If overrightarrow{a}+overrightarrow{b}+overri | vedclass

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Question

(D) Given,$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ $	herefore \overrightarrow{c}=-(\overrightarrow{a}+\overright

Vedclass · Accepted Answer

$\pi / 3$

Vedclass · Answer

(D) Given,$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ $	herefore \overrightarrow{c}=-(\overrightarrow{a}+\overrightarrow{b})$ Squaring both sides,we get: $|\overrightarrow{c}|^2 = (\overrightarrow{a}+\overrightarrow{b}) \cdot (\overrightarrow{a}+\overrightarrow{b})$ $|\overrightarrow{c}|^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{b})$ Since $\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}||\overrightarrow{b}| \cos 	heta$,where $	heta$ is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$: $|\overrightarrow{c}|^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + 2|\overrightarrow{a}||\overrightarrow{b}| \cos 	heta$ 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}$.

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

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