If for vectors vec{a}, vec{b}, and vec{c},vec | vedclass

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(A) Given $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.We need the angle between $\vec{b}$ and $\vec{c}$,so we write $\vec{b} + \vec{c} = -\vec{a}$.Squarin

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$60$

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(A) Given $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.We need the angle between $\vec{b}$ and $\vec{c}$,so we write $\vec{b} + \vec{c} = -\vec{a}$.Squaring both sides,we get $|\vec{b} + \vec{c}|^2 = |-\vec{a}|^2$.$|\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{b} \cdot \vec{c}) = |\vec{a}|^2$.Substituting the given magnitudes: $5^2 + 3^2 + 2(\vec{b} \cdot \vec{c}) = 7^2$.$25 + 9 + 2(\vec{b} \cdot \vec{c}) = 49$.$34 + 2(\vec{b} \cdot \vec{c}) = 49$.$2(\vec{b} \cdot \vec{c}) = 15 \implies \vec{b} \cdot \vec{c} = \frac{15}{2}$.Let $	heta$ be the angle between $\vec{b}$ and $\vec{c}$.Then $\cos 	heta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|} = \frac{15/2}{5 	imes 3} = \frac{15/2}{15} = \frac{1}{2}$.Since $\cos 	heta = \frac{1}{2}$,$	heta = 60^o$.

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

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