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

  • question_answer
    A plane which passes through the point (3, 2, 0) and the line \[\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\]is [AIEEE 2002]

    A)            \[x-y+z=1\]

    B)            \[x+y+z=5\]

    C)            \[x+2y-z=0\]

    D)            \[2x-y+z=5\]

    Correct Answer: A

    Solution :

                       Plane passing through (3, 2, 0) is            \[A(x-3)+B(y-2)+c(z-0)=0\]                               ?..(i)            Plane (i) is passing through the line,                             \[\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\]            \[\therefore \]  \[A(3-3)+B(6-2)+C(4-0)=0\]            \[0.A+4B+4C=0\]                                  ?..(ii)            and also 1.A + 5B + 4C = 0                 ?..(iii)            Solving (ii) and (iii), we get \[x-y+z=1\].            Trick: Required plane is \[\left| \,\begin{matrix}    x-3 & y-6 & z-4  \\    3-3 & 2-6 & 0-4  \\    1 & 5 & 4  \\ \end{matrix}\, \right|\,=0\]                    Solving, we get \[x-y+z=1.\]


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