If a,b,c and p,q,r are reciprocal systemm of vectors, then `axxp+bxxq+cxxr` is equal to

Options

A. `[abc]`
B. `[p+q+r]`
C. `0`
D. a+b+c

Answer

To solve the problem, we need to show that \( ax \times p + b \times q + c \times r = 0 \) given that \( a, b, c \) and \( p, q, r \) are reciprocal systems of vectors.

### Step-by-step Solution:

1. **Understanding Reciprocal Systems**:
- Vectors \( p, q, r \) are defined as reciprocal to vectors \( a, b, c \). This means that the scalar triple product of \( a, b, c \) with \( p, q, r \) is equal to 1:
\[ a \cdot (b \times c) = 1, \quad b \cdot (c \times a) = 1, \quad c \cdot (a \times b) = 1 \]

2. **Expressing \( p, q, r \)**:
- We can express \( p, q, r \) in terms of the vectors \( a, b, c \):
\[ p = \frac{b \times c}{\text{det}(a, b, c)}, \quad q = \frac{c \times a}{\text{det}(a, b, c)}, \quad r = \frac{a \times b}{\text{det}(a, b, c)} \]
- Here, \(\text{det}(a, b, c)\) is the scalar triple product \( a \cdot (b \times c) \).

3. **Substituting \( p, q, r \) into the expression**:
- Now substituting \( p, q, r \) into the expression \( ax \times p + b \times q + c \times r \):
\[ ax \times p = a \times \left(\frac{b \times c}{\text{det}(a, b, c)}\right) = \frac{a \times (b \times c)}{\text{det}(a, b, c)} \]
\[ b \times q = b \times \left(\frac{c \times a}{\text{det}(a, b, c)}\right) = \frac{b \times (c \times a)}{\text{det}(a, b, c)} \]
\[ c \times r = c \times \left(\frac{a \times b}{\text{det}(a, b, c)}\right) = \frac{c \times (a \times b)}{\text{det}(a, b, c)} \]

4. **Combining the terms**:
- Now, we combine all these terms:
\[ ax \times p + b \times q + c \times r = \frac{1}{\text{det}(a, b, c)} \left( a \times (b \times c) + b \times (c \times a) + c \times (a \times b) \right) \]

5. **Using the vector triple product identity**:
- Using the vector triple product identity \( x \times (y \times z) = (x \cdot z)y - (x \cdot y)z \):
- We can simplify each term:
\[ a \times (b \times c) = (a \cdot c)b - (a \cdot b)c \]
\[ b \times (c \times a) = (b \cdot a)c - (b \cdot c)a \]
\[ c \times (a \times b) = (c \cdot b)a - (c \cdot a)b \]

6. **Summing these results**:
- When we sum these results, we notice that each term cancels out:
\[ (a \cdot c)b - (a \cdot b)c + (b \cdot a)c - (b \cdot c)a + (c \cdot b)a - (c \cdot a)b = 0 \]

7. **Final Result**:
- Thus, we conclude that:
\[ ax \times p + b \times q + c \times r = 0 \]

### Final Answer:
\[ ax \times p + b \times q + c \times r = 0 \]