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 find the expression \( axxp + bxxq + cxxr \) given that \( a, b, c \) and \( p, q, r \) are reciprocal systems of vectors.

### Step-by-Step Solution:

1. **Understanding Reciprocal Vectors**:
Since \( a, b, c \) and \( p, q, r \) are reciprocal systems of vectors, we can express \( p, q, r \) in terms of \( a, b, c \):
\[ p = \frac{b \times c}{a \cdot (b \times c)}, \quad q = \frac{c \times a}{b \cdot (c \times a)}, \quad r = \frac{a \times b}{c \cdot (a \times b)} \]
Here, \( a \cdot (b \times c) \) is the scalar triple product.

2. **Finding Each Cross Product**:
We need to calculate \( a \times p \), \( b \times q \), and \( c \times r \):
- For \( a \times p \):
\[ a \times p = a \times \left(\frac{b \times c}{a \cdot (b \times c)}\right) = \frac{a \times (b \times c)}{a \cdot (b \times c)} \]
Using the vector triple product identity, \( a \times (b \times c) = (a \cdot c)b - (a \cdot b)c \):
\[ a \times p = \frac{(a \cdot c)b - (a \cdot b)c}{a \cdot (b \times c)} \]

- For \( b \times q \):
\[ b \times q = b \times \left(\frac{c \times a}{b \cdot (c \times a)}\right) = \frac{b \times (c \times a)}{b \cdot (c \times a)} \]
Again using the vector triple product identity:
\[ b \times q = \frac{(b \cdot a)c - (b \cdot c)a}{b \cdot (c \times a)} \]

- For \( c \times r \):
\[ c \times r = c \times \left(\frac{a \times b}{c \cdot (a \times b)}\right) = \frac{c \times (a \times b)}{c \cdot (a \times b)} \]
Using the vector triple product identity:
\[ c \times r = \frac{(c \cdot b)a - (c \cdot a)b}{c \cdot (a \times b)} \]

3. **Adding the Cross Products**:
Now we add \( a \times p + b \times q + c \times r \):
\[ a \times p + b \times q + c \times r = \frac{(a \cdot c)b - (a \cdot b)c}{a \cdot (b \times c)} + \frac{(b \cdot a)c - (b \cdot c)a}{b \cdot (c \times a)} + \frac{(c \cdot b)a - (c \cdot a)b}{c \cdot (a \times b)} \]

After simplifying and combining terms, we notice that all terms will cancel out, leading to:
\[ a \times p + b \times q + c \times r = 0 \]

### Final Answer:
Thus, we conclude that:
\[ axxp + bxxq + cxxr = 0 \]