The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors `hat(a), hat(b), hat(c)` such that `hat(a)*hat(b)=hat(b)*hat(c)=hat(c)*hat(a)=(1)/(2).` Then, the volume of the parallelopiped is

Options

A. a) `(1)/(sqrt(2))` cu units
B. b) `(1)/(2sqrt(2))` cu units
C. c) `(sqrt(3))/(2)` cu units
D. d) `(1)/(sqrt(3))` cu units

Answer

To find the volume of the parallelepiped formed by the unit vectors \(\hat{a}\), \(\hat{b}\), and \(\hat{c}\) with the given conditions, we can follow these steps:

### Step 1: Understanding the Given Information
We are given that:
- \(\hat{a}\), \(\hat{b}\), and \(\hat{c}\) are unit vectors.
- The dot products between the vectors are:
\[ \hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = \frac{1}{2} \]

### Step 2: Volume of the Parallelepiped
The volume \(V\) of the parallelepiped formed by the vectors \(\hat{a}\), \(\hat{b}\), and \(\hat{c}\) can be calculated using the scalar triple product:
\[ V = |\hat{a} \cdot (\hat{b} \times \hat{c})| \]
Alternatively, the volume can also be expressed as the square root of the determinant of the matrix formed by the dot products of the vectors:
\[ V^2 = \begin{vmatrix} \hat{a} \cdot \hat{a} & \hat{a} \cdot \hat{b} & \hat{a} \cdot \hat{c} \\ \hat{b} \cdot \hat{a} & \hat{b} \cdot \hat{b} & \hat{b} \cdot \hat{c} \\ \hat{c} \cdot \hat{a} & \hat{c} \cdot \hat{b} & \hat{c} \cdot \hat{c} \end{vmatrix} \]

### Step 3: Construct the Matrix
Since \(\hat{a}\), \(\hat{b}\), and \(\hat{c}\) are unit vectors, we have:
\[ \hat{a} \cdot \hat{a} = 1, \quad \hat{b} \cdot \hat{b} = 1, \quad \hat{c} \cdot \hat{c} = 1 \]
And the dot products are:
\[ \hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = \frac{1}{2} \]
Thus, the matrix becomes:
\[ \begin{vmatrix} 1 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 1 \end{vmatrix} \]

### Step 4: Calculate the Determinant
Now we calculate the determinant:
\[ \text{Det} = 1 \left( 1 \cdot 1 - \frac{1}{2} \cdot \frac{1}{2} \right) - \frac{1}{2} \left( \frac{1}{2} \cdot 1 - \frac{1}{2} \cdot \frac{1}{2} \right) + \frac{1}{2} \left( \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} \cdot 1 \right) \]
Calculating each term:
1. First term: \(1 \cdot (1 - \frac{1}{4}) = 1 \cdot \frac{3}{4} = \frac{3}{4}\)
2. Second term: \(-\frac{1}{2} \left( \frac{1}{2} - \frac{1}{4} \right) = -\frac{1}{2} \cdot \frac{1}{4} = -\frac{1}{8}\)
3. Third term: \(\frac{1}{2} \left( \frac{1}{4} - \frac{1}{2} \right) = \frac{1}{2} \cdot -\frac{1}{4} = -\frac{1}{8}\)

Combining these results:
\[ \text{Det} = \frac{3}{4} - \frac{1}{8} - \frac{1}{8} = \frac{3}{4} - \frac{2}{8} = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \]

### Step 5: Volume Calculation
Thus, we have:
\[ V^2 = \frac{1}{2} \implies V = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

### Final Answer
Therefore, the volume of the parallelepiped is:
\[ \boxed{\frac{1}{\sqrt{2}} \text{ cubic units}} \]