• # question_answer Let $\vec{a},\text{ }\vec{b}$ and $\vec{c}$ be three unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}.$ If $\lambda \,=\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}.$ and $\vec{d}=\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}$ then the ordered pair, $\left( \lambda ,\vec{d} \right)$ is equal to                                 [JEE MAIN Held on 07-01-2020 Evening] A) $\left( \frac{3}{2},3\vec{a}\times \vec{c} \right)$     B) $\left( -\frac{3}{2},3\vec{c}\times \vec{b} \right)$ C) $\left( -\frac{3}{2},3\vec{a}\times \vec{b} \right)$ D) $\left( -\frac{3}{2},3\vec{b}\times \vec{c} \right)$

[c]  $\because \,\,\left| {\vec{a}} \right|=\left| {\vec{b}} \right|=\left| {\vec{c}} \right|=1$             and  $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ On squaring both sides $\,{{\left| {\vec{a}} \right|}^{2}}={{\left| {\vec{b}} \right|}^{2}}={{\left| {\vec{c}} \right|}^{2}}+2(\vec{a}.\,\vec{b}+\vec{a}.\,\vec{c}+\vec{b}.\,\vec{c})=0$ $\therefore \,\,\,\lambda =\vec{a}.\,\vec{b}+\vec{a}.\,\vec{c}+\vec{b}.\,\vec{c}=-\frac{3}{2}$ and $\vec{d}=\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}$ $=\,\,\vec{a}\times \vec{b}+\vec{b}\times (-\vec{a}-\vec{b})+(-\vec{a}-\vec{b})\times \vec{a}$ $=\,\,\vec{a}\times \vec{b}-\vec{b}\times \vec{a}-0-0-\vec{b}\times \vec{a}$ $=\,3\,(\vec{a}\,\times \vec{b})$ $\therefore \,\,\,\,\,\,\,(\lambda ,\vec{d})=\left( -\frac{3}{2},3(\vec{a}\times \vec{b}) \right)$