Distance between the two planes:


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

(A)2 units

(B)4 units

(C)8 units

(D) $\frac{2}{\sqrt{29}}$ units


The equations of the planes are

$2 x+3 y+4 z=4$                                ...(1)

$4 x+6 y+8 z=12$

$\Rightarrow 2 x+3 y+4 z=6$            ...(2)   

It can be seen that the given planes are parallel.

It is known that the distance between two parallel planes, $a x+b y+c z=d_{1}$ and $a x+b y+c z=d_{2}$, is given by,


$\Rightarrow D=\left|\frac{6-4}{\sqrt{(2)^{2}+(3)^{2}+(4)^{2}}}\right|$


Thus, the distance between the lines is $\frac{2}{\sqrt{29}}$ units.

Hence, the correct answer is D.

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