JCECE Engineering JCECE Engineering Solved Paper-2011

  • question_answer
    The vectors\[\mathbf{a}=x\mathbf{i}+(x+1)\mathbf{j}+(x+2)\mathbf{k}\],\[\mathbf{b}=(x+3)\mathbf{i}+(x+4)\mathbf{j}+(x+5)\mathbf{k}\] and\[\mathbf{c}=(x+6)\mathbf{i}+(x+7)\mathbf{j}+(x+8)\mathbf{k}\]are coplanar for

    A)  all values of\[x\]             

    B)  only\[x=1\]

    C)  only\[x=0\]                       

    D)  None of these

    Correct Answer: A

    Solution :

    \[\mathbf{a},\,\,\mathbf{b},\,\,\mathbf{c}\]are coplanar, if\[[\mathbf{a}\,\,\mathbf{b}\,\,\mathbf{c}]=0\] We have,\[[\mathbf{a}\,\,\mathbf{b}\,\,\mathbf{c}]=\left| \begin{matrix}    x & x+1 & x+2  \\    x+3 & x+4 & x+5  \\    x+6 & x+7 & x+8  \\ \end{matrix} \right|\]                 \[=\left| \begin{matrix}    x & x+1 & x+2  \\    3 & 3 & 3  \\    6 & 6 & 6  \\ \end{matrix} \right|\] \[[\because \]applying\[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},\,\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}]\] \[=0,\,\,\forall x[\because \,\,{{R}_{2}}\]and \[{{R}_{3}}\] are proportional ]


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