The intersection of the spheres x2 + y2 + z2 + 7x - 2y - z = 13 and x2...
Intersection of Spheres and Plane:
The intersection of the given spheres can be found by solving the two equations simultaneously. The resulting intersection will be a common point or set of points where both spheres intersect. This intersection can also be represented as the intersection of one of the spheres and a plane.
Given Equations:
1. x^2 + y^2 + z^2 + 7x - 2y - z = 13
2. x^2 + y^2 + z^2 - 3x + 3y + 4z = 8
Intersection of Spheres:
By solving the two equations, we get the point of intersection as (2, -1, 3). This point lies on both spheres.
Intersection of Sphere and Plane:
To represent the intersection as the intersection of one of the spheres and a plane, we can rewrite one of the sphere equations in the form of a plane equation.
Choosing the correct option:
Among the given options, the equation 2x - y - z = 1 can be obtained by subtracting equation 2 from equation 1. Therefore, 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 the sphere and the plane 2x - y - z = 1. Hence, the correct option is D.