What is the angle between diagonals of two faces on a cube originating at the same vertex?
What I have done:
Vector representations of the diagonals joining the points $(0,0,0)$ to $(1,1,1)$ and $(1,0,0)$ to $(011)$ are $\langle1,1,1\rangle$ and $\langle1,1,1\rangle$. Let $\theta$ be the angle between these two vectors. $\langle1,1,1\rangle\cdot \langle-1,1,1\rangle=-1+1+1=1$ This gives $\cos \theta = 1/3$. Am I right?
Best Answer
The vector that joins $(0,0,0)$ to $(1,1,1)$ is not the diagonal of a face, but a diagonal of the cube.
Note that if $AB$ and $AC$ are the diagonals of two faces that share a common vertex $A$, then $BC$ is also the diagonal of a face, and this implies that $ABC$ is equilateral.