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JEE Main & Advanced Sample Paper JEE Main Sample Paper-3

  • question_answer
    If \[\left| \begin{matrix}    a & {{a}^{2}} & 1+{{a}^{3}}  \\    b & {{b}^{2}} & 1+{{b}^{3}}  \\    c & {{c}^{2}} & 1+{{c}^{3}}  \\ \end{matrix} \right|=0\]and the vectors \[\overrightarrow{\text{A}}=(1,\,a,\,{{a}^{2}});\] \[\overrightarrow{\text{B}}=(1,\,b,\,{{b}^{2}});\,\overrightarrow{\text{C}}=\text{(1,}\,\text{c,}\,{{\text{c}}^{2}}\text{)}\]are non-coplanar then the product abc =

    A)  0                            

    B)  1

    C)  -1                                          

    D)  None

    Correct Answer: C

    Solution :

     \[\left| \begin{matrix}    a & {{a}^{2}} & 1+{{a}^{3}}  \\    b & {{b}^{2}} & 1+{{b}^{3}}  \\    c & {{c}^{2}} & 1+{{c}^{3}}  \\ \end{matrix} \right|\] \[=(1+abc)(a-b)(b-c)(c-a)=0\]          ?(1) also, \[\left| \begin{matrix}    1 & a & {{a}^{2}}  \\    1 & b & {{b}^{2}}  \\    1 & c & {{c}^{2}}  \\ \end{matrix} \right|\ne 0,\] since A, B, C are not coplanar \[=(a-b)(b-c)(c-a)\ne 0\]         ?(2) From (1) & (2), abc = - 1


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