If the vectors `veca, vecb, vecc` and `vecd` are coplanar vectors, then `(vecaxxvecb)xx(veccxxvecd)` is equal to

Options

A. `veca+vecb+vecc+vecd`
B. `vec0`
C. `veca+vecb=vecc+vecd`
D. none of these

Explanation

To solve the problem, we need to evaluate the expression \((\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d})\) given that the vectors \(\vec{a}, \vec{b}, \vec{c},\) and \(\vec{d}\) are coplanar.

### Step-by-Step Solution:

1. **Understand the Coplanarity Condition**:
- Since the vectors \(\vec{a}, \vec{b}, \vec{c},\) and \(\vec{d}\) are coplanar, any scalar triple product involving these vectors will be zero. This means:
\[ [\vec{a}, \vec{b}, \vec{c}] = 0 \quad \text{and} \quad [\vec{a}, \vec{b}, \vec{d}] = 0 \]
where \([\vec{x}, \vec{y}, \vec{z}]\) denotes the scalar triple product of vectors \(\vec{x}, \vec{y},\) and \(\vec{z}\).

2. **Apply the Vector Triple Product Identity**:
- We use the vector triple product identity:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \]
- In our case, let \(\vec{x} = \vec{a} \times \vec{b}\), \(\vec{y} = \vec{c}\), and \(\vec{z} = \vec{d}\):
\[ (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = (\vec{a} \times \vec{b} \cdot \vec{d}) \vec{c} - (\vec{a} \times \vec{b} \cdot \vec{c}) \vec{d} \]

3. **Evaluate the Scalar Products**:
- Since \(\vec{a}, \vec{b}, \vec{c},\) and \(\vec{d}\) are coplanar, we know:
\[ \vec{a} \times \vec{b} \cdot \vec{c} = 0 \quad \text{and} \quad \vec{a} \times \vec{b} \cdot \vec{d} = 0 \]
- Therefore, substituting these values into the expression gives:
\[ (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = 0 \cdot \vec{c} - 0 \cdot \vec{d} = \vec{0} \]

4. **Conclusion**:
- The final result is that the expression evaluates to the null vector:
\[ (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0} \]

### Final Answer:
\[ (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0} \]