What is the angle between the diagonal of one of the faces of the cube and the diagonal of the cube intersecting the diagonal of the face of the cube?

Options

A. `cos^(-1)(1//sqrt3)`
B. `cos^(-1)(2//sqrt3)`
C. `cos^(-1)(sqrt(2//3))`
D. `cos^(-1)(sqrt2//3)`

Explanation

To find the angle between the diagonal of one of the faces of a cube and the diagonal of the cube intersecting that face diagonal, 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 one of the faces, say face ABCD, can be taken as the diagonal AC.

### Step 2: Calculate the Vector Representations
The vector representation of diagonal AC can be calculated as:
- A to C: \( \vec{AC} = C - A = (1, 1, 0) - (0, 0, 0) = (1, 1, 0) \)

The diagonal of the cube, which connects vertex A to vertex G, can be represented as:
- A to G: \( \vec{AG} = G - A = (1, 1, 1) - (0, 0, 0) = (1, 1, 1) \)

### Step 3: Calculate the Magnitudes of the Vectors
Now, we need to calculate the magnitudes of both vectors:
- Magnitude of \( \vec{AC} \):
\[ |\vec{AC}| = \sqrt{(1)^2 + (1)^2 + (0)^2} = \sqrt{2} \]

- Magnitude of \( \vec{AG} \):
\[ |\vec{AG}| = \sqrt{(1)^2 + (1)^2 + (1)^2} = \sqrt{3} \]

### Step 4: Calculate the Dot Product of the Vectors
Next, we calculate the dot product of the two vectors:
\[ \vec{AC} \cdot \vec{AG} = (1)(1) + (1)(1) + (0)(1) = 1 + 1 + 0 = 2 \]

### Step 5: Use the Dot Product to Find the Cosine of the Angle
Using the dot product formula:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]
We can rearrange to find \( \cos \theta \):
\[ \cos \theta = \frac{\vec{AC} \cdot \vec{AG}}{|\vec{AC}| |\vec{AG}|} = \frac{2}{\sqrt{2} \cdot \sqrt{3}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3} \]

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

### Step 7: Conclusion
Thus, the angle between the diagonal of one of the faces of the cube and the diagonal of the cube intersecting that face diagonal is \( \theta = \cos^{-1} \left( \frac{\sqrt{6}}{3} \right) \).