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JEE Main & Advanced Mathematics Vector Algebra Question Bank Critical Thinking

  • question_answer
    If the vectors \[a\mathbf{i}+\mathbf{j}+\mathbf{k},\,\,\mathbf{i}+b\mathbf{j}+\mathbf{k}\] and \[\mathbf{i}+\mathbf{j}+c\mathbf{k}\] \[(a\ne b\ne c\ne 1)\] are coplanar, then the value of \[\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=\] [BIT Ranchi 1988; RPET 1987; IIT 1987; DCE 2001; MP PET 2004; Orissa JEE 2005]

    A) - 1

    B) \[-\frac{1}{2}\]

    C) \[\frac{1}{2}\]

    D) 1

    Correct Answer: D

    Solution :

    • Since \[\left| \begin{matrix}    a & 1 & 1  \\    1 & b & 1  \\    1 & 1 & c  \\ \end{matrix} \right|=0\]                   
    • Applying \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\] and \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}},\]                   
    • we get \[\left| \begin{matrix}    a & 1 & 1  \\    1-a & b-1 & 0  \\    1-a & 0 & c-1  \\ \end{matrix} \right|=0\]                   
    • On expanding, we get                   
    • \[a(b-1)(c-1)-(1-a)(c-1)-(1-a)(b-1)=0\]                   
    • On dividing by \[(1-a)(1-b)(1-c),\] we get \[\frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0\]                   
    • \[\Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=\frac{1}{1-a}-\frac{a}{1-a}=1.\]


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