Animated Solution for Physics - Kinematics: If $\mathbf{A}\times \mathbf{B}=\mathbf{B}\times \mathbf{A}$, then the angle between $\mathbf{A}$ and $\mathbf{B}$ is

In the world of vector algebra, the cross product is notoriously stubborn. Unlike regular scalar multiplication where $3\times 4$ is exactly the same as $4\times 3$, the cross product is strictly anti-commutative.

This means that the order of multiplication completely flips the outcome. If you swap the vectors, the resulting vector points in the exact opposite direction.

So, when a problem states that $\mathbf{A}\times \mathbf{B}=\mathbf{B}\times \mathbf{A}$, it should immediately trigger an alarm in your head. How can a vector be equal to its own opposite?

To visualize this, imagine holding your right hand out. If you curl your fingers from vector $\mathbf{A}$ towards vector $\mathbf{B}$, your thumb points in the direction of $\mathbf{A}\times \mathbf{B}$.

Now, if you reverse the order and curl your fingers from $\mathbf{B}$ to $\mathbf{A}$, your hand must flip upside down. Your thumb now points in the exact opposite direction, representing $\mathbf{B}\times \mathbf{A}$.

Mathematically, this geometric reality is written as the fundamental property: $\mathbf{B}\times \mathbf{A}=-(\mathbf{A}\times \mathbf{B})$. This negative sign is the mathematical manifestation of the right-hand rule.

Let us bring this property back to the strange condition given in our problem. We are told that $\mathbf{A}\times \mathbf{B}=\mathbf{B}\times \mathbf{A}$.

By substituting our anti-commutative property into the right side of the equation, we get a new relation. The equation transforms into $\mathbf{A}\times \mathbf{B}=-(\mathbf{A}\times \mathbf{B})$.

Now, let us move all the terms to one side of the equals sign. We add $(\mathbf{A}\times \mathbf{B})$ to both sides, resulting in $2(\mathbf{A}\times \mathbf{B})=0$.

Dividing by 2, we arrive at a profound conclusion: $\mathbf{A}\times \mathbf{B}=0$. The cross product must be the zero vector.

What does it physically mean for a cross product to be zero? To answer this, we must look at the magnitude of the cross product.

The magnitude is given by the formula $|\mathbf{A}\times \mathbf{B}|=AB\sin \theta$, where $\theta$ is the angle between the two vectors. Since the cross product is zero, its magnitude must also be zero, giving us $AB\sin \theta =0$.

Assuming that neither vector $\mathbf{A}$ nor vector $\mathbf{B}$ is a zero vector (otherwise the angle would be undefined), the only way this product can be zero is if $\sin \theta =0$.

For the sine of an angle to be zero, the angle $\theta$ must be either $0$ radians or $\pi$ radians.

Geometrically, an angle of $0$ means the vectors are perfectly parallel, pointing in the exact same direction. An angle of $\pi$ means they are anti-parallel, pointing in exactly opposite directions.

In either case, they lie on the same line, and the area of the parallelogram they form is zero. Looking at our given options, $\pi$ is the only matching answer. The mystery of the commuting cross product is solved!