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3 dimensional cube finding an angle

  1. Jun 4, 2013 #1
    1. The problem statement, all variables and given/known data
    A cube is positioned with its vertices at the following points:

    A=(0,0,0) C=(1,1,0) E=(0,0,1) G=(1,1,1)

    B=(1,0,0) D=(0,1,0) F=(1,0,1) H=(0,1,1)

    What is the angle of intersection of the planes formed by the triangles EBC and ECD

    2. Relevant equations



    3. The attempt at a solution

    I am stuck on this one. I drew a picture but I can seem to figure anything out. Could somebody give me a hint?
  2. jcsd
  3. Jun 5, 2013 #2
    To find the angle, you need to find the normal vectors to the planes.
  4. Jun 6, 2013 #3
    hi. you can solve by using scalar triple product * = cross product . = dot product
    [(A*B).C] = |A| |B| sin 90 |c| cos(θ) here theta angle made by intersection of planes
    if substitute all values you will get angle of intersection of two planes
    I got 55.24 degree
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