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 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$

Answer 1

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

Equation of two spheres are $x^2+y^2+z^2+7x-2y-z-13=0$ and $x^2+y^2+z^2-3x+3y+4z-8=0$.
Is these sphere intersect, then $S-S^{\prime }=0$ represent the equation of common plane of intersection.
$\therefore (x^2+y^2+z^2+7x-2y-z-13)-(x^2+y^2+z^2-3x+3y+4z-8)=0\Rightarrow x^2+y^2+z^2+7x-2y-z-13-x^2-y^2-z^2+3x-3y-4z+8=0\Rightarrow 10x-5y-5z-5=0\Rightarrow 2x-y-z=1$