question_answer

Simplify \[[\begin{matrix}    \vec{a}-b & \vec{b}-\vec{c} & \vec{c}-\vec{a}  \\ \end{matrix}].\]

Answer:

We have, \[[\begin{matrix}    \vec{a}-\vec{b} & \vec{b}-\vec{c} & \vec{c}-\vec{a}  \\ \end{matrix}]\] \[=\{(\vec{a}-\vec{b})\times (\vec{b}-\vec{c})\}\cdot (\vec{c}-\vec{a})\]    [by definition] \[=(\vec{a}\times \vec{b}-\vec{a}\times c-\vec{b}\times b+\vec{b}\times \vec{c})\cdot (\vec{c}-\vec{a})\]                                     [by distributive law] \[=(\vec{a}\times \vec{b}+\vec{c}\times \vec{a}+\vec{b}\times \vec{c})\cdot (\vec{c}-\vec{a})\] \[[\because \vec{b}\times \vec{b}=\vec{0}\,\,\text{and}\,\,\vec{a}\times \vec{c}=-(\vec{c}\times \vec{a})]\] \[=(\vec{a}\times \vec{b})\cdot \vec{c}-(\vec{a}\times \vec{b})\cdot \vec{a}+(\vec{c}\times \vec{a})\cdot \vec{c}\]             \[-\,(\vec{c}\times \vec{a})\cdot \vec{a}+(\vec{b}\times \vec{c})\cdot \vec{c}-(\vec{b}\times \vec{c})\cdot \vec{a}\]                                     [by distributive law] \[=[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]-[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{a}}  \\ \end{matrix}]+[\begin{matrix}    {\vec{c}} & {\vec{a}} & {\vec{c}}  \\ \end{matrix}]-[\begin{matrix}    {\vec{c}} & {\vec{a}} & {\vec{a}}  \\ \end{matrix}]\]                                    \[+[\begin{matrix}    {\vec{b}} & {\vec{c}} & {\vec{c}}  \\ \end{matrix}]-[\begin{matrix}    {\vec{b}} & {\vec{c}} & {\vec{a}}  \\ \end{matrix}]\] \[=[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]-[\begin{matrix}    {\vec{b}} & {\vec{c}} & {\vec{a}}  \\ \end{matrix}]\] [\[\because \] scalar triple product, when any two vectors are equal, is zero] \[=[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]-[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]=0\]                                                        \[[\begin{matrix}    \because [\vec{b} & {\vec{c}} & {\vec{a}}  \\ \end{matrix}]=[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]]\]