A) \[\mathbf{a}=\mathbf{b}=\mathbf{c}=\mathbf{0}\]
B) \[\mathbf{a}\,.\,\mathbf{b}=\mathbf{b}\,.\,\mathbf{c}=\mathbf{c}\,.\,\mathbf{a}\]
C) \[\mathbf{a}\times \mathbf{b}=\mathbf{b}\times \mathbf{c}=\mathbf{c}\times \mathbf{a}\]
D) None of these
Correct Answer: C
Solution :
Since \[\mathbf{a}+\mathbf{b}+\mathbf{c}=0\] \[\Rightarrow \mathbf{a}\times (\mathbf{a}+\mathbf{b}+\mathbf{c})=\mathbf{0}\Rightarrow \mathbf{a}\times \mathbf{a}+\mathbf{a}\times \mathbf{b}+\mathbf{a}\times \mathbf{c}=\mathbf{0}\] \[\Rightarrow \mathbf{a}\times \mathbf{b}=-\mathbf{a}\times \mathbf{c}=\mathbf{c}\times \mathbf{a}\] .....(i) Similarly, \[\mathbf{b}\times (\mathbf{a}+\mathbf{b}+\mathbf{c})=\mathbf{0}\]\[\Rightarrow \mathbf{a}\times \mathbf{b}=\mathbf{b}\times \mathbf{c}\] .....(ii) By (i) and (ii), we get \[\mathbf{a}\times \mathbf{b}=\mathbf{b}\times \mathbf{c}=\mathbf{c}\times \mathbf{a}.\]
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