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12th Class Mathematics Sample Paper Mathematics Olympiad - Sample Paper-1

  • question_answer
    The equation of a plane that passes through \[\left( \mathbf{2},-\mathbf{3},\mathbf{1} \right)\]and is perpendicular to the line joining the points \[\left( \mathbf{3},\mathbf{4},-\mathbf{1} \right)\] and \[\left( \mathbf{2},-\mathbf{1},\mathbf{5} \right)\] is:

    A)  \[x+5y-6z+19=0\]                   

    B)  \[x+5y-6z-19=0\]

    C)  \[x-5y-6z+19=0\]                   

    D)  \[x+5y+6z+19=0\]

    Correct Answer: A

    Solution :

    [a] Given points be \[\left( 3,4,-1 \right)\] and\[\left( 2,-1,5 \right)\]. So, the direction ratio of the line joining of these points be \[\left( 3-2,4+1,-1-5 \right)=\left( 1,5,-6 \right)\] The equation of the plane perpendicular to this line, whose d.r. be\[\left( 1,5,-6 \right)\], be \[x+5y-6z=\lambda \]       ...........(1) Since, the plane passes through the point \[\left( 2,-3,1 \right).\] \[\therefore 2+5(-3)-6\times 1=\lambda \] \[\Rightarrow 2-15-6=\lambda ~~~~~~~\Rightarrow \lambda =-19\] Hence, the required equation of the plane be \[x+5y-6z=-19\]    \[\Rightarrow x+5y-6z+19=0\] Hence, option [a] is correct.


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