Question

The plane of intersection of $x^2+y^2+z^2+2x+2y+2z+2=0$ and $4x^2+4y^2+4z^2+4x+4y+4z-1=0$ is:

• A. $x+y+z+9=0$
• B. $3x+3y+3z+4=0$
• C. $4x+4y+4z+9=0$
• D. They don't intesect

Answer 1

The correct option is C $4x+4y+4z+9=0$

Given: $S_1:x^2+y^2+z^2+2x+2y+2z+2=0$ and $S_2:4x^2+4y^2+4z^2+4x+4y+4z-1=0$
The plane of intersection is given by:
$S_1+\lambda S_2=0$
$x^2+y^2+z^2+2x+2y+2z+2+\lambda (4x^2+4y^2+4z^2+4x+4y+4z)=0\cdots \left(i\right)$
As the plane equation $\left(i\right)$ doesn't contain any $2nd$ order terms of $x,y,z$
So, $\lambda =\frac{-1}{4}$
Put in equation $\left(i\right)$
$\Rightarrow x^2+y^2+z^2+2x+2y+2z+2+\frac{-1}{4}(4x^2+4y^2+4z^2+4x+4y+4z-1)=0$
$\Rightarrow x^2+y^2+z^2+2x+2y+2z+2-(x^2+y^2+z^2+x+y+z-\frac{1}{4})=0$
$\Rightarrow x+y+z+\frac{9}{4}=0$
$\Rightarrow 4x+4y+4z+9=0$ is the required equation