The acute angle between a diagonal of a cube and a diagonal of one of its faces drawn from the same vertex is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\cos^{-1}\left(\displaystyle\frac{\sqrt{2}}{\sqrt{3}}\right)\)
Let the cube have side a, one vertex at the origin O and its edges along the axes.
The diagonal of the cube from O goes to (a, a, a): direction ratios 1, 1, 1.
A face diagonal from O goes, for example, to (a, a, 0): direction ratios 1, 1, 0.
\( \displaystyle \cos\theta = \frac{(1)(1)+(1)(1)+(1)(0)}{\sqrt{3}\,\sqrt{2}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{2}}{\sqrt{3}} \)
The other two face diagonals from O, to (a, 0, a) and (0, a, a), give the same value.
Why the other options are incorrect
Option (1) \(\cos^{-1}\left(\displaystyle\frac{1}{\sqrt{3}}\right)\) — Incorrect. This is the angle between a diagonal of the cube and an edge.
Option (3) \(\cos^{-1}\left(\displaystyle\frac{1}{3}\right)\) — Incorrect. This is the angle between two diagonals of the cube.
Option (4) \(\displaystyle\frac{\pi}{4}\) — Incorrect. This is the angle between a face diagonal and an edge of that face.