A) \[\vec{a}\]
B) \[\vec{c}\]
C) \[\vec{0}\]
D) \[\vec{a}+\vec{c}\]
Correct Answer: C
Solution :
\[\vec{a}+3\vec{b}=\lambda \vec{c}\] ??(1) \[\vec{b}+2\vec{c}=\mu \vec{a}\] ??(2) (1)-3(2) gives \[(1+3\mu )\vec{a}-(\lambda +6)\vec{c}=0\] As \[\vec{a}\] and \[\vec{c}\] are non collinear \[\therefore \]\[1+3\mu =0\]and\[\lambda +6=0\] From\[(1)\]\[\vec{a}+3\vec{b}+6\vec{c}=\vec{0}\]You need to login to perform this action.
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