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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Line and Plane

  • question_answer
    The equation of the plane passing through the line \[\frac{x-1}{5}=\frac{y+2}{6}=\frac{z-3}{4}\]and the point (4, 3, 7) is  [MP PET 2001]

    A)            \[4x+8y+7z=41\]

    B)            \[4x-8y+7z=41\]

    C)            \[4x-8y-7z=41\]

    D)            \[4x-8y+7z=39\]

    Correct Answer: B

    Solution :

                       Any plane through given line is                    \[A(x-1)+B(y+2)+C(z-3)=0\]     .....(i)            and             \[5A+6B+4C=0\]                                               ?..(ii)            Since, plane (i) passes through (4, 3, 7), we get                     \[3A+5B+4C=0\]                           .....(iii)            Solving (ii) and (iii), we get \[\frac{A}{4}=\frac{B}{-8}=\frac{C}{7}\]            \[\therefore \] Equation of required plane is \[4x-8y+7z=41\].


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