If `veca,vecb,vecc` and `vecp,vecq,vecr` are reciprocal system of vectors, then `vecaxxvecp+vecbxxvecq+veccxxvecc` equals

Options

A. `[veca vecb vecc]`
B. `vecp + vecq + vecr`
C. `vec0`
D. `veca + vecb + vecc`

Answer

To solve the problem, we need to find the value of the expression \( \vec{a} \times \vec{p} + \vec{b} \times \vec{q} + \vec{c} \times \vec{r} \), given that \( \vec{a}, \vec{b}, \vec{c} \) and \( \vec{p}, \vec{q}, \vec{r} \) are reciprocal systems of vectors.

### Step-by-Step Solution:

1. **Understanding Reciprocal Vectors**:
Since \( \vec{a}, \vec{b}, \vec{c} \) and \( \vec{p}, \vec{q}, \vec{r} \) are reciprocal systems, we can express \( \vec{p}, \vec{q}, \vec{r} \) in terms of \( \vec{a}, \vec{b}, \vec{c} \):
\[ \vec{p} = \frac{\vec{b} \times \vec{c}}{\text{box}(\vec{a}, \vec{b}, \vec{c})} \]
\[ \vec{q} = \frac{\vec{c} \times \vec{a}}{\text{box}(\vec{a}, \vec{b}, \vec{c})} \]
\[ \vec{r} = \frac{\vec{a} \times \vec{b}}{\text{box}(\vec{a}, \vec{b}, \vec{c})} \]

2. **Substituting the Values**:
Now, we substitute these expressions into the original equation:
\[ \vec{a} \times \vec{p} + \vec{b} \times \vec{q} + \vec{c} \times \vec{r} \]
becomes:
\[ \vec{a} \times \left(\frac{\vec{b} \times \vec{c}}{\text{box}(\vec{a}, \vec{b}, \vec{c})}\right) + \vec{b} \times \left(\frac{\vec{c} \times \vec{a}}{\text{box}(\vec{a}, \vec{b}, \vec{c})}\right) + \vec{c} \times \left(\frac{\vec{a} \times \vec{b}}{\text{box}(\vec{a}, \vec{b}, \vec{c})}\right) \]

3. **Factoring Out the Common Denominator**:
We can factor out \( \frac{1}{\text{box}(\vec{a}, \vec{b}, \vec{c})} \):
\[ = \frac{1}{\text{box}(\vec{a}, \vec{b}, \vec{c})} \left( \vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b}) \right) \]

4. **Using the Vector Triple Product Identity**:
We apply the vector triple product identity \( \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \):
- For \( \vec{a} \times (\vec{b} \times \vec{c}) \):
\[ = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]
- For \( \vec{b} \times (\vec{c} \times \vec{a}) \):
\[ = (\vec{b} \cdot \vec{a}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{a} \]
- For \( \vec{c} \times (\vec{a} \times \vec{b}) \):
\[ = (\vec{c} \cdot \vec{b}) \vec{a} - (\vec{c} \cdot \vec{a}) \vec{b} \]

5. **Combining the Terms**:
Combining all these results:
\[ = \left[(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}\right] + \left[(\vec{b} \cdot \vec{a}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{a}\right] + \left[(\vec{c} \cdot \vec{b}) \vec{a} - (\vec{c} \cdot \vec{a}) \vec{b}\right] \]

6. **Simplifying the Expression**:
Notice that the terms cancel out:
- \( -(\vec{a} \cdot \vec{b}) \vec{c} + (\vec{b} \cdot \vec{a}) \vec{c} = 0 \)
- \( -(\vec{b} \cdot \vec{c}) \vec{a} + (\vec{c} \cdot \vec{b}) \vec{a} = 0 \)
- \( -(\vec{c} \cdot \vec{a}) \vec{b} + (\vec{a} \cdot \vec{c}) \vec{b} = 0 \)

Therefore, we conclude:
\[ \vec{a} \times \vec{p} + \vec{b} \times \vec{q} + \vec{c} \times \vec{r} = 0 \]

7. **Final Result**:
Thus, the final answer is:
\[ \vec{a} \times \vec{p} + \vec{b} \times \vec{q} + \vec{c} \times \vec{r} = 0 \]