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

JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Self Evaluation Test - Three Dimensional Geometry

  • question_answer
    A variable plane at a distance of 1 unit form the origin cuts the coordinate axes at A, B and C. if the centroid \[D(x,y,z)\] of triangle ABC satisfies the relation \[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=k\], then the value of k is

    A) 3

    B) 1

    C) 1/3

    D) 9  

    Correct Answer: D

    Solution :

    [d] Let the equation of variable plane be
    \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.\]
    Which meets the axes at \[A(a,0,0),B(0,b,0)\] and\[C(0,0,c)\].
    The centroid of \[\Delta ABC\] is \[\left( \frac{a}{3},\frac{b}{3},\frac{c}{3} \right)\]and it satisfies the relation\[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=k\]. Thus,
    \[\frac{9}{{{a}^{2}}}+\frac{9}{{{b}^{2}}}+\frac{9}{{{c}^{2}}}=k\]
    or \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}=\frac{k}{9}\]   ??...(i)
    Also it is given that the distance of the plane \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\] from (0, 0, 0) is 1 unit. Therefore,
    \[\frac{1}{\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}}}=1\] or \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}=1\]   ?..(ii)
    From (i) and (ii), we get \[k/e=1,\] i.e., k=9


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