Angles in a Cube III
Let denote the vertices of a cube ABCD (for the bottom face) and A'B'C'D' (for the top face), with A' above A, etc. Let O stand for the center of the cube. What can be said about angle AOC, one of the central angles of the cube?
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Copyright © 1996-2018 Alexander BogomolnyAngles in a Cube III
Scott E. Brodie
Consider a 2 × 2 × 2 cube centered at the origin of a Cartesian coordinate system. The vertices are located at
Now consider a "central angle" of the tetrahedron, say, the angle AOC. By visualizing the tetrahedron in this way, we can readily compute the size of this angle - call it t:
Taking the dot product of the vectors from the origin to the two upper vertices of the tetrahedron, we obtain
√1² + 1² + 1² × √(-1)² + (-1)² + 1² × cos (t) = 1×(-1) + 1× (-1) + 1×1 = -1.
Or
cos (t) = -1/3,
t = arccos (-1/3) ~ 109.47122 degrees.
This angle is an important building block in many crystal structures, including those of diamonds and ice, as well as the basic structure of many organic molecules.
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While all central angles in a regular tetrahedron are equal, in a cube, there is another central angle distinct from the above: this is the angle that subtends an edge of a cube, e.g. ∠COC'. We denote it s. While t is obtuse, s is acute. Applying the same technique as for t, it is easily seen that
It is apparent that triangle ACC' is right, with legs 2 and 2√2 and hypotenuse 2√3. (To see that apply the Pythagorean theorem to a couple of right triangles.) The median to the hypotenuse is just a half of the latter, √3, drawing our attention once more to the fact that AC' is a straight line.
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|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny72106177