The angle between diagonal of a cube and diagonal
of a face of the cube will be

Options

A. `cos^(-1)sqrt(2/3)`
B. `sin^(-1)sqrt(2/3)`
C. `sin^(-1)(2/3)`
D. `cos^(-1)(1/sqrt(3))`

Answer

To find the angle between the diagonal of a cube and the diagonal of a face of the cube, we can follow these steps:

### Step 1: Define the cube and its diagonals
Let's consider a cube with vertices at the following coordinates:
- A(0, 0, 0)
- B(1, 0, 0)
- C(1, 1, 0)
- D(0, 1, 0)
- E(0, 0, 1)
- F(1, 0, 1)
- G(1, 1, 1)
- H(0, 1, 1)

The diagonal of the cube can be represented by the vector from A(0, 0, 0) to G(1, 1, 1). This vector is:
\[ \mathbf{u} = (1-0, 1-0, 1-0) = (1, 1, 1) \]

The diagonal of a face (let's take the face ABCD) can be represented by the vector from A(0, 0, 0) to C(1, 1, 0). This vector is:
\[ \mathbf{v} = (1-0, 1-0, 0-0) = (1, 1, 0) \]

### Step 2: Use the cosine formula to find the angle
To find the angle \( \theta \) between the two vectors \( \mathbf{u} \) and \( \mathbf{v} \), we can use the formula:
\[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} \]

### Step 3: Calculate the dot product \( \mathbf{u} \cdot \mathbf{v} \)
The dot product is calculated as follows:
\[ \mathbf{u} \cdot \mathbf{v} = (1)(1) + (1)(1) + (1)(0) = 1 + 1 + 0 = 2 \]

### Step 4: Calculate the magnitudes of \( \mathbf{u} \) and \( \mathbf{v} \)
The magnitude of \( \mathbf{u} \) is:
\[ |\mathbf{u}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]
The magnitude of \( \mathbf{v} \) is:
\[ |\mathbf{v}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2} \]

### Step 5: Substitute values into the cosine formula
Now substituting the values into the cosine formula:
\[ \cos \theta = \frac{2}{\sqrt{3} \cdot \sqrt{2}} = \frac{2}{\sqrt{6}} \]

### Step 6: Rationalize the denominator
To rationalize the denominator:
\[ \cos \theta = \frac{2}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3} \]

### Step 7: Find the angle
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{\sqrt{6}}{3}\right) \]

### Conclusion
The angle between the diagonal of the cube and the diagonal of a face of the cube is:
\[ \theta = \cos^{-1}\left(\frac{\sqrt{6}}{3}\right) \]