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JEE Main & Advanced Mathematics Vector Algebra Question Bank Critical Thinking

  • question_answer
    If three non-zero vectors are \[\mathbf{a}={{a}_{1}}\mathbf{i}+{{a}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{k},\] \[\mathbf{b}={{b}_{1}}\mathbf{i}+{{b}_{2}}\mathbf{j}+{{b}_{3}}\mathbf{k}\] and \[\mathbf{c}={{c}_{1}}\mathbf{i}+{{c}_{2}}\mathbf{j}+{{c}_{3}}\mathbf{k}.\] If c is the unit vector perpendicular to the vectors a and b and the angle between a and b is \[\frac{\pi }{6},\] then \[{{\left| \,\begin{matrix}  {{a}_{1}} & {{a}_{2}} & {{a}_{3}}  \\  {{b}_{1}} & {{b}_{2}} & {{b}_{3}}  \\    {{c}_{1}} & {{c}_{2}} & {{c}_{3}}  \\ \end{matrix}\, \right|}^{2}}\] is equal to    [IIT 1986]

    A) 0

    B) \[\frac{3\,(\Sigma a_{1}^{2})\,(\Sigma b_{1}^{2})\,(\Sigma c_{1}^{2})}{4}\]

    C) 1    

    D) \[\frac{(\Sigma a_{1}^{2})\,(\Sigma b_{1}^{2})}{4}\]

    Correct Answer: D

    Solution :

    • \[|\mathbf{c}|\,=1,\] we have \[|\mathbf{c}{{|}^{2}}=1\] or  \[c_{1}^{2}+c_{2}^{2}+c_{3}^{2}=1\]    .....(i)                   
    • Again, since \[\mathbf{c}\,\bot \,\mathbf{a}\] and \[\mathbf{c}\,\bot \,\mathbf{b},\]we have \[\mathbf{c}\,.\,\mathbf{a}=0\]                   
    • \[\Rightarrow {{a}_{1}}{{c}_{1}}+{{a}_{2}}{{c}_{2}}+{{a}_{3}}{{c}_{3}}=0\]                                 .....(ii)                   
    • and \[\mathbf{c}\,.\,\mathbf{b}=0\Rightarrow {{b}_{1}}{{c}_{1}}+{{b}_{2}}{{c}_{2}}+{{b}_{3}}{{c}_{3}}=0\]                 ...(iii)                   
    • Also since angle between \[\mathbf{a}\] and \[\mathbf{b}\] is \[\frac{\pi }{6},\] we have \[\mathbf{a}\,.\,\mathbf{b}={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}\]                   
    • Þ \[|\mathbf{a}|\,|\mathbf{b}|\cos \frac{\pi }{6}={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}\]                   
    • Þ \[\frac{3}{4}(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})(b_{1}^{2}+b_{2}^{2}+b_{3}^{2})\]\[={{({{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}})}^{2}}\]     .....(iv)                   
    • Now, \[{{\left| \begin{matrix}    {{a}_{1}} & {{a}_{2}} & {{a}_{3}}  \\    {{b}_{1}} & {{b}_{2}} & {{b}_{3}}  \\    {{c}_{1}} & {{c}_{2}} & {{c}_{3}}  \\\end{matrix} \right|}^{2}}=\left| \begin{matrix}    {{a}_{1}} & {{a}_{2}} & {{a}_{3}}  \\    {{b}_{1}} & {{b}_{2}} & {{b}_{3}}  \\    {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\end{matrix} \right|\,\left| \begin{matrix}  {{a}_{1}} & {{a}_{2}} & {{a}_{3}}  \\    {{b}_{1}} & {{b}_{2}} & {{b}_{3}}  \\    {{c}_{1}} & {{c}_{2}} & {{c}_{3}}  \\ \end{matrix} \right|\,\]                   
    • \[=\left| \begin{matrix}    a_{1}^{2}+a_{2}^{2}+a_{3}^{2} & {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}} & 0  \\   {{b}_{1}}{{a}_{1}}+{{b}_{2}}{{a}_{2}}+{{b}_{3}}{{a}_{3}} & b_{1}^{2}+b_{2}^{2}+b_{3}^{2} & 0  \\    0 & 0 & 1  \\ \end{matrix} \right|\]           
    • {Using (i), (ii) and (iii)}                   
    • \[=\frac{1}{4}(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})(b_{1}^{2}+b_{2}^{2}+b_{3}^{2})\], {Using (iv)}                   
    • \[=\frac{(\Sigma a_{1}^{2})(\Sigma b_{1}^{2})}{4}\],                   
    • where \[\Sigma a_{1}^{2}=a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\] and \[\Sigma b_{1}^{2}=b_{1}^{2}+b_{2}^{2}+b_{3}^{2}.\]


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