# Find the distance between the planes 2x – y + 2z = 5 and 5x – 2.5y + 5z = 20. [CBSE 2017]

Formula Used:

The distance between two planes: a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given by,

Let a = a1, b = b1, and c = c1

Explanation:

Let P(x1, y1, z1) be any point on the plane 2x – y + 2z = 5.

2x1 - y1 + 2z1 = 5

The length of perpendicular from P(x1, y1, z1) to the plane

5x - 2.5y + 5z = 20 is

= |-1|

= 1

Thus, the distance between the planes 2x – y + 2z = 5 and 5x - 2.5y + 5z = 20 is 1 unit.

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