Question 204508
A body diagonal of a cube is a line through the centre joining opposite verticies. Find the angles between the body diagonals of a cube. 
-------------
Put the Origin at the center of the cube.  The vertices are:
(1,1,1), (-1,1,1), (1,-1,1), etc.
All angles will be equal, finding one is sufficient.
--------------------
Find the angle between vector A i+j+k and B i+j-k
A dot B = |A|*|B|*cos(x)
|A| = |B| = sqrt(3)
A dot B = 1 + 1 - 1 = 1
cos = 1/sqrt(3)
Angle = arccos(sqrt(3)/3)
Angle = ~ 54.7356 degs