JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Sphere

  • question_answer
    The ratio in which the sphere \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=504\] divides the line segment  AB joining the points \[A\ (12,\ -4,\ 8)\] and \[(27,\ -9,\ 18)\] is given by

    A)            \[2:3\] externally

    B)            \[2:3\] internally

    C)            \[1:2\] externally

    D)            None of these

    Correct Answer: A

    Solution :

                       Let the ratio is \[\lambda :1\] so the point on sphere is \[\frac{27\lambda +12}{\lambda +1},\frac{-9\lambda -4}{\lambda +1},\frac{18\lambda +8}{\lambda +1}\]                                 Also, \[{{\left( \frac{27\lambda +12}{\lambda +1} \right)}^{2}}+{{\left( \frac{9\lambda +4}{\lambda +1} \right)}^{2}}+{{\left( \frac{18\lambda +8}{\lambda +1} \right)}^{2}}=504\]                                                               \ \[\lambda =\frac{-2}{3}\],  So that the ratio is 2 : 3 externally.


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