2. Solved Papers
  3. VIT Engineering

VIT Engineering VIT Engineering Solved Paper-2009

  • question_answer
    If \[{{m}_{1}},\] \[{{m}_{2}},\] \[{{m}_{3}}\]and \[{{m}_{4}}\] are respectively the magnitudes of the vectors \[{{\overrightarrow{a}}_{1}}=2\hat{i}-\hat{j}+\hat{k},\text{ }{{\overrightarrow{a}}_{2}}=3\hat{i}-4\hat{j}-4\hat{k},\] \[{{\overrightarrow{a}}_{3}}=\hat{i}+\hat{j}-\hat{k},\]and \[{{\overrightarrow{a}}_{4}}=-\hat{i}+3\hat{j}+\hat{k},\] 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, \[{{m}_{1}}=\left| \,{{{\vec{a}}}_{1}} \right|=\sqrt{{{2}^{2}}+{{(-1)}^{2}}+{{(1)}^{2}}}=\sqrt{6}\] \[{{m}_{2}}=\left| \,{{{\vec{a}}}_{2}} \right|=\sqrt{{{3}^{2}}+{{(-4)}^{2}}+{{(-4)}^{2}}}=\sqrt{41}\] \[{{m}_{3}}=\left| \,{{{\vec{a}}}_{3}} \right|=\sqrt{{{1}^{2}}+{{1}^{2}}+{{(-1)}^{2}}}=\sqrt{3}\] 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.
You will be redirected in 3 sec spinner