Cross product of two paralle or antiparallel vectors is a null vector. Is this statement true or false?

Answer

To determine whether the statement "the cross product of two parallel or antiparallel vectors is a null vector" is true or false, we can analyze the properties of the cross product.

### Step-by-Step Solution:

1. **Understanding the Cross Product**:
The cross product of two vectors **A** and **B** is defined mathematically as:
\[ \mathbf{A} \times \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \sin(\theta) \, \mathbf{n} \]
where:
- \(|\mathbf{A}|\) and \(|\mathbf{B}|\) are the magnitudes of vectors **A** and **B**,
- \(\theta\) is the angle between the two vectors,
- \(\mathbf{n}\) is the unit vector perpendicular to the plane formed by **A** and **B**.

2. **Analyzing Parallel Vectors**:
When two vectors are parallel, the angle \(\theta\) between them is \(0^\circ\). Therefore, we can substitute this value into the cross product formula:
\[ \mathbf{A} \times \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \sin(0^\circ) \, \mathbf{n} \]
Since \(\sin(0^\circ) = 0\), we have:
\[ \mathbf{A} \times \mathbf{B} = 0 \, \mathbf{n} = \mathbf{0} \]
This means the cross product is a null vector.

3. **Analyzing Antiparallel Vectors**:
For antiparallel vectors, the angle \(\theta\) between them is \(180^\circ\). We again substitute this value into the cross product formula:
\[ \mathbf{A} \times \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \sin(180^\circ) \, \mathbf{n} \]
Since \(\sin(180^\circ) = 0\), we have:
\[ \mathbf{A} \times \mathbf{B} = 0 \, \mathbf{n} = \mathbf{0} \]
Thus, the cross product is also a null vector in this case.

4. **Conclusion**:
Since both for parallel and antiparallel vectors the cross product results in a null vector, we conclude that the statement is **true**.

### Final Answer:
The statement "the cross product of two parallel or antiparallel vectors is a null vector" is **true**.