If `veca, vecb, vecc` and `veca', vecb', vecc'` form a reciprocal system of vectors then `[(veca', vecb', vecc')]=`

Options

A. `[(veca, vecb, vecc)]`
B. `1/([(veca, vecb, vecc)])`
C. `[(veca, vecb, vecc)]^(2)`
D. `(-1)/([(veca, vecb, vecc)])`

Explanation

To solve the problem, we need to find the scalar triple product of the vectors \( \vec{a}', \vec{b}', \vec{c}' \) given that \( \vec{a}, \vec{b}, \vec{c} \) and \( \vec{a}', \vec{b}', \vec{c}' \) form a reciprocal system of vectors.

### Step-by-Step Solution:

1. **Understanding Reciprocal Vectors**:
Since \( \vec{a}', \vec{b}', \vec{c}' \) are the reciprocal vectors of \( \vec{a}, \vec{b}, \vec{c} \), we can express:
\[ \vec{a}' = \frac{\vec{b} \times \vec{c}}{[\vec{a}, \vec{b}, \vec{c}]} \]
\[ \vec{b}' = \frac{\vec{c} \times \vec{a}}{[\vec{a}, \vec{b}, \vec{c}]} \]
\[ \vec{c}' = \frac{\vec{a} \times \vec{b}}{[\vec{a}, \vec{b}, \vec{c}]} \]
where \( [\vec{a}, \vec{b}, \vec{c}] \) is the scalar triple product of \( \vec{a}, \vec{b}, \vec{c} \).

2. **Assuming Scalar Triple Product**:
Let \( [\vec{a}, \vec{b}, \vec{c}] = \frac{1}{\lambda} \). Then we can rewrite the reciprocal vectors as:
\[ \vec{a}' = \lambda (\vec{b} \times \vec{c}), \quad \vec{b}' = \lambda (\vec{c} \times \vec{a}), \quad \vec{c}' = \lambda (\vec{a} \times \vec{b}) \]

3. **Finding the Scalar Triple Product**:
We need to find \( [\vec{a}', \vec{b}', \vec{c}'] \):
\[ [\vec{a}', \vec{b}', \vec{c}'] = \vec{a}' \cdot (\vec{b}' \times \vec{c}') \]

4. **Substituting the Values**:
Substitute \( \vec{b}' \) and \( \vec{c}' \):
\[ [\vec{a}', \vec{b}', \vec{c}'] = \vec{a}' \cdot \left( \lambda (\vec{c} \times \vec{a}) \times \lambda (\vec{a} \times \vec{b}) \right) \]
This simplifies to:
\[ = \lambda^2 \vec{a}' \cdot \left( (\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b}) \right) \]

5. **Using the Vector Triple Product Identity**:
Using the identity \( \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \):
\[ = \lambda^2 \vec{a}' \cdot \left( (\vec{c} \cdot \vec{b}) \vec{a} - (\vec{c} \cdot \vec{a}) \vec{b} \right) \]

6. **Evaluating the Dot Product**:
Since \( \vec{a}' = \lambda (\vec{b} \times \vec{c}) \):
\[ = \lambda^2 \left( \lambda (\vec{b} \times \vec{c}) \cdot \left( (\vec{c} \cdot \vec{b}) \vec{a} - (\vec{c} \cdot \vec{a}) \vec{b} \right) \right) \]
The term \( \vec{b} \times \vec{c} \cdot \vec{a} \) will yield the scalar triple product \( [\vec{a}, \vec{b}, \vec{c}] \).

7. **Final Expression**:
Therefore, we have:
\[ [\vec{a}', \vec{b}', \vec{c}'] = \lambda^2 \cdot [\vec{a}, \vec{b}, \vec{c}] \]
Substituting \( [\vec{a}, \vec{b}, \vec{c}] = \frac{1}{\lambda} \):
\[ [\vec{a}', \vec{b}', \vec{c}'] = \lambda^2 \cdot \frac{1}{\lambda} = \lambda \]

8. **Conclusion**:
Since \( \lambda = [\vec{a}, \vec{b}, \vec{c}]^{-1} \):
\[ [\vec{a}', \vec{b}', \vec{c}'] = \frac{1}{[\vec{a}, \vec{b}, \vec{c}]^2} \]

### Final Answer:
\[ [\vec{a}', \vec{b}', \vec{c}'] = \frac{1}{[\vec{a}, \vec{b}, \vec{c}]^2} \]