The distance between the planes $4 x - 5 y + 3 z = 5$ and $4 x - 5 y + 3 z + 2 = 0$ is

\frac{7}{2 \sqrt{5}}

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7

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\frac{7}{5 \sqrt{2}}

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3

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Solution

The correct option is B $\frac{7}{5 \sqrt{2}}$
Given planes are $4 x - 5 y + 3 z - 5 = 0$ and $4 x - 5 y + 3 z + 2 = 0$
Since both the planes are parallel so distance between them is,
$= ∣ ∣ ∣ ∣ \frac{(- 5 - 2)}{\sqrt{4^{2} + 5^{2} + 3^{2}}} ∣ ∣ ∣ ∣ = \frac{7}{5 \sqrt{2}}$

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Similar questions

\frac{x + 3}{3} = \frac{y - 2}{- 2} = \frac{z + 1}{1}

4 x + 5 y + 3 z - 5 = 0

\frac{x}{1} = \frac{y}{2} = \frac{z}{3}

4 x + 5 y = 7

3 x + 4 y = 5

4 x - 5 y = - 7

4 x - y = 5

\frac{x + 2}{3} = \frac{2 y + 3}{4} = \frac{3 z + 4}{5}

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