Question

Distance between the two planes : $2x+3y+4z=4$ and $4x+6y+8z=12$ is

• A. $2$ units
• B. $4$ units
• C. $8$ units
• D. $\frac{2}{\sqrt{29}}$ units

Answer 1

The correct option is D $\frac{2}{\sqrt{29}}$ units

The equations of the planes are
$2x+3y+4z=\mathrm{4....}\left(1\right)$
$4x+6y+8z=12$
$\Rightarrow$ $2x+3y+4z=\mathrm{6.......}\left(2\right)$
It can be seen that the given planes are parallel.
Thus distance $\left(D\right)$ between them is given by.
$D=\left|\frac{d_2-d_1}{\sqrt{a^2+b^2+c^2}}\right|$
$\Rightarrow$ $D=\left|\frac{6-4}{\sqrt{2^2+{\left(3\right)}^2+{\left(4\right)}^2}}\right|=\frac{2}{\sqrt{29}}$