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(C) Let $	heta$ be the angle between unit vectors $\vec{a}$ and $\vec{b}$.Since $\vec{a}$ and $\vec{b}$ are unit vectors,$|\vec{a}| = 1$ and $|\vec{b

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$\pi /3$

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(C) Let $	heta$ be the angle between unit vectors $\vec{a}$ and $\vec{b}$.Since $\vec{a}$ and $\vec{b}$ are unit vectors,$|\vec{a}| = 1$ and $|\vec{b}| = 1$.Given that $(\vec{a} + 3\vec{b})$ is perpendicular to $(7\vec{a} - 5\vec{b})$,their dot product is zero:$(\vec{a} + 3\vec{b}) \cdot (7\vec{a} - 5\vec{b}) = 0$Expanding the dot product:$7(\vec{a} \cdot \vec{a}) - 5(\vec{a} \cdot \vec{b}) + 21(\vec{b} \cdot \vec{a}) - 15(\vec{b} \cdot \vec{b}) = 0$$7|\vec{a}|^2 + 16(\vec{a} \cdot \vec{b}) - 15|\vec{b}|^2 = 0$Substituting $|\vec{a}| = 1$,$|\vec{b}| = 1$,and $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos 	heta = \cos 	heta$:$7(1)^2 + 16 \cos 	heta - 15(1)^2 = 0$$7 + 16 \cos 	heta - 15 = 0$$16 \cos 	heta - 8 = 0$$16 \cos 	heta = 8$$\cos 	heta = \frac{8}{16} = \frac{1}{2}$Since $\cos 	heta = \frac{1}{2}$,the angle $	heta = \frac{\pi}{3}$.

If $\vec{a}$ and $\vec{b}$ are unit vectors such that the vector $\vec{a} + 3\vec{b}$ is perpendicular to $7\vec{a} - 5\vec{b}$,then find the angle between $\vec{a}$ and $\vec{b}$.

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