3 dimensional cube finding an angle

  • Thread starter Toranc3
  • Start date
  • #1
189
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Homework Statement


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


Homework Equations



AB=ABcosθ

Ab=Absinθ

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?
 

Answers and Replies

  • #2
6,054
391
To find the angle, you need to find the normal vectors to the planes.
 
  • #3
24
0
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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