Distance between parallel planes $2 x - 2 y + z + 3 = 0$ and $4 x - 4 y + 2 z + 5 = 0$ is

6

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\frac{1}{6}

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3

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\frac{1}{3}

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Solution

The correct option is C $\frac{1}{6}$
Given planes can be written as, $2 x - 2 y + z + 3 = 0$ and $2 x - 2 y + z + \frac{5}{2} = 0$
Hence distance between them is $= ∣ ∣ ∣ ∣ ∣ \frac{3 - \frac{5}{2}}{\sqrt{2^{2} + (- 2)^{2} + 1^{2}}} ∣ ∣ ∣ ∣ ∣ = \frac{1}{6}$

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