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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Mock Test - Three Dimensional Geometry

  • question_answer
    The point of intersection of the line passing through (0, 0, 1) and intersecting the lines \[x+2y+z=1\], \[-x+y-2z=2\] and \[x+y=2,\,\,x+z=2\] with xy plane is

    A) \[\left( \frac{5}{3},-\frac{1}{3},0 \right)\]

    B) (1, 1, 0)

    C) \[\left( \frac{2}{3},-\frac{1}{3},0 \right)\] 

    D) \[\left( -\frac{5}{3},\frac{1}{3},0 \right)\]

    Correct Answer: A

    Solution :

    [a] Equation of line. \[x+2y+z-1+\lambda (-x+y-2z-2)=0\]      ...(i) \[x+y-2+\mu (x+z-2)=0\]                       ...(ii) (0, 0, 1) lies on it \[\Rightarrow \lambda =0,\mu =-2\]         ...(iii) For point of intersection, z=0 and solve (i) and (ii).


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