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
JEE · Math · previous-year question
- A.$$2x-y-z=1$$correct
- B.$$x-2y-z=1$$
- C.$$x-y-2z=1$$
- D.$$x-y-z=1$$
Answer
A. $$2x-y-z=1$$
Explanation
The equation of spheres are $${S_1}:{x^2} + {y^2} + {z^2} + 7x - 2y - z - 13 = 0$$ and $${S_2}:{x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8 = 0$$ Their plane of intersection is $${S_1} - {S_2} = 0$$ $$ \Rightarrow 10x - 5y - 5z - 5 = 0$$ $$ \Rightarrow 2x - y - z = 1$$
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