2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Three Dimensional Geometry

JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Critical Thinking

  • question_answer
    P is a fixed point \[(a,\,a,\,a)\] on a line through the origin equally inclined to the axes, then any plane through P perpendicular to OP, makes intercepts on the axes, the sum of whose reciprocals is equal to

    A) a

    B) \[\frac{3}{2a}\]

    C) \[\frac{3a}{2}\]

    D) None of these

    Correct Answer: D

    Solution :

    • Since the line is equally inclined to the axes and passes through the origin, its direction ratios are 1, 1, 1. So its equation is \[\frac{x}{1}=\frac{y} {1}=\frac{z}{1}.\]           
    • A point P on it is given by (a, a, a). So equation of the plane through P (a, a, a) and perpendicular to OP is \[1\,(x-a)+1(y-a)+1\,(z-a)=0\]              
    • \[[\,\because \,\,\,OP\] is normal to the plane]           
    • i.e. \[x+y+z=3a\] or \[\frac{x}{3a}+\frac{y}{3a}+\frac{z}{3a}=1\]           
    • Intercepts on axes are \[3a,\,\,3a,\,\,3a.\] Therefore sum of reciprocals of these intercepts           
    • \[=\frac{1}{3a}+\frac{1}{3a}+\frac{1}{3a}=\frac{3}{3a}=\frac{1}{a}\].


You need to login to perform this action.
You will be redirected in 3 sec spinner