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

  • question_answer
    The equation of the plane through the point \[(2,-1,-3)\]and parallel to the lines \[\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}\] and \[\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}\] is [Kerala (Engg.) 2005]

    A)  \[8x+14y+13z+37=0\]

    B)            \[8x-14y+13z+37=0\]

    C)            \[8x+14y-13z+37=0\]

    D)            \[8x+14y+13z-37=0\]

    E) (e)       \[8x-14y-13z-37=0\]

    Correct Answer: A

    Solution :

                       Equation of  plane passing through the point (2, ?1, ?3) is,            Also, \[A(x-2)+B(y+1)+C(z+3)=0\]             Also, \[3A+2B-4C=0\] and \[2A-3B+2C=0\]                    \ \[\frac{A}{-8}=\frac{B}{-14}=\frac{C}{-13}=k\], (Let)                    So, \[A=-8k,B=-14k,C=-13k\]                    Equation of required plane is,                    \[-k[8(x-2)+14(y+1)+13(z+3)]=0\]            i.e., \[8x+14y+13z+37=0\].


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