JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Line

  • question_answer
    The equation of straight line passing through the point (a, b, c) and parallel to z- axis, is         [MP PET 1995; Pb. CET 2000]

    A)                                                               \[\frac{x-a}{1}=\frac{y-b}{1}=\frac{z-c}{0}\]

    B)                                                               \[\frac{x-a}{0}=\frac{y-b}{1}=\frac{z-c}{1}\]

    C)                                                               \[\frac{x-a}{1}=\frac{y-b}{0}=\frac{z-c}{0}\]

    D)                                                               \[\frac{x-a}{0}=\frac{y-b}{0}=\frac{z-c}{1}\]

    Correct Answer: D

    Solution :

                     The line through \[(a,\,\,b,\,\,c)\] is \[\frac{x-a}{l}=\frac{y-b}{m}=\frac{z-c}{n}\]                                                                                                 ?..(i)                 Since the line is parallel to z-axis, therefore any point on this line will be of the form \[(a,\,\,b,\,\,{{z}_{1}}).\]                 Also any point on line (i) is \[(lr+a,\,\,mr+b,\,\,nr+c).\]                 Hence  \[\begin{matrix}    lr+a=a  \\    mr+b=b  \\ \end{matrix}\Rightarrow \,\,l=m=0\]                                                                       Hence the line will be \[\frac{x-a}{0}=\frac{y-b}{0}=\frac{z-c}{1}\].


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