If vec{a} + vec{b} + vec{c} = 0,|vec{a}| = 3, | vedclass

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(A) Given $\vec{a} + \vec{b} + \vec{c} = 0$,we can write $\vec{a} + \vec{b} = -\vec{c}$.Taking the dot product of both sides with themselves:$(\vec{a}

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$\frac{\pi}{3}$

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(A) Given $\vec{a} + \vec{b} + \vec{c} = 0$,we can write $\vec{a} + \vec{b} = -\vec{c}$.Taking the dot product of both sides with themselves:$(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (-\vec{c}) \cdot (-\vec{c})$$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{c}|^2$Since $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 	heta$,where $	heta$ is the angle between $\vec{a}$ and $\vec{b}$:$3^2 + 5^2 + 2(3)(5) \cos 	heta = 7^2$$9 + 25 + 30 \cos 	heta = 49$$34 + 30 \cos 	heta = 49$$30 \cos 	heta = 15$$\cos 	heta = \frac{15}{30} = \frac{1}{2}$Therefore,$	heta = \frac{\pi}{3}$.

If $\vec{a} + \vec{b} + \vec{c} = 0$,$|\vec{a}| = 3$,$|\vec{b}| = 5$,and $|\vec{c}| = 7$,then the angle between $\vec{a}$ and $\vec{b}$ is

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