Find the distance between the planes `2x-y+2z=4 and 6x-3y+6z=2.`

Explanation

Given planes are

`2x –y+2z-4=0 ldots (1)`

And `6x-3y+6z-2=0 ldots (2)`

We find that `a_1/a_2=b_1/b_2=c_1/c_2`

Hence planes `(1)` and `(2)` are parallel.

Plane `(2)` may be written as `2x-y+2z-2/3=0 ldots (3)`

`therefore` Required distance between the planes

`|4-2/3|/sqrt(2^2 + ((-1)^2) + 2^2) = 10/(3*3)`

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