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If three unit vectors $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}} \quad$ satisfy $\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{0}}$, then the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is
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Verified Answer
The correct answer is:
$\frac{2 \pi}{3}$
Given, condition is
$\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$ ...(i)
and $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}$ are the unit vectors.
Then $\quad|\vec{a}|=|\vec{b}|=|\vec{c}|=1$
Let the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is $\theta$.
Now, from Eq. (i)
$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathbf{c}}=0$
$(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})=-\overrightarrow{\mathbf{c}}$
Squaring on both sides
$(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})^2=(\overrightarrow{\mathbf{c}})^2 \quad\left[\because(\overrightarrow{\mathbf{c}})^2=|\overrightarrow{\mathbf{c}}|^2\right]$
$\Rightarrow(\overrightarrow{\mathbf{a}})^2+(\overrightarrow{\mathbf{b}})^2+2(\overrightarrow{\mathbf{a}}) \cdot(\overrightarrow{\mathbf{b}})=|\overrightarrow{\mathbf{c}}|^2$
$\Rightarrow \quad|\overrightarrow{\mathbf{a}}|^2+|\overrightarrow{\mathbf{b}}|^2+2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=|\overrightarrow{\mathbf{c}}|^2$
$\Rightarrow \quad 1+1+2\{|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \cos \theta\}=1$
$[\because \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \cos \theta]$
$\Rightarrow \quad 2\{1 \cdot 1 \cdot \cos \theta\}=-1$
$\Rightarrow \quad \cos \theta=\frac{-1}{2}$
$\Rightarrow \quad \cos \theta=\cos \left(\frac{2 \pi}{3}\right)$
$\Rightarrow \quad \theta=\frac{2 \pi}{3}$
$\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$ ...(i)
and $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}$ are the unit vectors.
Then $\quad|\vec{a}|=|\vec{b}|=|\vec{c}|=1$
Let the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is $\theta$.
Now, from Eq. (i)
$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathbf{c}}=0$
$(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})=-\overrightarrow{\mathbf{c}}$
Squaring on both sides
$(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})^2=(\overrightarrow{\mathbf{c}})^2 \quad\left[\because(\overrightarrow{\mathbf{c}})^2=|\overrightarrow{\mathbf{c}}|^2\right]$
$\Rightarrow(\overrightarrow{\mathbf{a}})^2+(\overrightarrow{\mathbf{b}})^2+2(\overrightarrow{\mathbf{a}}) \cdot(\overrightarrow{\mathbf{b}})=|\overrightarrow{\mathbf{c}}|^2$
$\Rightarrow \quad|\overrightarrow{\mathbf{a}}|^2+|\overrightarrow{\mathbf{b}}|^2+2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=|\overrightarrow{\mathbf{c}}|^2$
$\Rightarrow \quad 1+1+2\{|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \cos \theta\}=1$
$[\because \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \cos \theta]$
$\Rightarrow \quad 2\{1 \cdot 1 \cdot \cos \theta\}=-1$
$\Rightarrow \quad \cos \theta=\frac{-1}{2}$
$\Rightarrow \quad \cos \theta=\cos \left(\frac{2 \pi}{3}\right)$
$\Rightarrow \quad \theta=\frac{2 \pi}{3}$
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