Distance between parallel planes?

[LaraCroft]

Homework Statement

How do I determine the distance between the parallel planes
2x-4y+ 2z = 5 and -2x+4y-2z= -9: ___?

Homework Equations

The Attempt at a Solution

I got 1/4, but I think that is wrong. Any help would be great!

 

[Dr.D]

Why don't you show how you got 1/4, and let's see what we can do to help.

 

[nlightner]

The distance between parallel planes can be determined by finding the shortest distance between any two points on the two planes. This can be done by finding the perpendicular distance from one plane to the other. To do this, we can use the formula d = |ax + by + cz + d| / √(a^2 + b^2 + c^2), where (a,b,c) is the normal vector to one of the planes and d is the constant term.

In this case, the normal vectors for the two planes are (2,-4,2) and (-2,4,-2), respectively. So, the distance between the two planes can be calculated as d = |(2)(-2) + (-4)(4) + (2)(-2) + 5| / √(2^2 + (-4)^2 + 2^2) = 14 / √24 = √(14/6) = √(7/3) = 1.527 units.

Therefore, the distance between the two planes is 1.527 units.