A) \[a=1,\,\,b=c\]
B) \[c=1,\,\,a=1\]
C) \[b=2,\,\,c=2a\]
D) \[b=1,\,\,c=a\]
Correct Answer: D
Solution :
\[\mathbf{a}=\mathbf{b}\times \mathbf{c}\] and \[\mathbf{a}\times \mathbf{b}=\mathbf{c}\] \[\therefore \] a is perpendicular to both b and \[\mathbf{c}\] and \[\mathbf{c}\] is perpendicular to both \[\mathbf{a}\] and b. \ \[\mathbf{a},\,\mathbf{b},\,\mathbf{c}\]are mutually perpendicular Now, \[\mathbf{a}=\mathbf{b}\times \mathbf{c}=\mathbf{b}\times (\mathbf{a}\times \mathbf{b})=(\mathbf{b}\,.\,\mathbf{b})\mathbf{a}-(\mathbf{b}\,.\,\mathbf{a})\mathbf{b}\] or \[\mathbf{a}={{b}^{2}}\mathbf{a}-(\mathbf{b}\,.\,\mathbf{a})\mathbf{b}={{b}^{2}}\mathbf{a}\], \[\left\{ \because \,\mathbf{a}\,\bot \,\mathbf{b} \right\}\] \[\Rightarrow 1={{b}^{2}}\], \[\therefore \,\mathbf{c}=\mathbf{a}\times \mathbf{b}=ab\sin 90{}^\circ \,\mathbf{\hat{n}}\] Take moduli of both sides, then \[c=ab\], but \[b=1\Rightarrow c=a\].
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