A) \[4\sqrt{2}\]
B) \[3\sqrt{2}\]
C) \[5\sqrt{2}\]
D) \[3\sqrt{3}\]
Correct Answer: C
Solution :
Since, \[\vec{a} and \vec{b}+\vec{c}\] are mutually perpendicular. \[\therefore ~\,\,\,\vec{a}\left( \vec{b}\,+\vec{c} \right)=\,\,0 \,\,\Rightarrow \,\,\,\vec{a}.\vec{b}\,\,+\vec{c}.\vec{a}\,\,=\,\,0\] ... (i) Similarly, \[\vec{b} .\vec{c} + \vec{a}. \vec{b}=0\] ... (ii) and \[\vec{c}.\vec{a}\,+\vec{b}.\vec{c}=0\] ... (iii) On adding eqs. (i), (ii) and (iii), we get \[2\left( \vec{a}.\vec{b}+\vec{b}\,.\,\vec{c}+\vec{c}\,.\,\vec{a} \right)=0\] Now, \[{{\left| \vec{a}+\vec{b}+\vec{c} \right|}^{2}}=\,\,{{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}+{{\left| {\vec{c}} \right|}^{2}}\] \[+\,\,2\left( \vec{a}.\vec{b}\,\,+\vec{b}.\vec{c}\,+\vec{c}.\vec{a} \right)\] \[=\,\,\,{{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}+{{\left| {\vec{c}} \right|}^{2}}\] \[=\,\,\,9+16+25~\,\,\,\left( \because \,\left| {\vec{a}} \right|=3,\left| {\vec{b}} \right|=4,\,\,\left| {\vec{c}} \right|=5 \right)\] \[=\,\,50\] \[\Rightarrow \,\,\,\,\,\left| \vec{a}+\vec{b}+\vec{c} \right|=5\sqrt{2}\]
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