A) 3
B) -3
C) 0
D) Cannot be evaluated
Correct Answer: B
Solution :
[b] \[\vec{a}+\vec{b}=\lambda \vec{c}\] and \[\vec{b}+\vec{c}=\mu \vec{a}\] \[\therefore (\lambda \vec{c}-\vec{a})+\vec{c}=\mu \vec{a}(putting\,\vec{b}=\lambda \vec{c}-\vec{a})\] \[\Rightarrow (\lambda +1)\vec{c}=(\mu +1)\vec{a}\] \[\Rightarrow \lambda =\mu =-1\] \[\Rightarrow \vec{a}+\vec{b}+\vec{c}=0\] \[\Rightarrow |\vec{a}{{|}^{2}}+|\vec{b}{{|}^{2}}+|\vec{c}{{|}^{2}}+2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a})=0\] \[\Rightarrow \vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}=-3\]You need to login to perform this action.
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