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

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Question

(D) Given that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$.We can write this as $\overrightarrow{a}+\overrightarrow{

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(D) Given that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$.We can write this as $\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}$.Squaring both sides,we get $(\overrightarrow{a}+\overrightarrow{b})^2=(-\overrightarrow{c})^2$.Expanding the dot product,we have $|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2|\overrightarrow{a}||\overrightarrow{b}| \cos 	heta=|\overrightarrow{c}|^2$,where $	heta$ is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.Substituting the given values: $3^2+4^2+2(3)(4) \cos 	heta = (\sqrt{37})^2$.$9+16+24 \cos 	heta = 37$.$25+24 \cos 	heta = 37$.$24 \cos 	heta = 37-25 = 12$.$\cos 	heta = \frac{12}{24} = \frac{1}{2}$.Since $\cos 	heta = \frac{1}{2}$,we have $	heta = \frac{\pi}{3}$.

If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ and $|\overrightarrow{a}|=3, |\overrightarrow{b}|=4$ and $|\overrightarrow{c}|=\sqrt{37}$,then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is:

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