Question

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 a the intersection of one of the sphere and plane

• A. $x-y-2z=1$
• B. $x-2y-z=1$
• C. $x-y-z=1$
• D. $2x-y-z=1$

Answer 1

The correct option is D $2x-y-z=1$

Given two spheres are

$x_2+y^2+z^2+7x-2y-z=13\to S_1$

$x^2+y^2+z^2-3x+3y+4z=8\to S_2$

Intersection of plane of shperes is $S_1-S_2=0$

Substract $S_1$ and $S_2$

$10x-5y-5z=52x-y-z=1$

Intersection of spheres is $2x-y-z=1$