqrt{4a^2 \cos^2(	heta/2)} = 2a \cos(	heta/2)$.The magnitude of the difference is given by $|\overrightarrow{A} - \overrightarrow{B}| = \sqrt{a^2 + a^2 - 2a^2 \cos 	heta} = \sqrt{2a^2(1 - \cos 	heta)} = \sqrt{4a^2 \sin^2(	heta/2)} = 2a \sin(	heta/2)$.According to the problem,$|\overrightarrow{A} + \overrightarrow{B}| = 2|\overrightarrow{A} - \overrightarrow{B}|$.Substituting the expressions,we get $2a \cos(	heta/2) = 2(2a \sin(	heta/2))$.This simplifies to $\cos(	heta/2) = 2 \sin(	heta/2)$,or $	an(	heta/2) = 1/2$.Using the identity $\cos 	heta = \frac{1 - 	an^2(	heta/2)}{1 + 	an^2(	heta/2)}$,we substitute $	an(	heta/2) = 1/2$:$\cos 	heta = \frac{1 - (1/2)^2}{1 + (1/2)^2} = \frac{1 - 1/4}{1 + 1/4} = \frac{3/4}{5/4} = \frac{3}{5}$.Therefore,$	heta = \cos^{-1}(3/5)$.

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