Find the angle between the main diagonal of a cube and a skew diagonal of a face of the cube I was told it was $90$ degrees, but then others say it is about $35.26$ degrees. Now I am unsure which one it is.

 A: It depends on what you mean by the skew diagonal.
Consider the cube with corners at $(x,y,z)$ where each element is either zero or one.
In particular, one diagonal is $0=(0,0,0)$ to $u=(1,1,1)$.
Now it depends on what you mean by a "skew diagonal." 
If $0$ to $v=(1,1,0)$ is the other diagonal you are looking form, then the cosine of the angle is $$\frac{u\cdot v}{|u||v|}=\frac{2}{\sqrt{3}\sqrt{2}}$$
The inverse cosine of that value is approximately $35.26$ degrees.
On the other hand, if you mean a skew diagonal such as the diagonal from $(1,0,0)$ and $(0,1,0)$, then that vector of that diagonal is $w=(-1,1,0)$, and $u\cdot w=0$ so the two angles are perpendicular.
A: Take your cube
$\hspace{100pt}$
and make three more copies arranged like this:
$\hspace{50pt}$
Observe, that two red and two blue edges form a parallelogram, but it is symmetric
(find the appropriate symmetry yourself) and as such it is a rectangle, hence, the angle between red and blue edge is $90^\circ$.
I hope this helps ;-)
A: If we assume the cube has unit side length and lies in the first octant with faces parallel to the coordinate planes and one vertex at the origin, then the the vector $(1,1,0)$ describes a diagonal of a face, and the vector $(1,1,1)$ describes the skew diagonal.
The angle between two vectors $u$ and $v$ is given by: 
$$\cos(\theta)=\frac{u\cdot v}{|u||v|}$$
In our case, we have 
$$\cos(\theta)=\frac{2}{\sqrt{6}}\quad\Longrightarrow\quad\theta\approx 35.26$$
A: the answer will be 90 one vector is (1,1,1) and other is (1,0,-1)
