If `veca, vecb, vecc` are three non-coplanar mutually perpendicular unit vectors, then `[(veca, vecb, vecc)]` is

Options

A. `+-1`
B. `0`
C. `-2`
D. `2`

Explanation

To solve the problem, we need to find the scalar triple product of three non-coplanar mutually perpendicular unit vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). The scalar triple product is denoted as \([\vec{a}, \vec{b}, \vec{c}]\) and can be calculated using the formula:

\[ [\vec{a}, \vec{b}, \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) \]

### Step-by-Step Solution:

1. **Understanding the Vectors**:
- We have three unit vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).
- They are mutually perpendicular, which means:
\[ \vec{a} \cdot \vec{b} = 0, \quad \vec{b} \cdot \vec{c} = 0, \quad \vec{c} \cdot \vec{a} = 0 \]
- Since they are unit vectors, their magnitudes are:
\[ |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \]

2. **Calculating the Cross Product**:
- The cross product \(\vec{b} \times \vec{c}\) gives a vector that is perpendicular to both \(\vec{b}\) and \(\vec{c}\).
- The magnitude of the cross product can be calculated as:
\[ |\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin(\theta) \]
- Here, \(\theta\) is the angle between \(\vec{b}\) and \(\vec{c}\). Since they are perpendicular, \(\theta = 90^\circ\) and \(\sin(90^\circ) = 1\):
\[ |\vec{b} \times \vec{c}| = 1 \cdot 1 \cdot 1 = 1 \]

3. **Finding the Dot Product**:
- Now, we need to calculate \(\vec{a} \cdot (\vec{b} \times \vec{c})\).
- Since \(\vec{a}\) is also perpendicular to both \(\vec{b}\) and \(\vec{c}\), the angle between \(\vec{a}\) and \(\vec{b} \times \vec{c}\) is either \(0^\circ\) or \(180^\circ\) (i.e., they are parallel or anti-parallel).
- Therefore, we can write:
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = |\vec{a}| |\vec{b} \times \vec{c}| \cos(\phi) \]
- Here, \(\phi\) is the angle between \(\vec{a}\) and \(\vec{b} \times \vec{c}\). Since both vectors are unit vectors, we have:
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = 1 \cdot 1 \cdot \cos(\phi) = \cos(\phi) \]
- The value of \(\cos(\phi)\) can be either \(1\) or \(-1\), leading to:
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = \pm 1 \]

4. **Conclusion**:
- Therefore, the scalar triple product \([\vec{a}, \vec{b}, \vec{c}]\) is:
\[ [\vec{a}, \vec{b}, \vec{c}] = \pm 1 \]

### Final Answer:
The scalar triple product \([\vec{a}, \vec{b}, \vec{c}]\) is \(\pm 1\).