# Q. If $\overrightarrow{a}, \overrightarrow{b}$ and $\overrightarrow{c}$ are unit vectors, then $|\overrightarrow{a}-\overrightarrow{b} |^2+| \overrightarrow{b}-\overrightarrow{c}|^2+| \overrightarrow{c}-\overrightarrow{a}|^2$ does not exceed

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Solution:

## Now, $(\overrightarrow{a}+\overrightarrow{b}\overrightarrow{c})^2=\boldsymbol{\Sigma}\overrightarrow{a}^2+2\boldsymbol{\Sigma}\overrightarrow{a}.\overrightarrow{b}\ge 0$ $\Rightarrow \hspace25mm 2\boldsymbol{\Sigma}\overrightarrow{a}.\overrightarrow{b} \ge -3 [\because |\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1]$ Now, $\boldsymbol{\Sigma}|\overrightarrow{a}-\overrightarrow{b}|^2=2\boldsymbol{\Sigma}\overrightarrow{a}^2-2\boldsymbol{\Sigma}\overrightarrow{a}.\overrightarrow{b}\ge 2(3)+3=9$

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