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

  • question_answer
    The shortest distance from the plane \[12x+4y+3z=327\] to the sphere \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+\] \[4x-2y-6z=155\] is [AIEEE 2003]

    A) 26

    B) \[11\frac{4}{13}\]

    C) 13

    D) 39

    Correct Answer: C

    Solution :

    • \[\because \] Shortest distance = Perpendicular distance ? r Now, perpendicular distance           
    • \[=\,\left| \frac{-2\times 12+4\times 1+3\times 3-327}{\sqrt{144+9+16}} \right|\,=\,26\]           
    • \ Shortest distance \[=26-\sqrt{4+1+9+155}\],   \[[\because \,\,26-r]\]                                                 
    • \[=26-13=13\].


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