Thanks to this wonderful article I have been able to prove that the cross product follows the right hand rule when the cross product has a non-zero z component. In this case we show that the z component is the same for the vectors and their projections onto the xy plane, and that this component is positive for positivily oriented vectors and negative for negatively oriented ones.

The article (in less detail) describes how to deal with cases where the vectors $u$ and $v$ have the same ratio of $y$ to $x$. In this case the z component of the cross product would be equal to zero, and so we project the vectors onto the zy-plane instead, and calculate the x-component of the cross product in the same manner as above.

There is however one more case where the cross product has a z-component equal to zero - when the vectors $u$ and $v$ lack an y-component, and span the $xz$ plane. In this case the cross product will be limited to only a y component.

This time we won't have to project anything onto the $xz$ plane (the vectors are already there).

I try to apply the same reasoning as the article. If $u=(a_1, b_1, c_1)$ and $v=(a_2, b_2, c_2)$ then this would mean that

$v \times u=\begin{bmatrix}0c_2-c_1 0\\-(a_1 c_2-c_1 a_2)\\a_1 0-0a_2 \end{bmatrix}$

If we measure our angles from the positive x axis, and if we let $\alpha$ denote the angle for $u$ and $\beta$ the angle for $v$, we should be able to write this as:

$$c_1a_2-a_1c_2=|u|sin\alpha |v|cos \beta-|u|cos \alpha |v|sin \beta$$

According to the sine subtraction identity, we can write this as:

$$|u|sin\alpha |v|cos \beta-|u|cos \alpha |v|sin \beta = |u||v|sin(\alpha-\beta)$$

This is where things get confusing. In the other two cases we had to make the component of the cross product equal to $sin (\beta-\alpha)$, the minus sign switches things around I guess.

If $(u,v)$ are positively oriented there are two options, either $\alpha$ is smaller than $\beta$ (figure to the left) or vice versa (figure to the right). In both cases the $v$ must be less than 180 degrees conterclockwise from $u$.

enter image description here

(picture from article, all credit to Eric Thurschwell)

In the first case, $\alpha-\beta$ would give us the negative difference between $\beta$ and $\alpha$, which by definition must be less than 180 degrees. The sine of something a value between 0 and -180 must be negative, which gives us a negative y component.

In the second case, $\alpha-\beta$ would give us the positive difference between $\alpha$ and $\beta$, but since the vectors are positively oriented, it means this value must be greater that 180 degrees (and smaller than 360 degrees) by definition, making the sine of that value negative.

...Which would mean the cross product doesn't follow the right hand rule at all, or at least not for these kinds of vectors.

Would someone mind helping me out? I've been working on this for two weeks. This is the last thing I need to straighten out before I can move on to other topics...


1 Answer 1


To start with, you have transcribed the rule incorrectly.

In your source, the formula is given this way: if $\mathbf a = \langle a_1,a_2,a_3\rangle$ and $\mathbf b = \langle b_1,b_2,b_3\rangle$, their cross product is denoted $\mathbf a\times \mathbf b$ and defined analytically as the vector $\langle a_2b_3 − a_3b_2, a_3b_1 − a_1b_3, a_1b_2 − a_2b_1\rangle.$

Transcribing this into your notation, that is, $\mathbf a \to v,$ $\mathbf b \to u,$ $a_1 \to a_2,$ $a_2 \to b_2,$ $a_3 \to c_2,$ $b_1 \to a_1,$ $b_2 \to b_1,$ and $b_3 \to c_1$:

If $v=(a_2, b_2, c_2)$ and $u=(a_1, b_1, c_1),$ then $$ v \times u = \begin{bmatrix} b_2 c_1 - c_2 b_1 \\ c_2 a_1 - a_2 c_1 \\ a_2 b_1 - b_2 a_1 \end{bmatrix} . $$

You set the second coordinate to $a_2 c_1 - c_2 a_1,$ opposite the correct sign. On the other hand I'm not sure you actually meant to write $v \times u$; if you change the left-hand side of your equation to $u \times v$ then the right-hand side you wrote will be correct as it is.

The next thing that is problematic in your analysis is that if you look "down" at the $xz$ plane from a place far along the positive $y$ axis, the positive $z$ axis should be on the right and the positive $x$ axis on the left, that is, what you see should be this:

enter image description here

The convention I would expect to follow is that angles are measured counterclockwise from the $z$ axis in this view, as the figure above illustrates for the vector $u.$

By saying that $c_1 = \lvert u\vert \sin \alpha,$ in effect you have decided to measure your angles clockwise from the positive $x$ axis. Reversing rotation direction from counterclockwise to clockwise is equivalent to flipping the direction of the normal vector. Note that $\langle 1,0,0\rangle \times \langle 0,0,1\rangle = \langle 0,-1,0\rangle,$ that is, $\hat x \times \hat z = -\hat y.$

There are so many inadvertent sign changes in your analysis that I find it hard to count them. But there are apparently an odd number of them, because they don't all cancel out to give you the expected answer.

Now for a personal opinion about the linked article, https://www.researchgate.net/publication/305759205_An_Even_Simpler_Proof_of_the_Right-Hand_Rule. I find this article unsatisfying. Setting aside the failure to give any guidance about handling two vectors in the $x,z$ plane, where is the argument that projecting both vectors onto a coordinate plane will preserve the right-hand rule? Such an argument possibly could be made, but it would make this "proof" not nearly as simple as it claims to be.

If you want a proof that is as simple as the one claimed in the linked paper, I think you will have to accept as given that the cross-product formula is correct up to possible reversals of the signs of the three resulting coordinates. That is, we assume that the magnitude of each coordinate of the cross-product is correct in the given formula, and we merely need to show that the sign is correct for the right-hand rule rather than for the left-hand rule.

In order to show this for the $y$ axis, it is sufficient to take any two vectors such that these two vectors and the $y$ axis do not all lie in one plane and show that the resulting $y$ coordinate has the correct sign. In particular, we can take the vectors $\hat x = \langle 1,0,0\rangle$ and $\hat z = \langle 0,0,1\rangle.$ The right-hand rule requires that $\hat z \times \hat x = \hat y.$ We see that the formula satisfies this requirement, so it is correct for the $y$ component of the cross product.

  • $\begingroup$ Thanks for pointing out my incorrect axes, you're right in that I meant to write $u \times v$. This proof actually works intuitively for me and to me that's the most important part. By the time I can see the flaw in the argument I will hopefully be able to work through more complicated proofs. $\endgroup$
    – Magnus
    Mar 8, 2021 at 7:01
  • $\begingroup$ I did try to work through the proof by Gao but it's quite densly written and essentially I hit a brick wall quite early in the process math.stackexchange.com/questions/4044199/… $\endgroup$
    – Magnus
    Mar 8, 2021 at 7:05

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