BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer If the vectors \[\vec{a}=\hat{i}+a\hat{j}+{{a}^{2}}\hat{k},\vec{b}=\hat{i}+b\hat{j}+{{b}^{2}}\hat{k},\]\[\vec{c}=\hat{i}+c\hat{j}+{{c}^{2}}\hat{k}\]are three non-coplanar vectors and \[\left| \begin{matrix}    a & {{a}^{2}} & 1+{{a}^{3}}  \\    b & {{b}^{2}} & 1+{{b}^{3}}  \\    c & {{c}^{2}} & 1+{{c}^{3}}  \\ \end{matrix} \right|=0,\] then the value of aback is

    A)  0                            

    B)         1                            

    C)  2                            

    D)         - 1

    Correct Answer: D

    Solution :

    Since, \[\vec{a}=\hat{i}+a\hat{j}+{{a}^{2}}\hat{k},\vec{b}=\hat{i}+b\hat{j}+{{b}^{2}}\hat{k}\] and        \[\vec{c}=\hat{i}+c\hat{j}+{{c}^{2}}\hat{k}\]                 are coplanar vectors                 \[\therefore \]  \[[\vec{a}\,\vec{b}\,\vec{c}]\ne 0\]                 \[\Rightarrow \]               \[\left| \begin{matrix}    1 & a & {{a}^{2}}  \\    1 & b & {{b}^{2}}  \\    1 & c & {{c}^{2}}  \\ \end{matrix} \right|\ne 0\]                 \[\Rightarrow \]               \[\Delta \ne 0\]                 where   \[\Delta =\left| \begin{matrix}    1 & a & {{a}^{2}}  \\    1 & b & {{b}^{2}}  \\    1 & c & {{c}^{2}}  \\ \end{matrix} \right|\]                 Now,     \[\left| \begin{matrix}    a & {{a}^{2}} & 1+{{a}^{3}}  \\    b & {{b}^{2}} & 1+{{b}^{3}}  \\    c & {{c}^{2}} & 1+{{c}^{3}}  \\ \end{matrix} \right|=0\]                 \[\Rightarrow \]               \[\left| \begin{matrix}    a & {{a}^{2}} & 1  \\    b & {{b}^{2}} & 1  \\    c & {{c}^{2}} & 1  \\ \end{matrix} \right|+\left| \begin{matrix}    a & {{a}^{2}} & {{a}^{3}}  \\    b & {{b}^{2}} & {{b}^{3}}  \\    c & {{c}^{2}} & {{c}^{3}}  \\ \end{matrix} \right|=0\]                 \[\Rightarrow \]               \[\left| \begin{matrix}    a & {{a}^{2}} & 1  \\    b & {{b}^{2}} & 1  \\    c & {{c}^{2}} & 1  \\ \end{matrix} \right|+abc\left| \begin{matrix}    1 & a & {{a}^{2}}  \\    1 & b & {{b}^{2}}  \\    1 & c & {{c}^{2}}  \\ \end{matrix} \right|=0\]                 \[\Rightarrow \]               \[\Delta +abc\,\Delta =0\]                 \[\Rightarrow \]               \[\Delta (1+abc)=0\]                 \[\Rightarrow \]               \[1+abc=0\]                        \[(\because \,\Delta \ne 0)\] \[\Rightarrow \]               \[abc=-1\]

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