JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    The intersection of the spheres \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+7x-2y-z=13\] and \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3x+3y+4z=8\] is the same as the intersection of one of the sphere and the plane

    A) \[x-y-z=1\]         

    B) \[x-2y-z=1\]

    C)                        \[x-y-2z=1\]   

    D)        \[2x-y-z=1\]

    Correct Answer: D

    Solution :

    Equation of plane of intersection of two spheres S and S' is\[S-S'=0\]. Equation of two spheres are \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+7x-2y-z-13=0\] and \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3x+3y+4z-8=0\]. If these spheres intersect, then\[S-S'=0\] represents the equation of common plane of intersection. \[\therefore \] \[({{x}^{2}}+{{y}^{2}}+{{z}^{2}}+7x-2y-z-13)\] \[-({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3x+3y+4z-8)=0\] \[\Rightarrow \]\[10x-5y-5z-5=0\] \[\Rightarrow \]\[2x-y-z=1\]

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