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(C) Given that $|\vec{a}|=3, |\vec{b}|=4$ and $|\vec{a}+\vec{b}|=\sqrt{37}$.Using the identity $|\vec{a}+\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\ve

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$3$

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(C) Given that $|\vec{a}|=3, |\vec{b}|=4$ and $|\vec{a}+\vec{b}|=\sqrt{37}$.Using the identity $|\vec{a}+\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}$:$37 = 3^2 + 4^2 + 2\vec{a} \cdot \vec{b}$$37 = 9 + 16 + 2\vec{a} \cdot \vec{b}$$37 = 25 + 2\vec{a} \cdot \vec{b} \Rightarrow 2\vec{a} \cdot \vec{b} = 12 \Rightarrow \vec{a} \cdot \vec{b} = 6$.Since $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos 	heta$, we have $6 = 3 \cdot 4 \cos 	heta \Rightarrow \cos 	heta = \frac{6}{12} = \frac{1}{2}$.Thus, $	heta = 60^{\circ}$ and $\sin 	heta = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$.Now, using $|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}$:$k^2 = 9 + 16 - 12 = 13$.Finally, calculate $\frac{4}{13}(k \sin 	heta)^2 = \frac{4}{13} \cdot k^2 \sin^2 	heta = \frac{4}{13} \cdot 13 \cdot (\frac{\sqrt{3}}{2})^2 = 4 \cdot \frac{3}{4} = 3$.

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=3, |\vec{b}|=4, |\vec{a}+\vec{b}|=\sqrt{37}, |\vec{a}-\vec{b}|=k$ and the angle between $\vec{a}$ and $\vec{b}$ is $\theta$, then find the value of $\frac{4}{13}(k \sin \theta)^2$.

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