Q9 of 36 Page 1

What is the distance (in units) between the two planes 3x + 5y + 7z = 3 and 9x + 15y + 21z = 9?

Given: The two planes 3x + 5y + 7z = 3 and 9x + 15y + 21z = 9


To Find: Distance between the planes.


If we take common 3 from the 2nd plane we get the 1st plane, i.e., 3x + 5y + 7z = 3. That means both the planes are parallel to each other as the coefficients of x, y, z are equal.


We know that, for two parallel planes Ax + By + Cz = d1 and Ax + By + Cz = d2, the distance between two parallel planes,



Here the planes are, 3x + 5y + 7z = 3 and 3x + 5y + 7z = 3 [After taking common from the 2nd plane].


So, A = 3, B = 5, C = 7, d1 = 3 and d2 = 3.


So, the distance is,

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