Q4 of 45 Page 1

Find the distance between the planes 2x – y + 2z = 5 and 5x – 2.5y + 5z = 20.

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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