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Question:

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$, find the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} .$

Solution:

It is given that $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$.

$\therefore \vec{a} \cdot(\vec{a}+\vec{b}+\vec{c})=\vec{a} \cdot \overrightarrow{0}$

$\Rightarrow \vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=\vec{a} \cdot \overrightarrow{0}$                   [Distributivity of scalar product over addition]

$\Rightarrow 1+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=0$           ...(1) $\left[\begin{array}{l}\vec{a} \cdot \vec{a}=|\vec{a}| \cdot|\vec{a}| \cos 0^{\circ}=1 \\ (\vec{a} \text { is unit vector } \Rightarrow|\vec{a}|=1)\end{array}\right]$

$\therefore \vec{b} \cdot(\vec{a}+\vec{b}+\vec{c})=\vec{b} \cdot \overrightarrow{0}$

$\Rightarrow \vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{b} \cdot \vec{c}=\vec{b} \cdot \overrightarrow{0}$

$\Rightarrow \vec{b} \cdot \vec{a}+1+\vec{b} \cdot \vec{c}=0$             ...(2)                   $[\vec{b} \cdot \vec{b}=1]$

$\therefore \vec{c} \cdot(\vec{a}+\vec{b}+\vec{c})=\vec{c} \cdot \overrightarrow{0}$

$\Rightarrow \vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}+\vec{c} \cdot \vec{c}=\vec{c} \cdot \overrightarrow{0}$

$\Rightarrow \vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}+1=0$                ...(3)                   $[\vec{c} \cdot \vec{c}=1]$

$(1+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c})+(\vec{b} \cdot \vec{a}+1+\vec{b} \cdot \vec{c})+(\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}+1)=0+0+0$

$\Rightarrow(3+\vec{a} \cdot \vec{b}+\vec{c} \cdot \vec{a})+(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c})+(\vec{c} \cdot \vec{a}+\vec{b} \cdot \vec{c})=0$                                                                                                                           [Scalar product is commutative]

$\Rightarrow 3+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$

$\Rightarrow \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=\frac{-3}{2}$

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