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, \[{{m}_{1}}=|{{\overrightarrow{\mathbf{a}}}_{1}}|=\sqrt{{{2}^{2}}+{{(-1)}^{2}}+{{(1)}^{2}}}=\sqrt{6}\] \[{{m}_{2}}=|{{\overrightarrow{\mathbf{a}}}_{2}}|=\sqrt{{{3}^{2}}+{{(-4)}^{2}}+{{(-4)}^{2}}}=\sqrt{41}\] \[{{m}_{3}}=|{{\overrightarrow{\mathbf{a}}}_{3}}|=\sqrt{{{1}^{2}}+{{1}^{2}}+{{(-1)}^{2}}}=\sqrt{3}\] and \[{{m}_{4}}=|{{\overrightarrow{\mathbf{a}}}_{4}}|=\sqrt{{{(-1)}^{2}}+{{(3)}^{2}}+{{(1)}^{2}}}=\sqrt{11}\] \[\therefore \] \[{{m}_{3}}<{{m}_{1}}<{{m}_{4}}<{{m}_{2}}\]You need to login to perform this action.
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