A) \[{{m}_{3}}<{{m}_{1}}<{{m}_{4}}<{{m}_{2}}\]
B) \[{{m}_{3}}<{{m}_{1}}<{{m}_{2}}<{{m}_{4}}\]
C) \[{{m}_{3}}<{{m}_{4}}<{{m}_{1}}<{{m}_{2}}\]
D) \[{{m}_{3}}<{{m}_{4}}<{{m}_{2}}<{{m}_{1}}\]
Correct Answer: A
Solution :
Given, \[{{m}_{1}}=\left| \,{{{\vec{a}}}_{1}} \right|=\sqrt{{{2}^{2}}+{{(-1)}^{2}}+{{(1)}^{2}}}=\sqrt{6}\] \[{{m}_{2}}=\left| \,{{{\vec{a}}}_{2}} \right|=\sqrt{{{3}^{2}}+{{(-4)}^{2}}+{{(-4)}^{2}}}=\sqrt{41}\] \[{{m}_{3}}=\left| \,{{{\vec{a}}}_{3}} \right|=\sqrt{{{1}^{2}}+{{1}^{2}}+{{(-1)}^{2}}}=\sqrt{3}\] and \[{{m}_{4}}=\left| \,{{{\vec{a}}}_{4}} \right|=\sqrt{{{(-1)}^{2}}+{{(3)}^{2}}+{{(1)}^{2}}}=\sqrt{11}\] \[\therefore \] \[{{m}_{3}}<{{m}_{1}}<{{m}_{4}}<{{m}_{2}}\]You need to login to perform this action.
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