3-D cube

Take any corner of the cube as the origin and assign an arbitrary right-handed three-dimensional Cartesian coordinate system along the sides of the cube. By doing so, the space diagonals are reduced to two vectors , for example, $\vec{D_1}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{D_2}=\hat{i}-\hat{j}+\hat{k}$ .

To find the angle between these vectors, we take the dot product :

$\vec{D_1}\cdot \vec{D_2}=|\vec{D_1}||\vec{D_2}|\cos{\theta}$

Substituting and solving for $\theta$ , we obtain $\theta = \cos^{-1}{\dfrac{1}{3}}$

From proper view, both diagonals are just on a rectangle of side length $\sqrt2$ and side length $1$ .

$t = \tan \frac {\theta}{2} = \frac {1}{\sqrt2}$ {For an acute angle to stand for angle of lines.}

$\Rightarrow \cos \theta = \frac{1 - t^2}{1 + t^2}= \frac {1 - \frac12}{1 + \frac12} = \frac13$

$\Rightarrow \theta = cos^{-1} \frac13$

$tan(\frac{\alpha}{2})=\sqrt{2} \rightarrow \alpha=2tan^{-1}(\sqrt{2}) \rightarrow cos(\alpha)=cos(2tan^{-1}(\sqrt{2}))$ $cos(\alpha)=cos^2(tan^{-1}(\sqrt{2}))-sin^2(tan^{-1}(\sqrt{2})) \rightarrow \frac{3}{2} - \frac{1}{3}$ $\alpha=cos^{-1}(\frac{1}{3})$