question_answer

            Prove that   \[\vec{a}.\,(\vec{b}\,+\vec{c})\times (\vec{a}\,+2\,\vec{b}\,+3\,\vec{c})=[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}].\]

Answer:

We have, \[\vec{a}.\,(\overrightarrow{b}\,+\vec{c})\times (\vec{a}\,+2\,\overrightarrow{b}\,+3\,\vec{c})\] \[=\vec{a}.\{(\vec{b}\,+\vec{c})\times (\vec{a}\,+2\vec{b}\,+3\vec{c})\}\]          ­\[=\vec{a}.\,\{\overrightarrow{b}\times \vec{a}+2(\overrightarrow{b}\times \overrightarrow{b})+3\,(\overrightarrow{b}\times \vec{c})+\vec{c}\times \vec{a}\]                         \[+\,2(\vec{c}\times \overrightarrow{b})\,+3(\vec{c}\times \vec{c})\}\]             \[=\vec{a}.\,\{\overrightarrow{b}\times \vec{a}+3\,(\overrightarrow{b}\times \vec{c})+\vec{c}\times \vec{a}\,-2(\overrightarrow{b}\times \vec{c})\}\] \[=\vec{a}.\,\{-\,(\vec{a}\,\times \overrightarrow{b})+\overrightarrow{b}\times \vec{c}+\vec{c}\times \vec{a}\}\] \[=-\,\vec{a}.\,(\vec{a}\,\times \overrightarrow{b})+\vec{a}.(\overrightarrow{b}\times \vec{c})+\vec{a}.(\vec{c}\,\times \vec{a})\] \[=0+[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]+0=[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]\] Hence proved.