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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


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    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


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


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    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.
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