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

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(A) Given that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.This implies $\vec{a} + \vec{b} = -\vec{c}$.Squaring both sides,we get $|\vec{a} + \vec{b}|^2 =

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$\pi/3$

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(A) Given that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.This implies $\vec{a} + \vec{b} = -\vec{c}$.Squaring both sides,we get $|\vec{a} + \vec{b}|^2 = |-\vec{c}|^2$.$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{c}|^2$.Substituting the given values $|\vec{a}| = 3$,$|\vec{b}| = 5$,and $|\vec{c}| = 7$:$3^2 + 5^2 + 2|\vec{a}||\vec{b}| \cos 	heta = 7^2$.$9 + 25 + 2(3)(5) \cos 	heta = 49$.$34 + 30 \cos 	heta = 49$.$30 \cos 	heta = 49 - 34$.$30 \cos 	heta = 15$.$\cos 	heta = \frac{15}{30} = \frac{1}{2}$.Since $\cos 	heta = \frac{1}{2}$,the angle $	heta = \frac{\pi}{3}$.

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

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