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(C) Given that $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are unit vectors,so $|\overrightarrow{a}|=|\overrightarrow{b}|=|\overright

Vedclass · Accepted Answer

$-3$

Vedclass · Answer

(C) Given that $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are unit vectors,so $|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1$. Given $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$. Squaring both sides: $(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}) \cdot (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}) = 0$. $|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+|\overrightarrow{c}|^2+2(\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a}) = 0$. $1+1+1+2(\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a}) = 0$. $3+2(\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a}) = 0 \implies \overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a} = -3/2$. Since $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$,the vectors form an equilateral triangle,so the angle between any two vectors is $120^{\circ}$. Thus,$\overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{b} \cdot \overrightarrow{c} = \overrightarrow{c} \cdot \overrightarrow{a} = 1 	imes 1 	imes \cos(120^{\circ}) = -1/2$. Substituting these values: $3(-1/2)+2(-1/2)+(-1/2) = -3/2 - 2/2 - 1/2 = -6/2 = -3$.

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