If three unit vectors overrightarrow{a}, over | vedclass

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(A) Given, the condition is $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ ...$(i)$ Since $\overrightarrow{a}, \overrig

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$\frac{2 \pi}{3}$

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(A) Given, the condition is $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ ...$(i)$ Since $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are unit vectors, we have $|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1$. Let the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ be $	heta$. From equation $(i)$, we can write: $\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}$ Squaring both sides, we get: $(\overrightarrow{a}+\overrightarrow{b})^2=(-\overrightarrow{c})^2$ $|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2(\overrightarrow{a} \cdot \overrightarrow{b})=|\overrightarrow{c}|^2$ Substituting the values $|\overrightarrow{a}|=1, |\overrightarrow{b}|=1, |\overrightarrow{c}|=1$: $1^2+1^2+2(\overrightarrow{a} \cdot \overrightarrow{b})=1^2$ $2+2(\overrightarrow{a} \cdot \overrightarrow{b})=1$ $2(\overrightarrow{a} \cdot \overrightarrow{b})=-1$ $\overrightarrow{a} \cdot \overrightarrow{b}=-\frac{1}{2}$ Using the definition of the dot product, $\overrightarrow{a} \cdot \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}| \cos 	heta$: $1 \cdot 1 \cdot \cos 	heta = -\frac{1}{2}$ $\cos 	heta = -\frac{1}{2}$ Since $\cos 	heta = -\frac{1}{2}$, the angle $	heta = \frac{2 \pi}{3}$.

If three unit vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ satisfy $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is

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