Question

The intersection of the spheres $x^2+y^2+z^2+7x-2y-z=1$ and $x^2+y^2+z^2-3x+3y+4z=-4$ is same as the intersection of one of the spheres and the plane is

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

Answer 1

The correct option is A

$2x-y-z=1$

The explanation for the correct option

The given equations of the spheres: $S_1:x^2+y^2+z^2+7x-2y-z=1$ and $S_2:x^2+y^2+z^2-3x+3y+4z=-4$.

Thus, the equation of the intersecting plane of two spheres can be given by, ${\mathit{S}}_{\mathbf{1}}\mathbf{-}{\mathit{S}}_{\mathbf{2}}\mathbf{=}\mathbf{0}$

$$\begin{gathered} \Rightarrow \left(x^2+y^2+z^2+7x-2y-z\right)-\left(x^2+y^2+z^2-3x+3y+4z\right)=1-\left(-4\right) \\ \Rightarrow \left(7+3\right)x+\left(-2-3\right)y+\left(-1-4\right)z=1+4 \\ \Rightarrow 10x-5y-5z=5 \\ \Rightarrow 5\left(2x-y-z\right)=5 \\ \Rightarrow 2x-y-z=1 \end{gathered}$$

Hence, the intersecting plane is $2x-y-z=1$, so, option (A) is the correct answer.