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

Distance between two parallel planes Ax + By + Cz = d_{1} and Ax + By + Cz = d_{2} is

⇒

Given -

First Plane is

2x + 3y + 4z = 4

Comparing with Ax + By + Cz = d_{1}, we get -

A = 2, B = 3, C = 4, d_{1} = 4

Second Plane is

4x + 6y + 8z = 12

After Dividing by 2,

2x + 3y + 4z = 6

Comparing with Ax + By + Cz = d_{1}, we get -

A = 2, B = 3, C = 4, d_{2} = 6

So,

Distance between two planes

= 2/√29

Hence, (D) is the correct option.

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