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

  • question_answer
     The distance of the  point (?1, ?5, ?10) from the point of intersection of the line \[\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\] and the plane \[x-y+z=5\], is [AISSE 1985; DSSE 1984; MP PET 2002]

    A)            10

    B)            11

    C)            12

    D)            13

    Correct Answer: D

    Solution :

                       Any point on the line \[\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=t\] is \[(3t+2,\,4t-1,\,12t+2)\]                    This lies on \[x-y+z=5\]                    \[\therefore \] \[3t+2-4t+1+12t+2=5\] i.e.,  \[11t=0\Rightarrow t=0\]                    \[\therefore \] Point is \[(2,\,-1,\,2)\]. Its distance from \[(-1,\,-5,\,-10)\] is,                =\[\sqrt{{{(2+1)}^{2}}+{{(-1+5)}^{2}}+{{(2+10)}^{2}}}\]=\[\sqrt{9+16+144}=13\].


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