The acute angle between any two diagonals of a cube is

The correct answer is Option (2) → \(\cos^{-1}\left(\displaystyle\frac{1}{3}\right)\)

Let the cube have side a, with one corner at the origin O and edges along the axes. Two of its diagonals are

    \(O(0, 0, 0)\) to \((a, a, a)\): direction ratios 1, 1, 1

    \((0, 0, a)\) to \((a, a, 0)\): direction ratios 1, 1, −1

The angle between the lines:

    \( \displaystyle \cos\theta = \frac{|1+1-1|}{\sqrt{3}\,\sqrt{3}} = \frac{1}{3} \)

By symmetry every pair of diagonals gives the same value, so the angle is \(\cos^{-1}\left(\frac{1}{3}\right)\).

Why the other options are incorrect

• Option (1) \(\displaystyle\frac{\pi}{3}\) — Incorrect. That needs \(\cos\theta = \frac{1}{2}\), but the value is \(\frac{1}{3}\).

• Option (3) \(\cos^{-1}\left(\displaystyle\frac{1}{\sqrt{3}}\right)\) — Incorrect. This is the angle between a diagonal and an edge of the cube, for example between 1, 1, 1 and the x-axis.

• Option (4) \(\cos^{-1}\left(\displaystyle\frac{2}{3}\right)\) — Incorrect. The cosine is \(\frac{1}{3}\), not \(\frac{2}{3}\).