Question

The distance between the parallel planes $x+2y-3\sigma =2$ and $2x+4y-6z+7=0$ is

• A. $\frac{2}{\sqrt{14}}$
• B. $\frac{11}{\sqrt{56}}$
• C. $\frac{7}{\sqrt{56}}$
• D. None of these

Answer 1

Answer: (B)

$\frac{11}{\sqrt{56}}$

Let us assume a point on the first plane be taken as $(2,0,0)$
Required distance between the given planes$=$ Distance of $(2,0,0)$ from the plane $2x+4y-6z+7=0$
$=\frac{2\left(2\right)+4\left(0\right)-6\left(0\right)+7}{\sqrt{4+16+36}}$
$=\frac{11}{\sqrt{56}}$