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

  • question_answer
    The point where the line \[\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}\] meets the plane \[2x+4y-z=1\], is [DSSE 1981]

    A)            (3, ?1, 1)

    B)            (3, 1, 1)

    C)            (1, 1, 3)

    D)            (1, 3, 1)

    Correct Answer: A

    Solution :

                        Let point be (a, b, c),  then \[2a+4b-c=1\]    .....(i)                    and \[a=2k+1,\,\,b=-3k+2\] and \[c=4k-3\],                                                                              (where k is constant)                    Substituting these values in (i), we get            \[2\,(2k+1)+4\,(-3k+2)-(4k-3)=1\,\,\Rightarrow \,\,k=1\]            Hence required point is (3, ?1, 1).            Trick : The point must satisfy the lines and plane. Obviously (3, ? 1, 1) satisfies.


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