The edges of a parallelopiped are of unit length and a parallel to non-coplanar unit vectors `hata, hatb, hatc` such that `hata.hatb=hatb.hatc=hatc.veca=1//2`. Then the volume of the parallelopiped in cubic units is

Options

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

Explanation

To find the volume of the parallelepiped formed by the unit vectors \(\hat{a}\), \(\hat{b}\), and \(\hat{c}\), we will use the scalar triple product formula. The volume \(V\) of a parallelepiped defined by three vectors \(\vec{u}\), \(\vec{v}\), and \(\vec{w}\) is given by:

\[ V = |\vec{u} \cdot (\vec{v} \times \vec{w})| \]

In this case, the vectors are \(\hat{a}\), \(\hat{b}\), and \(\hat{c}\). The volume can also be expressed in terms of the determinant of the matrix formed by these vectors:

\[ V = |\det(\hat{a}, \hat{b}, \hat{c})| \]

### Step 1: Write the dot products in matrix form
Given the dot products:
- \(\hat{a} \cdot \hat{b} = \frac{1}{2}\)
- \(\hat{b} \cdot \hat{c} = \frac{1}{2}\)
- \(\hat{c} \cdot \hat{a} = \frac{1}{2}\)

We can represent the vectors in a matrix form as follows:

\[ \begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 1 \end{bmatrix} \]

### Step 2: Calculate the determinant of the matrix
To find the volume, we need to calculate the determinant of the above matrix:

\[ D = \begin{vmatrix} 1 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 1 \end{vmatrix} \]

Using the determinant formula for a \(3 \times 3\) matrix:

\[ D = 1 \cdot \begin{vmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{vmatrix} - \frac{1}{2} \cdot \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1 \end{vmatrix} + \frac{1}{2} \cdot \begin{vmatrix} \frac{1}{2} & 1 \\ \frac{1}{2} & \frac{1}{2} \end{vmatrix} \]

Calculating each of these \(2 \times 2\) determinants:

1. \(\begin{vmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{vmatrix} = 1 \cdot 1 - \frac{1}{2} \cdot \frac{1}{2} = 1 - \frac{1}{4} = \frac{3}{4}\)

2. \(\begin{vmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1 \end{vmatrix} = \frac{1}{2} \cdot 1 - \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}\)

3. \(\begin{vmatrix} \frac{1}{2} & 1 \\ \frac{1}{2} & \frac{1}{2} \end{vmatrix} = \frac{1}{2} \cdot \frac{1}{2} - 1 \cdot \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}\)

Substituting back into the determinant expression:

\[ D = 1 \cdot \frac{3}{4} - \frac{1}{2} \cdot \frac{1}{4} + \frac{1}{2} \cdot -\frac{1}{4} \]
\[ D = \frac{3}{4} - \frac{1}{8} - \frac{1}{8} = \frac{3}{4} - \frac{2}{8} = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \]

### Step 3: Calculate the volume
The volume \(V\) is the absolute value of the determinant:

\[ V = |D| = \left|\frac{1}{2}\right| = \frac{1}{2} \]

Thus, the volume of the parallelepiped is \(\frac{1}{2}\) cubic units.