Intuition for the directional relationship in $\vec{A}\times\vec{B}=-\vec{B}\times\vec{A}$

Question

How do we "prove" the directional relationship illustrated in $\vec{A}\times\vec{B}=-\vec{B}\times\vec{A}$? In other words, how do we know that $\vec{A}\times\vec{B}$ always points in the opposite direction as $\vec{B}\times\vec{A}$? We can obviously see that it holds true if we apply the right-hand rule to 2 specific vectors. However, I am wondering if there's an intuition/proof/reasoning that shows why that directional relationship holds true for general cases of $\vec{A}$ & $\vec{B}$.

EDIT: Thanks to all the comments and answers so far, I understand how to prove this mathematically. Now, another thing confuses me - why does the right-hand rule correspond so well to when there is/isn't a negative sign?

I don't know whether you are familiar with this but an elegant way is via the Levi-Civita symbol: $(\vec{A} \times \vec{B})^i = \epsilon_{ijk}A^j B^k$ and $\epsilon_{ijk} = -\epsilon_{ikj}$. Otherwise you could proof it by writing out the components explicitly. — Mathphys meister, May 20 '21 at 16:43
Why the right hand rule? Because we made an arbitrary choice to set up the cross product so that $\mathbf i\times\mathbf j=+\mathbf k$ instead of $-\mathbf k$. — Mark S., May 20 '21 at 17:33
@MarkS. Yes, but I am also wondering why the right-hand rule happens to allow j×i=-k at the same time...? — Claire, May 20 '21 at 17:38
@Cheryl that follows from the answers to this question you just asked: just set $\vec A$ and $\vec B$ to be $\mathbf i$ and $\mathbf j$. — Mark S., May 20 '21 at 17:43
I think you may be interested in Why does the cross product follow the right hand rule?. — Mark S., May 20 '21 at 17:46

score 3 · Answer 1 · answered May 20 '21 at 16:56

If you are not familiar with Levi-Civita symbol, you can use the computation which abuses of the determinant notation. If you set $\vec{A}\times \vec{B}$, then $\vec{B}\times \vec{A}$ is swapping two rows, and therefore, a minus sign shows up.

If you don't like that approach, you must know that $$\vec{D}\times \vec{D} = \vec{0}, \forall \vec{D}$$ And then, apply this result to $\vec{D}= \vec{B} + \vec{A}$:

$$(\vec{B} + \vec{A}) \times (\vec{B} + \vec{A}) = \vec{0}$$ $$\vec{B}\times \vec{B} + \vec{B}\times\vec{A} + \vec{A} \times \vec{B} + \vec{A}\times \vec{A} = \vec{0}$$ $$\vec{A}\times \vec{B} = - \vec{B}\times \vec{A}$$

score 1 · Answer 2 · answered May 20 '21 at 17:16

$\vec{A}\times \vec{B}$ is, in some texts, defined as "The vector with length $|\vec{A}||\vec{B}|sin(\theta)$ ($\theta$ is the angle between $\vec{A}$ and $\vec{B}$) and direction determined by the "right hand rule"- "curve your fingers from the direction of $\vec{A}$ to $\vec{B}$. The direction of the vector product is the direction of your thumb. Swapping $\vec{A}$ and $\vec{B}$ reverses the direction of your fingers so reverses the direction of your thumb.

Other texts define $\vec{A}\times\vec{B}$ as $\left|\begin{array}{cc} \vec{i} & \vec{j} & \vec{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{array}\right|$ and, as said before, swapping two row of a determinant multiplies the determinant by -1.

Intuition for the directional relationship in $\vec{A}\times\vec{B}=-\vec{B}\times\vec{A}$

2 Answers2