Question

The distance between the planes $2x+y+2z=8$ and $4x+2y+4z+5=0$ is

• A. $\frac{3}{2}$
• B. $\frac{5}{2}$
• C. $\frac{7}{2}$
• D. $\frac{9}{2}$

Answer 1

Answer: (C)

$\frac{7}{2}$

Given planes are $2x+y+2z=8$ and $4x+2y+4z+5=0$

Multiplying first equation by 2 we get,

$4x+2y+4z=16,4x+2y+4z=-5$

Distance betweeb the planes $=\frac{21}{\sqrt{4^2+2^2+4^2}}$

$\therefore \frac{21}{6}=\frac{7}{2}$