A) \[4\]
B) \[7\]
C) \[8\]
D) \[9\]
Correct Answer: B
Solution :
Given, \[\vec{a}+2\vec{b}+4\vec{c}=\vec{0}\] Now, \[\vec{a}\times (\vec{a}+2\vec{b}+4\vec{c})=\vec{0}\] \[\Rightarrow \] \[2(\vec{a}\times \vec{b})=4(\vec{a}\times \vec{c})=\vec{0}\] \[\Rightarrow \] \[2(\vec{a}\times \vec{b})=4(\vec{c}\times \vec{a})/2\] \[\Rightarrow \] \[(\vec{a}\times \vec{b})/4=(\vec{c}\times \vec{a})/2\] ?..(i) Again, \[\vec{b}\times (\vec{a}+2\vec{b}+4\vec{c})=\vec{0}\] \[\Rightarrow \] \[\vec{b}\times \vec{a}+4(\vec{b}\times \vec{c})=\vec{0}\] \[\Rightarrow \] \[\vec{b}\times \vec{c}=(\vec{a}\times \vec{b})/4\] From Eqs. (i) and (ii), \[(\vec{a}\times \vec{b})/4=\vec{b}\times \vec{c}=(\vec{c}\times \vec{a})/2=p(let)\] \[\therefore \] \[\vec{a}\times \vec{b}=4p,\,\vec{b}\times \vec{c}=p\] and \[\vec{c}\times \vec{a}=2p\] So, \[(\vec{a}\times \vec{b})+(\vec{b}\times \vec{c})+(\vec{c}\times \vec{a})=4p+p+2p\] \[=7p=7(\vec{b}\times \vec{c})\] \[\therefore \] \[\lambda =7\]
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