question_answer

Let \[\vec{a},\,\,\vec{b}\] and \[\vec{c}\] be non-zero, non-coplanar vectors. Prove that \[\vec{a}-2\vec{b}+3\vec{c},\] \[-\,2\vec{a}+3\vec{b}\,-4\vec{c}\] and \[\vec{a}-3\vec{b}\,+5\vec{c}\] are coplanar vectors.

Answer:

Given \[\vec{a},\] \[\vec{b}\] and \[\vec{c}\] are non-zero, non-coplanar vectors. To show \[\vec{a}-2\vec{b}+3\vec{c},\] \[-\,2\vec{a}+3\vec{b}-4\vec{c}\] and \[\vec{a}-3\vec{b}+5\vec{c}\] are coplanar vectors. Since, \[\vec{a},\] \[\vec{b}\] and \[\vec{c}\] are non-zero, non-coplanar vectors therefore we have,  \[[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]\ne 0\]                                ?(i) Now, let \[\vec{x}=\vec{a}-2\vec{b}+3\vec{c}\] \[\vec{y}=-\,2\vec{a}+3\vec{b}-4\vec{c}\] and  \[\vec{z}=\vec{a}-3\vec{b}+5\vec{c}\] Then, \[\vec{y}\times \vec{z}=(-\,2\vec{a}+3\vec{b}-4\vec{c})\times (\vec{a}-3\vec{b}+5\vec{c})\]\[=-\,2(\vec{a}\times \vec{a})+6(\vec{a}\times \vec{b})-10(\vec{a}\times \vec{c})+3(\vec{b}\times \vec{a})\]      \[-9(\vec{b}\times \vec{b})+15(\vec{b}\times \vec{c})-\,4(\vec{c}\times \vec{a})\] \[+12(\vec{c}\times \vec{b})-20(\vec{c}\times \vec{c})\] [\[\because \] cross product is distributive over addition and subtraction]              \[=6(\vec{a}\times \vec{b})+10(\vec{c}\times \vec{a})-3(\vec{a}\times \vec{b})+15(\vec{b}\times \vec{c})\] \[-\,4(\vec{c}\times \vec{a})-12(\vec{b}\times \vec{c})\] \[[\because \vec{a}\times \vec{a}=\vec{b}\times \vec{b}=\vec{c}\times \vec{c}=0\,\,\]and for vectors\[\vec{u}\]and\[\vec{v},\vec{u}\times \vec{v}=-(\vec{v}\times \vec{u})]\] \[=3(\vec{a}\times \vec{b})+6(\vec{c}\times \vec{a})+3(\vec{b}\times \vec{c})\] \[\therefore \] \[\vec{x}\cdot (\vec{y}\times \vec{z})=(\vec{a}\,-2\vec{b}+3\vec{c}).\]                         \[3\{(\vec{a}\,\,\times \,\,\vec{b})+2(\vec{c}\,\,\times \,\,\vec{a})+(\vec{b}\,\,\times \,\,\vec{c})\}\] \[=3[(\vec{a}-2\vec{b}+3\vec{c})\cdot \{(\vec{a}\times \vec{b})+(\vec{b}\times \vec{c})+2(\vec{c}\times \vec{a})\}\] \[=3[\vec{a}\cdot (\vec{a}\times \vec{b})+\vec{a}\cdot (\vec{b}\times \vec{c})+2\vec{a}\cdot (\vec{c}\times \vec{a})-\,2\vec{b}\cdot (\vec{a}\times \vec{b})\] \[-\,2b\cdot (\vec{a}\times \vec{b})-\,2\vec{b}\cdot (\vec{b}\times \vec{c})-\,4\vec{b}\cdot (\vec{c}\times \vec{a})+3\vec{c}\cdot (\vec{a}\times \vec{b})\] \[+3\vec{c}\cdot (\vec{b}\times \vec{c})+6\vec{c}\cdot (\vec{c}\times \vec{a})]\] \[=3\cdot \{[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]-\,4\,[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]+3\,[\begin{matrix}    {\vec{c}} & {\vec{a}} & {\vec{b}}  \\ \end{matrix}]\}\] If any two of [\[\because \]if any two of \[\vec{u},\,\,\vec{v}\] and \[\vec{w}\] are same then \[[\begin{matrix}    {\vec{u}} & {\vec{v}} & {\vec{w}}  \\ \end{matrix}]=0\]] = \[3\cdot \{[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]-\,4\,[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]+3\,[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]\}\] \[[\because \,\,\,[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]=[\begin{matrix}    {\vec{b}} & {\vec{c}} & {\vec{a}}  \\ \end{matrix}]=[\begin{matrix}    {\vec{c}} & {\vec{a}} & {\vec{b}}  \\ \end{matrix}]]\] = 3.0 = 0 Hence, the given vectors are coplanar.