Find the distance between parallel planes

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MarkSA

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Sep 8, 2007
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Hello,

Q: Find the distance between parallel planes x-5y-z=1 and 5x-25y-5z=-3. I know these are parallel since the normal vector of the 2nd plane is a multiple of the normal vector of the 1st plane.

For this problem, I believe I need to use an equation from the book. D = |comp b onto n| = |n dot b|/|n|
This is I believe the scalar projection of b onto n. n is supposed to be the normal vector (yet I have two in this case due to the two planes above? would I just use the one with the multiple factored out <1,-5,-1> or does it not matter?)
But I can't find any real definition of what 'b' is or how to derive it.

Any idea? Thanks.
 
Find a point P on one of the planes and a point Q on the other plane.
Use the vector \(\displaystyle \overrightarrow {PQ}\) as your vector b in the formula.
 
thanks for the reply. What would I use as the normal vector? (why?) I wish I could find a way to better visualize this, or understand it. It seems like in this chapter I am more memorizing what steps to do for each problem rather than actually learning why I should be doing them, which bothers me.
 
Two parallel planes have the same normal.
 
For ease of calculation (and visualization), I find the points on the planes at x = 0 and y = 0 (z calculated from the given equations).
 
Let a point in the first plane be(1,0,0), 1 = 1.

Then D = |ax+by+cz+d|/sqrt(a^2+b^2+c^2) (distance formula) = |5(1)+(-25)(0)+(-5)(0)+3|/sqrt[(5)^2+(-25)^2+(-5)^2]= 8[sqrt(3)]/45.
 
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