3 dimensional cube finding an angle

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To find the angle of intersection between the planes formed by triangles EBC and ECD, it's essential to calculate the normal vectors of these planes. The discussion suggests using the scalar triple product, which involves the cross product and dot product of the vectors. By substituting the appropriate values into the equation, the angle of intersection can be determined. One participant calculated the angle to be approximately 55.24 degrees. This method provides a systematic approach to solving the problem.
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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?
 
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To find the angle, you need to find the normal vectors to the planes.
 
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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