1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Distance between two parallel lines

  1. Nov 18, 2008 #1
    1. The problem statement, all variables and given/known data

    Determine the distance between the parallel planes –4x–4y+1z=–1 and 8x+8y–2z=12

    2. Relevant equations

    Proj_n_v = ((vn)/(nn))n

    3. The attempt at a solution

    I thought I understood how to do this, but I am not getting a correct answer for it. What I did was:
    I made the equations to be:
    –4x–4y+1z=–1 and
    and therefore got the normal to these two planes to be: (-4,-4,1)
    Then, I took a point P (0,0,-1) from plane 1. Then, took a point A on plane 2 to be (0,0,-12).

    From these two points, I got vector AP = (0,0,11).

    I projected vector AP onto the normal... so
    proj_n_AP = (((AP)n)/(nn))n

    And got the result: (11/33)[(-4,-4,1)]

    I then got the distance by ll(11/33)[(-4,-4,1)]ll to be (11/33)sqrt33

    Am I doing this totally wrong? Thanks!
  2. jcsd
  3. Nov 18, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    No, you are doing it right. You just made a booboo. Dividing 8x+8y–2z=12 by -2 gives –4x–4y+1z=–6, doesn't it?
  4. Nov 18, 2008 #3
    Aa! Can't believe I made such a stupid mistake. Thanks!!

    So, the answer would just be (5/33)(sqrt33), correct?...

    As I would keep P as (0,0,-1), but make A (0,0,-6); and get the vector AP = (0,0,5)

    So, projecting AP onto the normal, and then getting the distance of the projection would result in: (5/33)(sqrt33)
  5. Nov 18, 2008 #4


    User Avatar
    Science Advisor
    Homework Helper

    That seems right.
  6. Nov 19, 2008 #5


    User Avatar
    Science Advisor

    But why bother with projecting? You know that a point on plane 1 is (0, 0, -1) and that the normal vector is given by <-4, -4 , 1> so a line through that point, normal to the plane is x= -4t, y= -4t, z= -1+ t. Where does that line intersect plane 2? The distance between those two points is the distance between the planes.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook