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JEE Main & Advanced Mathematics Vector Algebra Question Bank Critical Thinking

  • question_answer
    If \[\mathbf{a}=\mathbf{i}+\mathbf{j}+\mathbf{k},\,\,\mathbf{a}\,.\,\mathbf{b}=1\] and \[\mathbf{a}\times \mathbf{b}=\mathbf{j}-\mathbf{k},\] then \[\mathbf{b}=\] [IIT Screening 2004]

    A) \[\mathbf{i}\]

    B) \[\mathbf{i}-\mathbf{j}+\mathbf{k}\]

    C) \[2\mathbf{j}-\mathbf{k}\]

    D) \[2\mathbf{i}\]

    Correct Answer: A

    Solution :

    • Let \[\mathbf{b}={{b}_{1}}\,\mathbf{i}+{{b}_{2}}\,\mathbf{j}+{{b}_{3}}\mathbf{k}\]                   
    • Now, \[\mathbf{j}-\mathbf{k}=\mathbf{a}\times \mathbf{b}=\left| \begin{matrix}    \mathbf{i} & \mathbf{j} & \mathbf{k}  \\    1 & 1 & 1  \\    {{b}_{1}} & {{b}_{2}} & {{b}_{3}}  \\ \end{matrix} \right|\]                   
    • Þ \[{{b}_{3}}-{{b}_{2}}=0,{{b}_{1}}-{{b}_{3}}=1,{{b}_{2}}-{{b}_{1}}=-1\]                   
    • Þ \[{{b}_{3}}={{b}_{2}},{{b}_{1}}={{b}_{2}}+1\]                   
    • Now, \[\mathbf{a}.\mathbf{b}=1\Rightarrow {{b}_{1}}+{{b}_{2}}+{{b}_{3}}=1\]                   
    • Þ \[3{{b}_{2}}+1=1\]\[\Rightarrow {{b}_{2}}=0\]Þ \[{{b}_{1}}=1,{{b}_{3}}=0\].                   
    • Thus \[\mathbf{b}=\mathbf{i}\].


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