If vec{a}, vec{b}, vec{c} are 3 vectors such | vedclass

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(A) Given $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$,we can write $\vec{a}+\vec{b}=-\vec{c}$.Squaring both sides,we get $|\vec{a}+\vec{b}|^2 = |-\ve

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$\cos ^{-1} \frac{2}{5}$

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(A) Given $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$,we can write $\vec{a}+\vec{b}=-\vec{c}$.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 magnitudes: $5^2 + 8^2 + 2|\vec{a}||\vec{b}| \cos 	heta = 11^2$,where $	heta$ is the angle between $\vec{a}$ and $\vec{b}$.$25 + 64 + 2(5)(8) \cos 	heta = 121$.$89 + 80 \cos 	heta = 121$.$80 \cos 	heta = 121 - 89 = 32$.$\cos 	heta = \frac{32}{80} = \frac{2}{5}$.Therefore,$	heta = \cos ^{-1} \left(\frac{2}{5}ight)$.

If $\vec{a}, \vec{b}, \vec{c}$ are $3$ vectors such that $|\vec{a}|=5, |\vec{b}|=8, |\vec{c}|=11$ and $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$,then the angle between the vectors $\vec{a}$ and $\vec{b}$ is

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