The cosine of the angle between any two diagonals of a cube is

Options

A. `1//3`
B. `1//2`
C. `2//3`
D. `1//sqrt(3)`

Answer

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

### Step 1: Define the Cube and its Vertices
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, a, a)
- E (0, 0, a)
- F (a, 0, a)
- G (a, a, a)

### Step 2: Identify Two Diagonals
We will consider the diagonals O to G and A to D. The coordinates for these points are:
- O (0, 0, 0) to G (a, a, a)
- A (a, 0, 0) to D (0, a, a)

### Step 3: Find the Vectors for the Diagonals
The vector for diagonal OG can be calculated as:
\[ \vec{OG} = G - O = (a, a, a) - (0, 0, 0) = (a, a, a) \]

The vector for diagonal AD can be calculated as:
\[ \vec{AD} = D - A = (0, a, a) - (a, 0, 0) = (-a, a, a) \]

### Step 4: Calculate the Dot Product of the Vectors
The dot product of vectors \(\vec{OG}\) and \(\vec{AD}\) is given by:
\[ \vec{OG} \cdot \vec{AD} = (a, a, a) \cdot (-a, a, a) = a(-a) + a(a) + a(a) = -a^2 + a^2 + a^2 = a^2 \]

### Step 5: Calculate the Magnitudes of the Vectors
The magnitude of vector \(\vec{OG}\) is:
\[ |\vec{OG}| = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3} \]

The magnitude of vector \(\vec{AD}\) is:
\[ |\vec{AD}| = \sqrt{(-a)^2 + a^2 + a^2} = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3} \]

### Step 6: Use the Dot Product to Find the Cosine of the Angle
Using the formula for the cosine of the angle \(\theta\) between two vectors:
\[ \cos(\theta) = \frac{\vec{OG} \cdot \vec{AD}}{|\vec{OG}| |\vec{AD}|} \]

Substituting the values we calculated:
\[ \cos(\theta) = \frac{a^2}{(a\sqrt{3})(a\sqrt{3})} = \frac{a^2}{3a^2} = \frac{1}{3} \]

### Conclusion
The cosine of the angle between any two diagonals of a cube is:
\[ \cos(\theta) = \frac{1}{3} \]