Find the distance between the given parallel planes. 2z = 6y − 4x, 3z = 1 − 6x + 9y?

Answer 1

The first plane includes the origin, so use the formula for distance from a  point (the origin) to a plane (the second given plane).

3z = 1 − 6x + 9y

6x - 9y + 3z - 1 = 0 ... general form

|6(0) - 9(0) + 3(0) - 1| / √(6² + 9² + 3²)

= 1 / √(126)

= √(14) / 42

Answer 2

Normal vector to either plane is

2i - 3j + k.

The first plane goes through the origin, so let's find the line through the origin and normal to the first plane:

r(t) = <2, -3, 1> t.

Where will this intersect the 2nd plane?

-...wherever 3(1t) = 1 - 6(2t) + 9(-3t) => 

3t = 1 - 12t - 27t =>

t = 1/42.

r(1/42) = <1/21, -1/14 + 1/42>.

Is it clear that 3(1/42) = 1 - 6/21 - 9/14 ?

Let's see, 3/42 = (42 - 12 - 27)/42 = 3/42.  Yup.

The distance is the distance between the planes is the distance from (0,0,0) to (1/21,-1/14,1/42),

which is sqrt(2^2 + 3^2 + 1^2)/42 = sqrt(14)/42.