Q. The intersection of the spheres $x^{2} + y^{2} + \mathcal{z}^{2} + 7x - 2y - \mathcal{z} = 13$ and $x^{2} + y^{2} + \mathcal{z}^{2} - 3x + 3y + 4\mathcal{z} = 8$ is the same as the intersection of one of the spheres and the plane
A) $x - y - \mathcal{z} = 1$
B) $x - 2y - \mathcal{z} = 1$
C) $x - y - 2\mathcal{z} = 1$
D) $2x - y - \mathcal{z} = 1$
Explanation
(d)
The given sphere are
$x^{2} + y^{2} + \mathcal{z}^{2} + 7x - 2y - \mathcal{z} - 13 = 0$ (i)
and $x^{2} + y^{2} + \mathcal{z}^{2} - 3x + 3y + 4\mathcal{z} - 8 = 0$ (ii)
Subtracting (ii) from (i), we get
$10x - 5y - 5\mathcal{z} - 5 = 0$
$\Rightarrow \ 2x - y - \mathcal{z} = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.