Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is


Distance between two parallel planes Ax + By + Cz = d1 and Ax + By + Cz = d2 is


Given -


First Plane is


2x + 3y + 4z = 4


Comparing with Ax + By + Cz = d1, we get -


A = 2, B = 3, C = 4, d1 = 4


Second Plane is


4x + 6y + 8z = 12


After Dividing by 2,


2x + 3y + 4z = 6


Comparing with Ax + By + Cz = d1, we get -


A = 2, B = 3, C = 4, d2 = 6


So,


Distance between two planes




= 2/√29


Hence, (D) is the correct option.

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