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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • 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

    A)  26                         

    B)         \[11\frac{4}{13}\]

    C)  13                         

    D)         39

    E)  None of these

    Correct Answer: C

    Solution :

    The centre of the given sphere is\[C\text{ }(-\text{ }2,1,\text{ }3)\]. The distance from the centre of sphere to the plane \[=\left| \frac{-2\times 12+4\times 1+3\times 3-327}{\sqrt{144+16+9}} \right|\] \[=\left| \frac{-24+4+9-327}{\sqrt{169}} \right|=\frac{338}{13}=26\] \[\therefore \]Shortest distance \[=26-\sqrt{4+1+9+155}\] \[=26-13=13\]


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