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

  • question_answer
    The equation of the plane containing the line \[\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\] and the point (0, 7, ?7) is [Roorkee 1999]

    A)            \[x+y+z=1\]

    B)            \[x+y+z=2\]

    C)            \[x+y+z=0\]

    D)            None of these

    Correct Answer: C

    Solution :

                       The equation of plane containing the line  \[\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\] is            \[a(x+1)+b(y-3)+c(z+2)=0\]                             .....(i)            where \[-3a+2b+c=0\]                                       .....(ii)            This passes through (0, 7, ?7)            \[\therefore \] \[a+4b-5c=0\]                                         ..?(iii)            From (ii) and (iii),\[\frac{a}{-14}=\frac{b}{-14}=\frac{c}{-14}\] or \[\frac{a}{1}=\frac{b}{1}=\frac{c}{1}\]                    Thus, the required plane is \[x+y+z=0\].


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