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] \[=[\begin{matrix} {\vec{a}} & {\vec{b}} & {\vec{c}} \\ \end{matrix}]-[\begin{matrix} {\vec{a}} & {\vec{b}} & {\vec{c}} \\ \end{matrix}]=0\] \[[\begin{matrix} \because [\vec{b} & {\vec{c}} & {\vec{a}} \\ \end{matrix}]=[\begin{matrix} {\vec{a}} & {\vec{b}} & {\vec{c}} \\ \end{matrix}]]\]
You need to login to perform this action.
You will be redirected in
3 sec