If `veca, vecb, vecc` and `veca', vecb', vecc'` form a reciprocal system of vectors then
`veca.veca'+vecb.vecb'+vecc.vecc'=`

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem, we need to show that if \(\vec{a}, \vec{b}, \vec{c}\) and \(\vec{a}', \vec{b}', \vec{c}'\) form a reciprocal system of vectors, then:

\[ \vec{a} \cdot \vec{a}' + \vec{b} \cdot \vec{b}' + \vec{c} \cdot \vec{c}' = 3 \]

### Step-by-Step Solution:

1. **Understanding Reciprocal Vectors**:
- The vectors \(\vec{a}', \vec{b}', \vec{c}'\) are defined as the reciprocals of \(\vec{a}, \vec{b}, \vec{c}\) respectively. This means:
\[ \vec{a}' = \frac{\vec{b} \times \vec{c}}{\vec{a} \cdot (\vec{b} \times \vec{c})} \]
\[ \vec{b}' = \frac{\vec{c} \times \vec{a}}{\vec{b} \cdot (\vec{c} \times \vec{a})} \]
\[ \vec{c}' = \frac{\vec{a} \times \vec{b}}{\vec{c} \cdot (\vec{a} \times \vec{b})} \]

2. **Calculating \(\vec{a} \cdot \vec{a}'\)**:
- Substitute \(\vec{a}'\):
\[ \vec{a} \cdot \vec{a}' = \vec{a} \cdot \left(\frac{\vec{b} \times \vec{c}}{\vec{a} \cdot (\vec{b} \times \vec{c})}\right) \]
- Using the property of the scalar triple product, we know:
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = \text{Volume of the parallelepiped} = \vec{a} \cdot \vec{b} \times \vec{c} \]
- Therefore:
\[ \vec{a} \cdot \vec{a}' = \frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{\vec{a} \cdot (\vec{b} \times \vec{c})} = 1 \]

3. **Calculating \(\vec{b} \cdot \vec{b}'\)**:
- Substitute \(\vec{b}'\):
\[ \vec{b} \cdot \vec{b}' = \vec{b} \cdot \left(\frac{\vec{c} \times \vec{a}}{\vec{b} \cdot (\vec{c} \times \vec{a})}\right) \]
- Again, using the scalar triple product:
\[ \vec{b} \cdot (\vec{c} \times \vec{a}) = 1 \]
- Thus:
\[ \vec{b} \cdot \vec{b}' = 1 \]

4. **Calculating \(\vec{c} \cdot \vec{c}'\)**:
- Substitute \(\vec{c}'\):
\[ \vec{c} \cdot \vec{c}' = \vec{c} \cdot \left(\frac{\vec{a} \times \vec{b}}{\vec{c} \cdot (\vec{a} \times \vec{b})}\right) \]
- Using the scalar triple product:
\[ \vec{c} \cdot (\vec{a} \times \vec{b}) = 1 \]
- Therefore:
\[ \vec{c} \cdot \vec{c}' = 1 \]

5. **Final Calculation**:
- Now, we can sum up all the results:
\[ \vec{a} \cdot \vec{a}' + \vec{b} \cdot \vec{b}' + \vec{c} \cdot \vec{c}' = 1 + 1 + 1 = 3 \]

### Conclusion:
Thus, we conclude that:

\[ \vec{a} \cdot \vec{a}' + \vec{b} \cdot \vec{b}' + \vec{c} \cdot \vec{c}' = 3 \]