Why is the product of two vectors not commutative?
The product of two vectors can be defined in two ways: the scalar product (dot product) and the vector product (cross product).
$1$. The scalar product is defined as $\vec{A} \cdot \vec{B} = AB \cos \theta$. Since $\cos \theta = \cos(-\theta)$,the scalar product is commutative,i.e.,$\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$.
$2$. The vector product is defined as $\vec{A} \times \vec{B} = AB \sin \theta \hat{n}$,where $\hat{n}$ is a unit vector perpendicular to the plane containing $\vec{A}$ and $\vec{B}$ determined by the right-hand rule.
$3$. According to the right-hand rule,the direction of $\vec{A} \times \vec{B}$ is opposite to the direction of $\vec{B} \times \vec{A}$.
$4$. Therefore,$\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$.
$5$. Because the direction changes,the vector product is anti-commutative,not commutative.
$1$. The scalar product is defined as $\vec{A} \cdot \vec{B} = AB \cos \theta$. Since $\cos \theta = \cos(-\theta)$,the scalar product is commutative,i.e.,$\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$.
$2$. The vector product is defined as $\vec{A} \times \vec{B} = AB \sin \theta \hat{n}$,where $\hat{n}$ is a unit vector perpendicular to the plane containing $\vec{A}$ and $\vec{B}$ determined by the right-hand rule.
$3$. According to the right-hand rule,the direction of $\vec{A} \times \vec{B}$ is opposite to the direction of $\vec{B} \times \vec{A}$.
$4$. Therefore,$\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$.
$5$. Because the direction changes,the vector product is anti-commutative,not commutative.
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