A) \[\left| \,\begin{matrix} \mathbf{a} & \mathbf{b} & \mathbf{c} \\ \mathbf{b} & \mathbf{c} & \mathbf{a} \\ \mathbf{c} & \mathbf{a} & \mathbf{b} \\ \end{matrix}\, \right|=\mathbf{0}\]
B) \[\left| \,\begin{matrix} \mathbf{a} & \mathbf{b} & \mathbf{c} \\ \mathbf{a}\,.\,\mathbf{a} & \mathbf{a}\,.\,\mathbf{b} & \mathbf{a}\,.\,\mathbf{c} \\ \mathbf{b}\,.\,\mathbf{a} & \mathbf{b}\,.\,\mathbf{b} & \mathbf{b}\,.\,\mathbf{c} \\ \end{matrix}\, \right|=\mathbf{0}\]
C) \[\left| \,\begin{matrix} \mathbf{a} & \mathbf{b} & \mathbf{c} \\ \mathbf{c}\,.\,\mathbf{a} & \mathbf{c}\,.\,\mathbf{b} & \mathbf{c}\,.\,\mathbf{c} \\ \mathbf{b}\,.\,\mathbf{a} & \mathbf{b}\,.\,\mathbf{c} & \mathbf{b}\,.\,\mathbf{b} \\ \end{matrix}\, \right|=\mathbf{0}\]
D) \[\left| \,\begin{matrix} \mathbf{a} & \mathbf{b} & \mathbf{c} \\ \mathbf{a}\,.\,\mathbf{b} & \mathbf{a}\,.\,\mathbf{a} & \mathbf{a}\,.\,\mathbf{c} \\ \mathbf{c}\,.\,\mathbf{a} & \mathbf{c}\,.\,\mathbf{c} & \mathbf{c}\,.\,\mathbf{b} \\ \end{matrix}\, \right|=\mathbf{0}\]
Correct Answer: B
Solution :
Since \[\mathbf{a},\,\,\mathbf{b}\] and \[\mathbf{c}\] are coplanar, therefore there exists \[(x,\,y,\,z\]not all zero) such that \[x\mathbf{a}+y\mathbf{b}+z\mathbf{c}=0\] .....(i) Multiply be \[\mathbf{a}\] scalarly, we get \[x(\mathbf{a}\,.\,\mathbf{a})+(\mathbf{a}\,.\,\mathbf{b})+z(\mathbf{a}\,.\,\mathbf{c})=0\] ......(ii) and \[x(\mathbf{a}\,.\,\mathbf{b})+y(\mathbf{b}\,.\,\mathbf{b})+z(\mathbf{b}\,.\,\mathbf{c})=0\] .....(iii) Eliminating \[x,\,y\] and \[z\] from (i), (ii) and (iii), we get \[\left| \,\begin{matrix} \mathbf{a} & \mathbf{b} & \mathbf{c} \\ \mathbf{a}\,.\,\mathbf{a} & \mathbf{a}\,.\,\mathbf{b} & \mathbf{a}\,.\,\mathbf{c} \\ \mathbf{a}\,.\,\mathbf{b} & \mathbf{b}\,.\,\mathbf{b} & \mathbf{b}\,.\,\mathbf{c} \\ \end{matrix}\, \right|=0\]. Note: Students should remember this question as a formula.
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