A) 9
B) 6
C) 5
D) 4
Correct Answer: B
Solution :
Since, \[\vec{a},\,\,\vec{b}\,\,and\,\,\vec{c}\] are three vectors with magnitude \[\left| {\vec{a}} \right|=\left| {\vec{b}} \right|=4 and\,\left| {\vec{c}} \right|=2,\] As \[\vec{a}\] is perpendicular to \[\left( b +c \right)\] \[\Rightarrow \,\,\vec{a}.\left( \vec{b} +\vec{c} \right)=0 or\,\,\vec{a}.\vec{b}\,\,+\vec{a}.\vec{c}=0\] ... (i) \[\vec{b}\] is perpendicular to \[(\vec{c} +\vec{a})\] \[\Rightarrow \,\,\,\vec{b}.(\vec{c}\,+\vec{a})=0\,\,or\,\,\vec{b}.\vec{c}\,+~\vec{b}.\vec{a}\,=0\] ... (ii) \[\vec{c}\] is perpendicular to \[\left( \vec{a} +\vec{b} \right)\] \[\Rightarrow \, \vec{c}\left( \vec{a} +\vec{b} \right)=0 or \vec{c}.\vec{a} +\vec{c}.\vec{b} =\,\,0\] ... (iii) From equations (i), (ii) and (iii), we get \[\Rightarrow \,\,\,\,2(\vec{a}.\vec{b} +\vec{b}.\vec{c}+\vec{c}.\vec{a})=0\] Further we know that \[{{\left| \vec{a}+\vec{b}+\vec{c} \right|}^{2}}=\,\,{{\left| {\vec{a}} \right|}^{2}}+\,\,{{\left| {\vec{b}} \right|}^{2}}\,\,+\,\,{{\left| {\vec{c}} \right|}^{2}}\] \[+\,\,\overrightarrow{2a}.\vec{b}+\overrightarrow{2b}\,.\,\vec{c}+\overrightarrow{2c}\,.\,\vec{a}\] \[\Rightarrow \,\,\,\,|\vec{a}+\vec{b}+\vec{c}|2={{4}^{2}}+{{4}^{2}}+{{2}^{2}}+0=36\] or \[\left| \vec{a} + \vec{b}+\vec{c} \right|= 6\]
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