The vertices of a cube are `(0,0,0),(2,0,0),(0,0,2),(0,2,0),(2,0,2),(0,2,2),(2,2,0),(2,2,2) `respectively
What is the angle between any two diagonals of the cube?

Options

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

Answer

To find the angle between any two diagonals of a cube, we can follow these steps:

### Step 1: Identify the diagonals of the cube
The vertices of the cube are given as:
- \( A(0, 0, 0) \)
- \( B(2, 0, 0) \)
- \( C(0, 0, 2) \)
- \( D(0, 2, 0) \)
- \( E(2, 0, 2) \)
- \( F(0, 2, 2) \)
- \( G(2, 2, 0) \)
- \( H(2, 2, 2) \)

We can choose two diagonals:
1. Diagonal 1: From \( A(0, 0, 0) \) to \( H(2, 2, 2) \)
2. Diagonal 2: From \( B(2, 0, 0) \) to \( F(0, 2, 2) \)

### Step 2: Find the direction vectors of the diagonals
For Diagonal 1:
- The vector \( \mathbf{u} = H - A = (2 - 0, 2 - 0, 2 - 0) = (2, 2, 2) \)

For Diagonal 2:
- The vector \( \mathbf{v} = F - B = (0 - 2, 2 - 0, 2 - 0) = (-2, 2, 2) \)

### Step 3: Calculate the dot product of the two vectors
The dot product \( \mathbf{u} \cdot \mathbf{v} \) is calculated as follows:
\[ \mathbf{u} \cdot \mathbf{v} = (2)(-2) + (2)(2) + (2)(2) = -4 + 4 + 4 = 4 \]

### Step 4: Find the magnitudes of the vectors
The magnitude of \( \mathbf{u} \):
\[ |\mathbf{u}| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} \]

The magnitude of \( \mathbf{v} \):
\[ |\mathbf{v}| = \sqrt{(-2)^2 + 2^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} \]

### Step 5: Use the formula to find \( \cos \theta \)
Using the formula:
\[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} \]
Substituting the values:
\[ \cos \theta = \frac{4}{(2\sqrt{3})(2\sqrt{3})} = \frac{4}{12} = \frac{1}{3} \]

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

### Final Answer
The angle between any two diagonals of the cube is:
\[ \theta = \cos^{-1}\left(\frac{1}{3}\right) \]