Question

The distance between the planes $2x+3y+4z=4$ and $4x+6y+8z=12$ is

$\left(a\right)\frac{2}{\sqrt{29}}\text{units}\left(b\right)\frac{10}{\sqrt{29}}\text{units}$
$\left(c\right)4\text{units}\left(d\right)2\text{units}$

Answer 1

$2x+3y+4z=4$
$4x+6y+8z=12\Rightarrow 2x+3y+4z=6$
Both the planes are parallel.
$\text{Required distance}=\frac{|4-6|}{\sqrt{2^2+3^2+4^2}}=\frac{2}{\sqrt{29}}$

Option $\left(a\right)$ is correct.