A) \[-8\]
B) \[8\]
C) \[-13\]
D) \[13\]
Correct Answer: C
Solution :
Given, \[4|\vec{a}|=12|\vec{b}|=3|\vec{c}|=12\] \[\Rightarrow \] \[|\vec{a}|=3,\,\,\,|\vec{b}|=1\] and \[|\vec{c}|=4\] and \[\vec{a}+\vec{b}+\vec{c}=\vec{0}\] \[\Rightarrow \] \[{{(\vec{a}+\vec{b}+\vec{c})}^{2}}=0\] \[\Rightarrow \] \[|\vec{a}{{|}^{2}}+|\vec{b}{{|}^{2}}+|\vec{c}{{|}^{2}}+2(\vec{a}.\vec{b}+\vec{b}.\vec{c}\] \[+\vec{c}.\vec{a})=0\] \[\Rightarrow \] \[2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a})=-(9+1+16)\] \[\Rightarrow \] \[\vec{a}.\,\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}=\frac{-26}{2}=-13\]
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