Question

The equation of the plane which is equidistant from the two parallel planes $2x-2y+z+3=0$ and $4x-4y+2z+9=0$ is :

• A. $8x-8y+2z+15=0$
• B. $8x-8y+4z+15=0$
• C. $8x-8y+4z+3=0$
• D. $8x-8y+4z-3=0$
• E. $8x-8y+4z+4=0$

Answer 1

The correct option is B $8x-8y+4z+15=0$

Let the equation of plane parallel between $2x-2y+z+3=0$
and $2x-2y+z+\frac{9}{2}=0$ is
$2x-2y+z+c=0$
Now, distance between two given parallel lines
$=\frac{\frac{9}{2}-3}{\sqrt{2^2+2^2+1^2}}=\frac{3/2}{3}=\frac{1}{2}$
Since required plane is equidistant from given planes
Thus $\frac{c-3}{\sqrt{2^2+2^2+1^2}}=\frac{1/2}{2}$
$\Rightarrow \frac{c-3}{3}=\frac{1}{4}$
$\Rightarrow c=\frac{3}{4}+3$
$\Rightarrow c=\frac{15}{4}$
Therefore, required equation of plane is
$2x-2y+z+\frac{15}{4}=0$
$\Rightarrow 8x-8y+4z+15=0$