Question

Distance between the parallel planes $2x-3y+4z-1=0$ and $4x-6y+8z+8=0$ is

• A. $\frac{5}{\sqrt{29}}$
• B. $\frac{9}{2\sqrt{29}}$
• C. $\frac{1}{\sqrt{29}}$
• D. $\frac{9}{\sqrt{29}}$

Answer 1

Answer: (A)

$\frac{5}{\sqrt{29}}$

Consider the given line

Let,

$2x-3y+4z-1=0$ ----- $\left(1\right)$

And,

$4x-6y+8z+8=0$

$2x-3y+4z+4=0$ ---- $\left(2\right)$

Now,

Distance between plane $1$ and $2$

$d=|\frac{d_1-d_2}{\sqrt{a^2+b^2+c^2}}|$

$=|\frac{-1-4}{\sqrt{2^2+(-3{)}^2+4^2}}|=\frac{5}{\sqrt{4+9+16}}$

$=\frac{5}{\sqrt{29}}$

Hence, distance between the planes is $\frac{5}{\sqrt{29}}$

So,

Option $A$ is correct.