The intersection of the spheres x2 + y2 + z2 + 7x − 2y − z = 13 and x2 + y2 + z2 − 3x + 3y + 4z = 8 is the same as the intersection of one of the sphere and the plane

Question

The intersection of the spheres x2 + y2 + z2 + 7x − 2y − z = 13 and x2 + y2 + z2 − 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

Tardigrade Admin · Accepted Answer

Required plane is S1 - S2 = 0 where S1 = x2 + y2 + z2 + 7x - 2y - z - 13 = 0 and S2 = x2 + y2 + z2 - 3x + 3y + 4z - 8 = 0 ⇒ 2x - y - z = 1.

Q. 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 - 2y - z = 1

x - y - 2z = 1

2x - y - z = 1

Required plane is $S_{1} - S_{2} = 0$
where $S_{1} = x_{2} + y_{2} + z_{2} + 7 x - 2 y - z - 13 = 0$ and
$S_{2} = x_{2} + y_{2} + z_{2} - 3 x + 3 y + 4 z - 8 = 0$
$\Rightarrow 2 x - y - z = 1.$

Q. The intersection of the spheres x2+y2+z2+7x−2y−z=13 and x2+y2+z2−3x+3y+4z=8 is the same as the intersection of one of the sphere and the plane