If a,b,c and d be four vectors, then `(a xxb). (c xx d) + ( b xx c) . (a xx d) + ( c xx a). ( bxxd) = `

Options

A. `a.b + c.d `
B. 0
C. `a.c +b.d`
D. none

Answer

To solve the problem, we need to evaluate the expression:

\[ (a \times b) \cdot (c \times d) + (b \times c) \cdot (a \times d) + (c \times a) \cdot (b \times d) \]

We will use the Lagrange Identity, which states that for any vectors \( \mathbf{u}, \mathbf{v}, \mathbf{w}, \mathbf{x} \):

\[ (\mathbf{u} \times \mathbf{v}) \cdot (\mathbf{w} \times \mathbf{x}) = (\mathbf{u} \cdot \mathbf{w})(\mathbf{v} \cdot \mathbf{x}) - (\mathbf{u} \cdot \mathbf{x})(\mathbf{v} \cdot \mathbf{w}) \]

### Step 1: Apply Lagrange Identity to the First Term
For the first term \( (a \times b) \cdot (c \times d) \):

Using Lagrange Identity:
\[ (a \times b) \cdot (c \times d) = (a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c) \]

### Step 2: Apply Lagrange Identity to the Second Term
For the second term \( (b \times c) \cdot (a \times d) \):

Using Lagrange Identity:
\[ (b \times c) \cdot (a \times d) = (b \cdot a)(c \cdot d) - (b \cdot d)(c \cdot a) \]

### Step 3: Apply Lagrange Identity to the Third Term
For the third term \( (c \times a) \cdot (b \times d) \):

Using Lagrange Identity:
\[ (c \times a) \cdot (b \times d) = (c \cdot b)(a \cdot d) - (c \cdot d)(a \cdot b) \]

### Step 4: Combine All Terms
Now, we combine all three results:

\[ (a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c) + (b \cdot a)(c \cdot d) - (b \cdot d)(c \cdot a) + (c \cdot b)(a \cdot d) - (c \cdot d)(a \cdot b) \]

### Step 5: Rearranging and Simplifying
Rearranging the terms, we have:

1. \( (a \cdot c)(b \cdot d) + (b \cdot a)(c \cdot d) + (c \cdot b)(a \cdot d) \)
2. \( - (a \cdot d)(b \cdot c) - (b \cdot d)(c \cdot a) - (c \cdot d)(a \cdot b) \)

### Step 6: Recognizing Cancellation
Notice that terms can cancel out. Specifically, if we assume that the vectors are such that:

- \( a \cdot b = b \cdot a \)
- \( a \cdot c = c \cdot a \)
- \( b \cdot d = d \cdot b \)

This leads to the conclusion that:

\[ (a \cdot c)(b \cdot d) + (b \cdot a)(c \cdot d) + (c \cdot b)(a \cdot d) = (a \cdot d)(b \cdot c) + (b \cdot d)(c \cdot a) + (c \cdot d)(a \cdot b) \]

Thus, the entire expression equals zero.

### Final Result
Therefore, we conclude that:

\[ (a \times b) \cdot (c \times d) + (b \times c) \cdot (a \times d) + (c \times a) \cdot (b \times d) = 0 \]