Question

The angle between two diagonals of a cube will be:

• A. \( \cos^{-1}\left(\frac{1}{3}\right) \)
• B. \( \sin^{-1}\left(\frac{1}{3}\right) \)
• C. \( \frac{\pi}{2} - \cos^{-1}\left(\frac{1}{3}\right) \)
• D. \( \frac{\pi}{2} - \sin^{-1}\left(\frac{1}{3}\right) \)

Hint:

For 3D geometry problems, use vector dot product and magnitude to calculate angles accurat

Answer 1

The Correct Option is A

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:

\[ \theta = \cos^{-1}\left(\frac{1}{3}\right). \]