Let `veca , vecb` and `vec c ` be non coplanar unit vectors equally inclined to one another at an acute angle `theta` . Then `|[veca vecb vec c]|` in terms of `theta` is equal to

Options

A. `(1 + cos theta) sqrt(cos 2theta)`
B. `(1+ cos theta) sqrt(1 - 2 cos 2theta)`
C. `(1 - cos theta) sqrt(1 + 2 cos theta)`
D. none of these

Answer

To solve the problem, we need to find the value of the scalar triple product \(|[\vec{a}, \vec{b}, \vec{c}]|\) in terms of the acute angle \(\theta\) between the non-coplanar unit vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

### Step-by-Step Solution:

1. **Understanding the Dot Products**:
Since \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are unit vectors and are equally inclined to one another at an angle \(\theta\), we can express the dot products as follows:
\[ \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = \cos \theta \]

2. **Setting Up the Determinant**:
The scalar triple product can be represented as the determinant of a matrix formed by the components of the vectors:
\[ |[\vec{a}, \vec{b}, \vec{c}]| = \begin{vmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{vmatrix} \]

3. **Using the Properties of Determinants**:
We can expand this determinant using the properties of dot products. The determinant can be expressed in terms of the dot products we calculated earlier:
\[ |[\vec{a}, \vec{b}, \vec{c}]|^2 = 1 - (\vec{a} \cdot \vec{b})^2 - (\vec{b} \cdot \vec{c})^2 - (\vec{c} \cdot \vec{a})^2 + 2(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c})(\vec{c} \cdot \vec{a}) \]

4. **Substituting the Values**:
Substituting \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = \cos \theta\) into the equation gives:
\[ |[\vec{a}, \vec{b}, \vec{c}]|^2 = 1 - 3\cos^2 \theta + 2\cos^3 \theta \]

5. **Simplifying the Expression**:
Rearranging the expression, we have:
\[ |[\vec{a}, \vec{b}, \vec{c}]|^2 = 2\cos^3 \theta - 3\cos^2 \theta + 1 \]

6. **Finding the Modulus**:
To find \(|[\vec{a}, \vec{b}, \vec{c}]|\), we take the square root:
\[ |[\vec{a}, \vec{b}, \vec{c}]| = \sqrt{2\cos^3 \theta - 3\cos^2 \theta + 1} \]

7. **Final Expression**:
We can express this in a more usable form:
\[ |[\vec{a}, \vec{b}, \vec{c}]| = (1 - \cos \theta) \sqrt{1 + 2\cos \theta} \]

### Final Answer:
\[ |[\vec{a}, \vec{b}, \vec{c}]| = (1 - \cos \theta) \sqrt{1 + 2\cos \theta} \]