overline{a}, overline{b} and overline{c} are | vedclass

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(C) Given $\overline{a}+\overline{b}+\overline{c}=\overline{0}$,we have $\overline{c}=-(\overline{a}+\overline{b})$.Let $	heta$ be the angle between

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(C) Given $\overline{a}+\overline{b}+\overline{c}=\overline{0}$,we have $\overline{c}=-(\overline{a}+\overline{b})$.Let $	heta$ be the angle between $\overline{a}$ and $\overline{b}$.Taking the square of both sides: $|\overline{c}|^2 = |-(\overline{a}+\overline{b})|^2 = |\overline{a}+\overline{b}|^2$.Expanding the right side: $|\overline{c}|^2 = |\overline{a}|^2 + |\overline{b}|^2 + 2(\overline{a} \cdot \overline{b})$.Using the definition of the dot product: $|\overline{c}|^2 = |\overline{a}|^2 + |\overline{b}|^2 + 2|\overline{a}||\overline{b}| \cos 	heta$.Substituting the given values: $7^2 = 3^2 + 5^2 + 2(3)(5) \cos 	heta$.$49 = 9 + 25 + 30 \cos 	heta$.$49 = 34 + 30 \cos 	heta$.$15 = 30 \cos 	heta$.$\cos 	heta = \frac{15}{30} = \frac{1}{2}$.Therefore,$	heta = \frac{\pi}{3}$.

$\overline{a}, \overline{b}$ and $\overline{c}$ are three vectors such that $\overline{a}+\overline{b}+\overline{c}=\overline{0}$ and $|\overline{a}|=3, |\overline{b}|=5, |\overline{c}|=7$. The angle between $\overline{a}$ and $\overline{b}$ is:

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