Two vectors vec{A} and vec{B} are defined as | vedclass

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(B) Given $\vec{A} = a \hat{i}$ and $\vec{B} = a \cos \omega t \hat{i} + a \sin \omega t \hat{j}$.$\vec{A} + \vec{B} = a(1 + \cos \omega t) \hat{i} +

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$2$

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(B) Given $\vec{A} = a \hat{i}$ and $\vec{B} = a \cos \omega t \hat{i} + a \sin \omega t \hat{j}$.$\vec{A} + \vec{B} = a(1 + \cos \omega t) \hat{i} + a \sin \omega t \hat{j}$.$|\vec{A} + \vec{B}|^2 = a^2(1 + \cos \omega t)^2 + a^2 \sin^2 \omega t = a^2(1 + 2 \cos \omega t + \cos^2 \omega t + \sin^2 \omega t) = a^2(2 + 2 \cos \omega t) = 2a^2(1 + \cos \omega t) = 4a^2 \cos^2(\omega t / 2)$.So,$|\vec{A} + \vec{B}| = 2a \cos(\omega t / 2)$.Similarly,$\vec{A} - \vec{B} = a(1 - \cos \omega t) \hat{i} - a \sin \omega t \hat{j}$.$|\vec{A} - \vec{B}|^2 = a^2(1 - \cos \omega t)^2 + a^2 \sin^2 \omega t = a^2(1 - 2 \cos \omega t + \cos^2 \omega t + \sin^2 \omega t) = a^2(2 - 2 \cos \omega t) = 4a^2 \sin^2(\omega t / 2)$.So,$|\vec{A} - \vec{B}| = 2a \sin(\omega t / 2)$.Given $|\vec{A} + \vec{B}| = \sqrt{3}|\vec{A} - \vec{B}|$,we have $2a \cos(\omega t / 2) = \sqrt{3} \cdot 2a \sin(\omega t / 2)$.$	an(\omega t / 2) = 1 / \sqrt{3}$.$\omega t / 2 = \pi / 6 \implies \omega t = \pi / 3$.Since $\omega = \pi / 6 	ext{ rad s}^{-1}$,we get $(\pi / 6) \cdot t = \pi / 3$.Therefore,$t = 2 	ext{ s}$.

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