A) \[\lambda \overrightarrow{a}\]
B) \[\lambda \overrightarrow{b}\]
C) \[\lambda \overrightarrow{c}\]
D) \[\overrightarrow{0}\]
Correct Answer: D
Solution :
Since, \[\overrightarrow{a}+2\overrightarrow{b}\] is collinear with \[\overrightarrow{c}\] \[\therefore \]\[\overrightarrow{a}+2\overrightarrow{b}=\lambda \overrightarrow{c},\forall \lambda \in R\] and \[\overrightarrow{b}+3\overrightarrow{c}\] is collinear with \[\overrightarrow{a}.\] \[\therefore \]\[\overrightarrow{b}+3\overrightarrow{c}=\mu \overrightarrow{a},\forall \mu \in R\] \[\Rightarrow \]\[\overrightarrow{a}+2\overrightarrow{b}+6\overrightarrow{c}=(\lambda +6)\overrightarrow{c}\]and\[\overrightarrow{a}+2\overrightarrow{b}+6\overrightarrow{c}=(2\mu +1)\overrightarrow{a}\] \[\Rightarrow \]\[(\lambda +6)\overrightarrow{c}=(2\mu +1)\overrightarrow{a}\] \[\Rightarrow \]\[\lambda +6=0\]and\[2\mu +1=0\] \[\Rightarrow \]\[\lambda =-6,\lambda =-\frac{1}{2}\] Hence, \[\overrightarrow{a}+2\overrightarrow{b}+6\overrightarrow{c}=\overrightarrow{0}\]You need to login to perform this action.
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