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Determining perpendicular planes

  1. Nov 9, 2008 #1
    Determine whether the planes are perpindicular

    (-2, 1, 4) . (x-1, y, z+3) = 0 (PLANE A)
    (1, -2, 1) . (x+3, y-5, z) = 0 (PLANE B)

    Here's what I have figured out so far:

    Plane A passes through (1,0,-3) and is perpendicular to (-2,1,4)
    Plane B passes through (-3, 5, 0) and is perpendicular to (1, -2, 1)

    I know that if I had to determine if they were parallel, (1,0,-3) and (-2,1,4) would have to be along the lines of (-1, 2, 4) and (2, -4, -8).

    I'm not sure if I'm on the right track and missing something right in front of me, or completely lost altogether.
  2. jcsd
  3. Nov 9, 2008 #2


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    Isn't the definition of two planes being perpendicular that their normal vectors are perpendicular?
  4. Nov 9, 2008 #3
    Yes I'm sure, but I'm not sure what the numbers of perpindicular normal vectors look like.

    Like, the example of parallel vectors I gave above - I recognize those as being parallel, but I'm not sure how to tell if something is perpendicular unless the vectors are drawn out.
  5. Nov 9, 2008 #4


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    Aren't (-2, 1, 4) and (1, -2, 1) the normal vectors?
  6. Nov 9, 2008 #5
    If they're perpendicular then the angle between then is 90 degrees. Then their dot product is |normal vector 1||normal vector 2|*cos90 = |normal vector 1||normal vector 2|*0 = 0. So they're perpendicular if their dot product is 0.
  7. Nov 10, 2008 #6
    Thanks Dick and JG, it makes sense! I think I was making it harder than it actually was :)
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