A) 0
B) \[\lambda \vec{b}\]
C) \[\lambda \vec{c}\]
D) \[\lambda \vec{a}\]
Correct Answer: C
Solution :
[c] Let \[\vec{a}+2\vec{b}=t\vec{c}\] and \[\vec{b}+3\vec{c}=s\vec{a}\], where t and s are scalars. Adding, we get \[\vec{a}+3\vec{b}+3\vec{c}=t\vec{c}+s\vec{a}\Rightarrow \vec{a}+2\vec{b}+6\vec{c}\] \[=t\vec{c}+s\vec{a}-\vec{b}+3\vec{c}\] \[=t\vec{c}+(\vec{b}+3\vec{c})-\vec{b}+3\vec{c}=(t+6)\vec{c}\] [using \[s\,\,\vec{a}=\vec{b}+3\vec{c}\]] \[=\lambda \vec{c}\], where \[\lambda =t+6\]You need to login to perform this action.
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