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

Options

A. `theta=30^@`
B. `theta=45^@`
C. `2costheta=1`
D. `3costheta=1`

Answer

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

### Step 1: Define the diagonals of the cube
Assume the cube has a side length \( a \). The diagonals can be represented as vectors. We can take two diagonals:
- Diagonal 1 (from vertex \( (0, 0, 0) \) to vertex \( (a, a, a) \)):
\[ \mathbf{A} = a \hat{i} + a \hat{j} + a \hat{k} \]
- Diagonal 2 (from vertex \( (0, 0, a) \) to vertex \( (a, a, 0) \)):
\[ \mathbf{B} = a \hat{i} + a \hat{j} - a \hat{k} \]

### Step 2: Calculate the dot product of the diagonals
The dot product \( \mathbf{A} \cdot \mathbf{B} \) is calculated as follows:
\[ \mathbf{A} \cdot \mathbf{B} = (a \hat{i} + a \hat{j} + a \hat{k}) \cdot (a \hat{i} + a \hat{j} - a \hat{k}) \]
This expands to:
\[ = a^2 + a^2 + (-a^2) = 2a^2 \]

### Step 3: Calculate the magnitudes of the diagonals
The magnitude of \( \mathbf{A} \) is:
\[ |\mathbf{A}| = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3} \]
The magnitude of \( \mathbf{B} \) is:
\[ |\mathbf{B}| = \sqrt{a^2 + a^2 + (-a)^2} = \sqrt{3a^2} = a\sqrt{3} \]

### Step 4: Use the dot product to find \( \cos \theta \)
Using the formula for the dot product:
\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta \]
Substituting the values:
\[ 2a^2 = (a\sqrt{3})(a\sqrt{3}) \cos \theta \]
This simplifies to:
\[ 2a^2 = 3a^2 \cos \theta \]
Dividing both sides by \( a^2 \) (assuming \( a \neq 0 \)):
\[ 2 = 3 \cos \theta \]
Thus, we have:
\[ \cos \theta = \frac{2}{3} \]

### Step 5: Find the value of \( \theta \)
To find \( \theta \), we can use the inverse cosine function:
\[ \theta = \cos^{-1}\left(\frac{2}{3}\right) \]

### Conclusion
The acute angle \( \theta \) between the diagonals of the cube is given by \( \cos \theta = \frac{2}{3} \).

### Final Answer
Among the options provided, the correct one is:
**Option C: \( 2 \cos \theta = 1 \)**