Question
$\text{If}\ \ \vec{a},\ \vec{b},\vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0},$ find the value of $\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec {a}.$
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Answer
Here
$\vec{a}+\vec{b}+\vec{c}=\vec{0}$
$\therefore\ \ \vec{a}+\vec{b}=-\vec{c}\ \ \ \ .....(1)$
$\therefore\ \ \vec{a}.(\vec{a}+\vec{b})=-\vec{a}.\vec{c}$
$\Rightarrow\ \ \vec{a}.\vec{a}+\vec{a}.\vec{b}=-\vec{a}.\vec{c}$
$\Rightarrow\ \big|\vec{a}\big|^2 +\ \vec{a}.\vec{b}+\vec{a}.\vec{c}=0$
$\therefore \ \ 1+\vec{a}.\vec{b}+\vec{a}.\vec{c}=0 $ $\ [\because\ \vec{a}\ \text{is a unit vector}]$
$\text{From}(1),\ \vec{b}.(\vec{a}+\vec{b})=-\vec{b}.\vec{c}\ \Rightarrow\ \ \vec{b}.\vec{a}+\vec{b}.\vec{b}=-\vec{b}.\vec{c}$
$\Rightarrow\ \vec{b}.\vec{a}+\big|\vec{b}\big|^2+\vec{b}.\vec{c}=0$
$\Rightarrow\ \ \vec{b}.\vec{a}+1+\vec{b}.\vec{c}=0$ $\ [\because\ \vec{b}\ \text{is a unit vector}]$
$\therefore\ \ \vec{b}.\vec{a}+\vec{b}.\vec{c}=-1\ \ ....(3)$
$\text{Again form(1)},\ \vec{c}.(\vec{a}+\vec{b})=-\vec{c}.\vec{c}$
$\therefore\ \vec{c}.\vec{a}+\vec{c}.\vec{b}=-\big|\vec{c}\big|^2 \ $ $\Rightarrow\ \ \vec{c}.\vec{a}+\vec{c}.\vec{b}+\big|\vec{c}\big|^2=0$
$\Rightarrow\ \ \vec{c}.\vec{a}+\vec{c}.\vec{b}+1=0$ $\ \ \ \ \ \ [\because\ \vec{c}\ \text{is a unit vector}]$
$\Rightarrow\ \ \vec{c}.\vec{a}+\vec{c}.\vec{b}=-1$
Adding (2), (3) and (4), we get
$2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a})=-3$ $\ \ \ \ \ [\because\ \vec{a}.\vec{b}=\vec{b}.\vec{a}\ \text{etc.}]$
$\therefore\ \vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}=-\frac{3}{2}$
$\vec{a}+\vec{b}+\vec{c}=\vec{0}$
$\therefore\ \ \vec{a}+\vec{b}=-\vec{c}\ \ \ \ .....(1)$
$\therefore\ \ \vec{a}.(\vec{a}+\vec{b})=-\vec{a}.\vec{c}$
$\Rightarrow\ \ \vec{a}.\vec{a}+\vec{a}.\vec{b}=-\vec{a}.\vec{c}$
$\Rightarrow\ \big|\vec{a}\big|^2 +\ \vec{a}.\vec{b}+\vec{a}.\vec{c}=0$
$\therefore \ \ 1+\vec{a}.\vec{b}+\vec{a}.\vec{c}=0 $ $\ [\because\ \vec{a}\ \text{is a unit vector}]$
$\text{From}(1),\ \vec{b}.(\vec{a}+\vec{b})=-\vec{b}.\vec{c}\ \Rightarrow\ \ \vec{b}.\vec{a}+\vec{b}.\vec{b}=-\vec{b}.\vec{c}$
$\Rightarrow\ \vec{b}.\vec{a}+\big|\vec{b}\big|^2+\vec{b}.\vec{c}=0$
$\Rightarrow\ \ \vec{b}.\vec{a}+1+\vec{b}.\vec{c}=0$ $\ [\because\ \vec{b}\ \text{is a unit vector}]$
$\therefore\ \ \vec{b}.\vec{a}+\vec{b}.\vec{c}=-1\ \ ....(3)$
$\text{Again form(1)},\ \vec{c}.(\vec{a}+\vec{b})=-\vec{c}.\vec{c}$
$\therefore\ \vec{c}.\vec{a}+\vec{c}.\vec{b}=-\big|\vec{c}\big|^2 \ $ $\Rightarrow\ \ \vec{c}.\vec{a}+\vec{c}.\vec{b}+\big|\vec{c}\big|^2=0$
$\Rightarrow\ \ \vec{c}.\vec{a}+\vec{c}.\vec{b}+1=0$ $\ \ \ \ \ \ [\because\ \vec{c}\ \text{is a unit vector}]$
$\Rightarrow\ \ \vec{c}.\vec{a}+\vec{c}.\vec{b}=-1$
Adding (2), (3) and (4), we get
$2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a})=-3$ $\ \ \ \ \ [\because\ \vec{a}.\vec{b}=\vec{b}.\vec{a}\ \text{etc.}]$
$\therefore\ \vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}=-\frac{3}{2}$
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