A) 0
B) \[\frac{1}{12}\]
C) \[\frac{5}{12}\]
D) \[3\]
Correct Answer: B
Solution :
Given, \[(a+\lambda b).[(b+3c)\times (c-4a)]=0\] \[\Rightarrow \]\[(a+\lambda b).[b\times c-4b\times a-12c\times a]=0\] \[\Rightarrow \]\[[a\,b\,c]-0-0+0+0-12\lambda [b\,c\,a]=0\] \[\Rightarrow \] \[[a\,b\,c]=12\lambda [b\,c\,a]\] \[\Rightarrow \] \[\lambda =\frac{1}{12}\]You need to login to perform this action.
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