1. Not finding help here? Sign up for a free 30min 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!

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

    AB=ABcosθ

    Ab=Absinθ

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted