A) \[\left( \frac{3}{2},3\vec{a}\times \vec{c} \right)\]
B) \[\left( -\frac{3}{2},3\vec{c}\times \vec{b} \right)\]
C) \[\left( -\frac{3}{2},3\vec{a}\times \vec{b} \right)\]
D) \[\left( -\frac{3}{2},3\vec{b}\times \vec{c} \right)\]
Correct Answer: C
Solution :
[c] \[\because \,\,\left| {\vec{a}} \right|=\left| {\vec{b}} \right|=\left| {\vec{c}} \right|=1\] and \[\vec{a}+\vec{b}+\vec{c}=\vec{0}\] On squaring both sides \[\,{{\left| {\vec{a}} \right|}^{2}}={{\left| {\vec{b}} \right|}^{2}}={{\left| {\vec{c}} \right|}^{2}}+2(\vec{a}.\,\vec{b}+\vec{a}.\,\vec{c}+\vec{b}.\,\vec{c})=0\] \[\therefore \,\,\,\lambda =\vec{a}.\,\vec{b}+\vec{a}.\,\vec{c}+\vec{b}.\,\vec{c}=-\frac{3}{2}\] and \[\vec{d}=\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}\] \[=\,\,\vec{a}\times \vec{b}+\vec{b}\times (-\vec{a}-\vec{b})+(-\vec{a}-\vec{b})\times \vec{a}\] \[=\,\,\vec{a}\times \vec{b}-\vec{b}\times \vec{a}-0-0-\vec{b}\times \vec{a}\] \[=\,3\,(\vec{a}\,\times \vec{b})\] \[\therefore \,\,\,\,\,\,\,(\lambda ,\vec{d})=\left( -\frac{3}{2},3(\vec{a}\times \vec{b}) \right)\]
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