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 - Mathematics MCQ AIEEE 2004 2026 | ExamTest.live

hardMCQAIEEE 20042026Mathematics

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

Solution & Step-by-step Explanation

The intersection of two spheres S_{1} and S_{2} lies on the plane given by S_{1} - S_{2} = 0 . S_{1} : x^{2} + y^{2} + z^{2} + 7 x - 2 y - z - 13 = 0 . S_{2} : x^{2} + y^{2} + z^{2} - 3 x + 3 y + 4 z - 8 = 0 . Subtracting S_{2} from S_{1} : (7 x - (- 3 x)) + (- 2 y - 3 y) + (- z - 4 z) + (- 13 - (- 8)) = 0 10 x - 5 y - 5 z - 5 = 0 Dividing by 5 : 2 x - y - z = 1 .

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

x - y - z = 1

x - 2 y - z = 1

x - y - 2 z = 1

2 x - y - z = 1

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