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

  • question_answer
    Find the equation of set points P such that \[P{{A}^{2}}+P{{B}^{2}}=2{{K}^{2}},\] where A and B are the points (3, 4, 5) and (-1, 3, -7), respectively:

    A) \[{{K}^{2}}-109\]

    B) \[2{{K}^{2}}-109\]

    C) \[3{{K}^{2}}-109\]

    D) \[4{{K}^{2}}-10\]

    Correct Answer: B

    Solution :

    [b] Let the coordinates of point P be (x, y, z). Here, \[P{{A}^{2}}={{(x-3)}^{2}}+{{(y-4)}^{2}}+{{(z-5)}^{2}}\] \[P{{B}^{2}}={{(x+1)}^{2}}+{{(y-3)}^{2}}+{{(z+7)}^{2}}\] By the given condition \[P{{A}^{2}}+P{{B}^{2}}=2{{K}^{2}}\] We have \[{{(x-3)}^{2}}+{{(y-4)}^{2}}+{{(z-5)}^{2}}+{{(x+1)}^{2}}\] \[+{{(y-3)}^{2}}+{{(z+7)}^{2}}=2{{K}^{2}}\] i.e. \[2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}-4x-14y+4z=2{{K}^{2}}-109\]


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