1. The coordinates of opposite vertices of a cube are \((0, 0, 0)\) and \((a, a, a)\). The diagonal of the cube is the line connecting these two points.
2. The diagonals of a cube form vectors:
\[ \vec{d}_1 = (a, a, a), \quad \vec{d}_2 = (-a, a, a). \]
3. The angle between two diagonals is given by:
\[ \cos \theta = \frac{\vec{d}_1 \cdot \vec{d}_2}{|\vec{d}_1||\vec{d}_2|}. \]
4. Compute:
\[ \vec{d}_1 \cdot \vec{d}_2 = -a^2 + a^2 + a^2 = a^2. \]
\[ |\vec{d}_1| = |\vec{d}_2| = \sqrt{3}a. \]
5. Substitute:
\[ \cos \theta = \frac{a^2}{3a^2} = \frac{1}{3}. \]
6. Therefore:
\[ \theta = \cos^{-1}\left(\frac{1}{3}\right). \]
A quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of: