Why do the face diagonals have different angles?

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Homework Help Overview

The discussion revolves around finding the angles of face diagonals in a unit cube, specifically addressing the confusion regarding the identification of face diagonals versus edges and inside diagonals.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the angle between face diagonals using the dot product but encounters confusion regarding which lines qualify as face diagonals.
  • Participants question the definitions and classifications of the lines involved, particularly distinguishing between edges, face diagonals, and inside diagonals.
  • Some participants seek to clarify the correct interpretation of face diagonals and the angles formed between them.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Clarifications about the definitions of face diagonals have been provided, but there is no explicit consensus on the original poster's calculations or the implications of the definitions.

Contextual Notes

There appears to be some confusion regarding the geometric relationships within the cube, particularly in identifying which lines are considered face diagonals and which are not. This may affect the calculations being discussed.

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Homework Statement


Find the angle of the face diagonals of a (unit) cube.
Screenshot - 06292015 - 07:59:34 AM.png
Screenshot - 06292015 - 07:59:45 AM.png

I agree with this solution, but I have a problem with another face diagonal: the face diagonal from the angle (0,0,1),(0,0,0), and (1,1,1).

Homework Equations


dot product
cos-1(a.b/ (|a||b|)

The Attempt at a Solution


From the solution, we have an angle given from the points (1,0,1), (0,0,0), and (0,1,1).
Using the def. of dot product, if A is the vector (0,0,0) to (1,0,1) and B is the vector (0,0,0) to (0,1,1)
cos-1(A.B/(|A||B|)), where A.B = 1, |A| = |B| = sqrt(2).
Thus cos-1(1/ [ (sqrt(2)sqrt(2) ]) = 60 deg. Ok.

the face diagonal from the angle (0,0,1),(0,0,0), and (1,1,1):
A is the vector (0,0,0) to (0,0,1) and B is the vector (0,0,0) to (1,1,1)
cos-1(A.B/(|A||B|)), where A.B = 1, |A| = 1, |B| = sqrt(3).
thus cos-1(1/ [ sqrt(3) ]) != 60 deg.

Why is this happening?
 
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The line from (0,0,0) to (0,0,1) is an edge, not a face diagonal. The line from (0,0,0) to (1,1,1) is an "inside" diagonal, not a face diagonal.
 
Isn't the line from (0,0,1) to (1,1,1) a face diagonal?
Fredrik said:
The line from (0,0,0) to (0,0,1) is an edge, not a face diagonal. The line from (0,0,0) to (1,1,1) is an "inside" diagonal, not a face diagonal.
The angle made with the edge and the inside diagonal is the angle of the face diagonal, assuming the face diagonal is (0,0,1) to (1,1,1), right?
 
raddian said:
Isn't the line from (0,0,1) to (1,1,1) a face diagonal?

The angle made with the edge and the inside diagonal is the angle of the face diagonal, assuming the face diagonal is (0,0,1) to (1,1,1), right?

Wrong: look again at the diagram. You want the angle between *two* face-diagonals originating at the same base-point and lying in two of the faces that meet at that point. So, the vector from (0,0,1) to (1,1,1) is a face diagonal. but the vector from (0,0,1) to (0,0,0) is not.
 
Ray Vickson said:
You want the angle between *two* face-diagonals originating at the same base-point and lying in two of the faces that meet at that point.
Oh so that's what face diagonals mean. :blushing: Thank you.
 
raddian said:
Isn't the line from (0,0,1) to (1,1,1) a face diagonal?
Yes it is. But you're looking for the angle between the two lines I mentioned, and they are not face diagonals.
 

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