A) \[\vec{a}\] and \[\vec{c}\] are parallel
B) \[\vec{a}\] and \[\vec{b}\] are parallel
C) \[\vec{b}\]and \[\vec{c}\] are parallel
D) \[\vec{a}\] and \[\vec{c}\]are perpendicular
Correct Answer: A
Solution :
Given that, \[(\vec{a}\times \vec{b})\times \vec{c}=\vec{a}\times (\vec{b}\times \vec{c})\] \[\Rightarrow \] \[-\vec{c}\times (\vec{a}\times \vec{b})=\vec{a}\times (\vec{b}\times \vec{c})\] \[\Rightarrow \] \[-\{(\vec{c}.\,\vec{b})\,\vec{a}-(\vec{c}.\vec{a})\,\vec{b}\}\] \[=(\vec{a}\,.\vec{c})\,\vec{b}-(\vec{a}.\vec{b})\vec{c}\,\,\,\,\,\,\,\{\because \,\,\vec{a}\,.\,\vec{c}=\vec{c}.\vec{a})\] \[(\vec{a}\,.\,\vec{b}=\vec{b}.\vec{a})\] \[(\vec{b}.\vec{c}=\vec{c}.\vec{b})\] \[\Rightarrow \] \[(\vec{a}\,.\,\vec{b})\,\vec{c}-(\vec{c}\,.\,\vec{b})\,\vec{a}=0\] \[\Rightarrow \] \[(\vec{b}\,.\,\vec{a})\,\vec{c}-(\vec{b}\,.\,\vec{c})\,\vec{a}=0\] \[\Rightarrow \] \[\vec{b}\,\times (\vec{c}\times \vec{a})=0\] \[\Rightarrow \] \[\vec{b}\] is parallel to \[\vec{c}\times \vec{a}\] But \[\vec{c}\times \vec{a}\] is perpendicular to \[\vec{c}\times \vec{a}\]
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