For two particular vectors `vec A and vec B` it is known that `vec A xx vec B =vec Bxx vec A`. What must be true about the two vectors?

Options

A. Atleast one of the two vectors must be the zero vector
B. `AtimesB=BtimesA` is true for any two vectors
C. One of the two vectors is a scalar multiple of the other vector
D. The two vectors must be perpendicular to each other

Answer

To solve the problem, we need to analyze the given condition that the cross product of two vectors \( \vec{A} \) and \( \vec{B} \) is equal to the cross product of \( \vec{B} \) and \( \vec{A} \).

### Step-by-step Solution:

1. **Understanding the Cross Product**:
The cross product of two vectors \( \vec{A} \) and \( \vec{B} \) is defined as:
\[ \vec{A} \times \vec{B} = |\vec{A}| |\vec{B}| \sin(\theta) \hat{n} \]
where \( \theta \) is the angle between the two vectors and \( \hat{n} \) is the unit vector perpendicular to the plane containing \( \vec{A} \) and \( \vec{B} \).

2. **Using the Property of Cross Products**:
We know that:
\[ \vec{A} \times \vec{B} = -(\vec{B} \times \vec{A}) \]
This means that the cross product is anti-commutative.

3. **Setting Up the Equation**:
Given the condition:
\[ \vec{A} \times \vec{B} = \vec{B} \times \vec{A} \]
We can substitute the anti-commutative property:
\[ \vec{A} \times \vec{B} = -(\vec{A} \times \vec{B}) \]

4. **Simplifying the Equation**:
This leads to:
\[ 2(\vec{A} \times \vec{B}) = 0 \]
Therefore, we have:
\[ \vec{A} \times \vec{B} = 0 \]

5. **Conclusion about the Vectors**:
The equation \( \vec{A} \times \vec{B} = 0 \) implies that the vectors \( \vec{A} \) and \( \vec{B} \) are either parallel or anti-parallel. This means:
\[ \vec{A} = k \vec{B} \]
where \( k \) is some scalar (which can be positive or negative).

### Final Answer:
The two vectors \( \vec{A} \) and \( \vec{B} \) must be parallel or anti-parallel.

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