question_answer

If \[\vec{a},\] \[\vec{b}\] and \[\vec{c}\] are three non-coplanar vectors, prove that \[[\begin{matrix}    \vec{a}+\vec{b}+\vec{c} & \vec{a}+\vec{b} & \vec{a}+\vec{c}  \\ \end{matrix}]=-[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}].\]

Answer:

\[LHS=[\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\,\overrightarrow{a}+\overrightarrow{b}\,\overrightarrow{a}+\overrightarrow{c}]\] \[=(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\cdot [(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}+\overrightarrow{c})]\] \[=(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\cdot [(\overrightarrow{a}\times \overrightarrow{a})+(\overrightarrow{a}\times \overrightarrow{c})+(\overrightarrow{b}\times \overrightarrow{c})]\] [by using distributive law] \[=(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\cdot [0-(\overrightarrow{c}\times \overrightarrow{a})-(\overrightarrow{a}\times \overrightarrow{b})+(\overrightarrow{b}\times \overrightarrow{c})]\] \[[\because \overrightarrow{a}\times \overrightarrow{a}=0,\,\,\overrightarrow{a}\times \overrightarrow{c}=-\,\overrightarrow{c}\times \overrightarrow{a},\,\,\overrightarrow{b}\times \overrightarrow{a}=-\,\overrightarrow{a}\times \overrightarrow{b}]\] \[=-\,\overrightarrow{a}\cdot (\overrightarrow{c}\times \overrightarrow{a})-\overrightarrow{a}\cdot (\overrightarrow{a}+\overrightarrow{b})+(\overrightarrow{a}\times \overrightarrow{b})+\overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{c})]\] \[-\,\overrightarrow{b}\cdot (\overrightarrow{c}\times \overrightarrow{a})-\overrightarrow{b}\cdot (\overrightarrow{a}\times \overrightarrow{b})+\overrightarrow{b}\cdot (\overrightarrow{b}\times \overrightarrow{c})\] \[-\,\overrightarrow{c}\cdot (\overrightarrow{c}+\overrightarrow{a})-\overrightarrow{c}\cdot (\overrightarrow{a}\times \overrightarrow{b})+\overrightarrow{c}\cdot (\overrightarrow{b}\times \overrightarrow{c})\] \[=-\,[\overrightarrow{a}\,\,\overrightarrow{c}\,\,\overrightarrow{a}]-[\overrightarrow{a}\,\,\overrightarrow{a}\,\,\overrightarrow{b}]+[\overrightarrow{a}\,\,\overrightarrow{b}\,\,\overrightarrow{c}]\] \[-\,\,[\overrightarrow{b}\,\,\overrightarrow{c}\,\,\overrightarrow{a}]-[\overrightarrow{b}\,\,\overrightarrow{a}\,\,\overrightarrow{b})+[\overrightarrow{b}\,\,\overrightarrow{b}\,\,\overrightarrow{c}]\] \[-\,\,[\overrightarrow{c}\,\,\overrightarrow{c}\,\,\overrightarrow{a}]-[\overrightarrow{c}\,\,\overrightarrow{a}\,\,\overrightarrow{b})+[\overrightarrow{c}\,\,\overrightarrow{b}\,\,\overrightarrow{c}]\] \[=-\,\,0-\,\,0+[\overrightarrow{a}\,\,\overrightarrow{b}\,\,\overrightarrow{c}]-[\overrightarrow{b}\,\,\overrightarrow{c}\,\,\overrightarrow{a})-0+0-0-[\overrightarrow{c}\,\,\overrightarrow{a}\,\,\overrightarrow{b}]+0\][\[\because \]scalar triple product of two equal vector is zero] \[=[\overrightarrow{a}\,\,\overrightarrow{b}\,\,\overrightarrow{c}]-[\overrightarrow{a}\,\,\overrightarrow{b}\,\,\overrightarrow{c})-[\overrightarrow{a}\,\,\overrightarrow{b}\,\,\overrightarrow{c}]\,\,[\because [\overrightarrow{b}\,\,\overrightarrow{c}\,\,\overrightarrow{a}]=[\overrightarrow{c}\,\,\overrightarrow{a}\,\,\overrightarrow{b}]=[\overrightarrow{a}\,\,\overrightarrow{b}\,\,\overrightarrow{c}]]\]            \[=-\,\,[\overrightarrow{a}\,\,\overrightarrow{b}\,\,\overrightarrow{c}]\] \[=R.H.S\]                    Hence proved.