The intersection of the spheres {x^2} + {y^2} | vedclass

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(A) The equation of the plane passing through the intersection of two spheres ${S_1} = 0$ and ${S_2} = 0$ is given by ${S_1} - {S_2} = 0$.Given the sp

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$2x - y - z = 1$

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(A) The equation of the plane passing through the intersection of two spheres ${S_1} = 0$ and ${S_2} = 0$ is given by ${S_1} - {S_2} = 0$.Given the spheres:${S_1}: {x^2} + {y^2} + {z^2} + 7x - 2y - z - 13 = 0$${S_2}: {x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8 = 0$Subtracting the two equations:$({x^2} + {y^2} + {z^2} + 7x - 2y - z - 13) - ({x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8) = 0$$(7x - (-3x)) + (-2y - 3y) + (-z - 4z) + (-13 - (-8)) = 0$$10x - 5y - 5z - 5 = 0$Dividing the entire equation by $5$:$2x - y - z = 1$Thus,the intersection of the two spheres lies on the plane $2x - y - z = 1$.

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 spheres and the plane:

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