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Find the angle between two planes

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the angle between the plane 3x+5y+7z = 1 and the plane z = 0.


    2. Relevant equations
    a.b=|a||b|cosθ



    3. The attempt at a solution
    Hi, I know that I need to have both these planes in the form (x,y,z) and then find the dot product to find the angle between them. The problem I am having is with putting them in that form, at first I assumed plane 1 would just be (3,5,7) and plane 2 would be (0,0,1), but I have also read that to find the angle between to planes I need the normal vector to each plane, and this has confused me. Using these vectors I came up with the answer 30.8°, but I don't know if what I did was right! Any help would be appreciated!
     
  2. jcsd
  3. Mar 9, 2013 #2

    Dick

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    (3,5,7) IS the normal vector to the plane, not the plane itself. Same for (0,0,1). It sounds like you are doing it correctly. I don't get the answer you got though. Can you show how your numbers worked?
     
  4. Mar 9, 2013 #3
    I think the answer should be 39.8
     
  5. Mar 9, 2013 #4
    can anyone help me with this question:
    Find the angle between
    (a) the line L1 given by the equations y = 2z, x = 0, and
    (b) the line L2 given by the equations x = 3z, y = 0.
     
  6. Mar 10, 2013 #5
    Sorry, working it out again I got 39.79;
    a.b=(3x0+5x0+7x1) = 7
    |a|=√83
    |b|=√1
    ∴θ=cos-1(a.b/|a||b|)= 39.79

    So for any equation ax+by+cz=d, will the normal vector always be (a,b,c)?
    Thanks for your replies:)
     
  7. Mar 10, 2013 #6

    HallsofIvy

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    AXidenT posted this same question at
    https://www.physicsforums.com/showthread.php?t=677426
     
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