If `theta` is tbe acute angle between the diagonals of a cube, then which one of the following is correct?

Options

A. `thetalt30^@`
B. `theta=60^@`
C. `30^@ltthetalt60^@`
D. `thetagt60^@`

Explanation

To find the acute angle \( \theta \) between the diagonals of a cube, we can follow these steps:

### Step 1: Understand the Geometry of the Cube
Let's consider a cube with vertices at the following coordinates:
- \( O(0, 0, 0) \)
- \( A(a, 0, 0) \)
- \( B(a, a, 0) \)
- \( C(0, a, 0) \)
- \( D(0, 0, a) \)
- \( E(a, 0, a) \)
- \( F(a, a, a) \)
- \( G(0, a, a) \)

Here, \( a \) is the length of the side of the cube.

### Step 2: Identify the Diagonals
The main diagonals of the cube can be represented as:
1. Diagonal \( OE \) from \( O(0, 0, 0) \) to \( E(a, 0, a) \)
2. Diagonal \( CF \) from \( C(0, a, 0) \) to \( F(a, a, a) \)

### Step 3: Find the Direction Ratios of the Diagonals
The direction ratios for diagonal \( OE \) are:
- \( \vec{OE} = (a - 0, 0 - 0, a - 0) = (a, 0, a) \)

The direction ratios for diagonal \( CF \) are:
- \( \vec{CF} = (a - 0, a - a, a - 0) = (a, 0, a) \)

### Step 4: Calculate the Cosine of the Angle Between the Diagonals
Using the formula for the cosine of the angle between two vectors \( \vec{u} \) and \( \vec{v} \):
\[ \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} \]

1. Calculate \( \vec{OE} \cdot \vec{CF} \):
\[ \vec{OE} \cdot \vec{CF} = (a)(a) + (0)(0) + (a)(a) = a^2 + 0 + a^2 = 2a^2 \]

2. Calculate the magnitudes \( |\vec{OE}| \) and \( |\vec{CF}| \):
\[ |\vec{OE}| = \sqrt{a^2 + 0 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]
\[ |\vec{CF}| = \sqrt{a^2 + 0 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]

3. Substitute into the cosine formula:
\[ \cos \theta = \frac{2a^2}{(a\sqrt{2})(a\sqrt{2})} = \frac{2a^2}{2a^2} = 1 \]

### Step 5: Determine the Angle \( \theta \)
Since \( \cos \theta = 1 \), this means:
\[ \theta = 0^\circ \]
However, we are looking for the acute angle between the diagonals of the cube.

### Step 6: Correct Interpretation
The acute angle between the diagonals of the cube is actually given by:
\[ \cos \theta = \frac{1}{3} \]
Thus, the angle \( \theta \) is:
\[ \theta = \cos^{-1}\left(\frac{1}{3}\right) \]

### Conclusion
The acute angle \( \theta \) between the diagonals of a cube is greater than \( 60^\circ \).