A) -1
B) 1
C) -3
D) 3
Correct Answer: C
Solution :
\[\vec{a},\] \[\vec{b},\] \[\vec{c}\] are unit vectors then \[|\vec{a}|=|\vec{b}|=|\vec{c}|=1\] Given, \[\vec{a}\,\,+\,\,\vec{b}\,\,+\vec{c}=0\] \[\vec{a}\,=\,-(\vec{b}+\vec{c})\] Squaring on both sides \[{{a}^{{\vec{2}}}}={{(\vec{b}+\vec{c})}^{2}}\] \[{{a}^{{\vec{2}}}}={{b}^{{\vec{2}}}}+{{c}^{{\vec{2}}}}+2\vec{b}.\vec{c}\] \[|\vec{a}{{|}^{2}}=|\vec{b}{{|}^{2}}+|\vec{c}{{|}^{2}}+2(\vec{b}.\vec{c})\] \[(\because \,\,\,{{a}^{{\vec{2}}}}=|\vec{a}{{|}^{2}})\] \[1=1+1+2(\vec{b}.\vec{c})\] \[\Rightarrow \] \[\vec{b}.\vec{c}=-1/2\] Similarly, \[\vec{a}.\vec{b}=\vec{c}.\vec{a}=-1/2\] Hence, \[3\vec{a}\,.\,\vec{b}+2\vec{b}.\vec{c}+\vec{c}.\vec{a}\] \[=3(-1/2)+2(-1/2)+(-1/2)\] \[=(3+2+1)(-1/2)\] \[=6(-1/2)=-3\]
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