If overline{a}+overline{b}+overline{c}=overli | vedclass

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(A) Given that $\overline{a}+\overline{b}+\overline{c}=\overline{0}$.We can write this as $\overline{a}+\overline{b}=-\overline{c}$.Squaring both side

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(A) Given that $\overline{a}+\overline{b}+\overline{c}=\overline{0}$.We can write this as $\overline{a}+\overline{b}=-\overline{c}$.Squaring both sides,we get $|\overline{a}+\overline{b}|^2 = |-\overline{c}|^2$,which implies $|\overline{a}+\overline{b}|^2 = |\overline{c}|^2$.Expanding the left side using the dot product property $|\overline{u}+\overline{v}|^2 = |\overline{u}|^2 + |\overline{v}|^2 + 2|\overline{u}||\overline{v}| \cos 	heta$,we have:$|\overline{a}|^2 + |\overline{b}|^2 + 2|\overline{a}||\overline{b}| \cos 	heta = |\overline{c}|^2$.Substituting the given values $|\overline{a}|=3, |\overline{b}|=5, |\overline{c}|=7$:$3^2 + 5^2 + 2(3)(5) \cos 	heta = 7^2$.$9 + 25 + 30 \cos 	heta = 49$.$34 + 30 \cos 	heta = 49$.$30 \cos 	heta = 49 - 34 = 15$.$\cos 	heta = \frac{15}{30} = \frac{1}{2}$.Since $\cos 	heta = \frac{1}{2}$,the angle $	heta = \frac{\pi}{3}$.

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

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