The distance between the parallel planes
$2 x - 3 y + 6 z + 1 = 0$ and
$4 x - 6 y + 12 z - 5 = 0$ is ___

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Solution

$2 x - 3 y + 6 z + 1 = 0$
$4 x - 6 y + 12 z - 5 = 0$
implise $2 x - 3 y + 6 y - \frac{5}{2} = 0$
Distance between the given planes $= \frac{| d_{1} - d_{2} |}{\sqrt{a^{2} + b^{2} + c^{2}}}$
$= \frac{∣ ∣ ∣ 1 + \frac{5}{2} ∣ ∣ ∣}{\sqrt{4 + 9 + 36}}$
$= \frac{1}{2}$ units

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