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Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    If\[{{m}_{1}},\,\,{{m}_{2}},\,\,{{m}_{3}}\]and\[{{m}_{4}}\]are respectively the magnitudes of the vectors \[{{\overrightarrow{\mathbf{a}}}_{1}}=2\widehat{\mathbf{i}}-\widehat{\mathbf{j}}+\widehat{\mathbf{k}},\,\,\,\,{{\overrightarrow{\mathbf{a}}}_{2}}=3\widehat{\mathbf{i}}-4\widehat{\mathbf{j}}-4\widehat{\mathbf{k}}\] \[{{\overrightarrow{\mathbf{a}}}_{3}}=\widehat{\mathbf{i}}+\widehat{\mathbf{j}}-\widehat{\mathbf{k}}\]and\[{{\overrightarrow{\mathbf{a}}}_{4}}=-\widehat{\mathbf{i}}+3\widehat{\mathbf{j}}+\widehat{\mathbf{k}}\] then the correct order of\[{{m}_{1}},\,\,{{m}_{2}},\,\,{{m}_{3}}\]and\[{{m}_{4}}\]is

    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        \[{{m}_{4}}=|{{\overrightarrow{\mathbf{a}}}_{4}}|=\sqrt{{{(-1)}^{2}}+{{(3)}^{2}}+{{(1)}^{2}}}=\sqrt{11}\] \[\therefore \]  \[{{m}_{3}}<{{m}_{1}}<{{m}_{4}}<{{m}_{2}}\]


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