A) \[2(\vec{b}\times \vec{c})\]
B) \[3(\vec{c}\times \vec{a})\]
C) \[\vec{0}\]
D) \[6(\vec{b}\times \vec{c})\]
Correct Answer: D
Solution :
Given, \[\vec{a}+2\vec{b}+3\vec{c}=\vec{0}\] ?..(i) Taking cross product with \[\vec{b},\] we get \[\vec{a}\times \vec{b}+2\vec{b}\times \vec{b}+3\vec{c}\times \vec{b}=\vec{0}\times \vec{b}\] \[\Rightarrow \] \[\vec{a}\times \vec{b}=3\vec{b}\times \vec{c}\] Again taking cross product with \[\vec{c}\] of Eq. (i), we get \[\vec{a}\times \vec{c}+2\vec{b}\times \vec{c}+3\vec{c}\times \vec{c}=\vec{0}\times \vec{c}\] \[\Rightarrow \] \[\vec{c}\times \vec{a}=2\vec{b}\times \vec{c}\] \[\therefore \] \[\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}\] \[=3\vec{b}\times \vec{c}+\vec{b}\times \vec{c}+2\vec{b}\times \vec{c}\] \[=6(\vec{b}\times \vec{c})\]
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