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