A) \[\mathbf{a}\,.\,\mathbf{b}=\mathbf{b}\,.\,\mathbf{c}=0\]
B) \[\mathbf{b}\,.\,\mathbf{c}=\mathbf{c}\,.\,\mathbf{a}=0\]
C) \[\mathbf{c}\,.\,\mathbf{a}=\mathbf{a}\,.\,\mathbf{b}=0\]
D) \[\mathbf{a}\,.\,\mathbf{b}=\mathbf{b}\,.\,\mathbf{c}=\mathbf{c}\,.\,\mathbf{a}=0\]
Correct Answer: D
Solution :
We have \[|(\mathbf{a}\times \mathbf{b}).\mathbf{c}|=|\mathbf{a}||\mathbf{b}||\mathbf{c}|\] \[\Rightarrow \left| |\mathbf{a}||\mathbf{b}|\sin \theta \,\mathbf{n}.\mathbf{c} \right|=|\mathbf{a}||\mathbf{b}||\mathbf{c}|\] \[\Rightarrow \left| |\mathbf{a}||\mathbf{b}||\mathbf{c}|\sin \theta \cos \alpha \right|=|\mathbf{a}||\mathbf{b}||\mathbf{c}|\] \[\Rightarrow \text{ }|\sin \theta ||\cos \alpha |=1\Rightarrow \theta =\frac{\pi }{2}\] and \[\alpha =0\] \[\Rightarrow \mathbf{a}\bot \mathbf{b}\] and \[\mathbf{c}||\mathbf{n}\] \[\Rightarrow \mathbf{a}\bot \mathbf{b}\] and \[\mathbf{c}\]is perpendicular to both \[\mathbf{a}\]and \[\mathbf{b}\] \ \[\mathbf{a},\,\mathbf{b},\,\mathbf{c}\] are mutually perpendicular Hence, \[\mathbf{a}.\mathbf{b}=\mathbf{b}.\mathbf{c}=\mathbf{c}.\mathbf{a}=0.\]
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