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Let $\mathbf{a}=\left(a_{1},a_{2},a_{3}\right)$ and $\mathbf{b}=\left(b_{1},b_{2},b_{3}\right)$ be vectors in $\mathbb{R}^3$. Then the only two distinct unit vectors that are perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ are those that point in the directions of:

$$\mathbf{u}=\left(\begin{array}{c} a_{2}b_{3}-a_{3}b_{2}\\ a_{3}b_{1}-a_{1}b_{3}\\ a_{1}b_{2}-a_{2}b_{1} \end{array}\right)$$

and

$$\mathbf{v}=\left(\begin{array}{c} a_{3}b_{2}-a_{2}b_{3}\\ a_{1}b_{3}-a_{3}b_{1}\\ a_{2}b_{1}-a_{1}b_{2} \end{array}\right)$$

where of course $\mathbf{u}=-\mathbf{v}$.

How do we prove that $\mathbf{u}$ satisfies the right-hand rule, while $\mathbf{v}$ satisfies the left-hand rule?


(A similar question has been asked several times before: e.g. 1, 2, 3, 4, 5. I have looked at all the answers and can't find a single one that I can understand and that I'm satisfied explains it. Hence, I'm posting this question again, worded slightly differently.

In particular, I've deliberately omitted any reference to the fact that by convention, we choose $\mathbf{u}$ and not $\mathbf{v}$ as the vector product $\mathbf{a}\times\mathbf{b}$. This fact seems to be the most popular "answer" but does not actually answer the above question.)

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For a definite answer, you need to define what a "right hand" is, and that's difficult.

However, note that we have a choice between two distinct cross products for every choice of non-collinear input vectors. For the particular choice that $\mathbf a$ is the first standard base vector $\mathbf e_x$, and $\mathbf b$ is the second standard base vector $\mathbf e_y$,, it turns out that $\mathbf u=\mathbf e_z$ and $\mathbf v=-\mathbf e_z$, its negative. So if we agree that $\mathbf e_x$, $\mathbf e_y$, $\mathbf e_z$ in that order form a right hand configuration, we must choose $\mathbf u$ instead of $\mathbf v$ at least in that situation. For all other cases, one can show that it is possible to continuously transform any given non-collinear $\mathbf a$ and $\mathbf b$ to $\mathbf e_x$ and $\mathbf e_y$ (without them being collinear at any intermediate point of time). In such a transformation, $\mathbf u$ and $\mathbf v$ are also continuous functions of the time and never coincide. So it akes perfect sense to make the consistent convention of picking $\mathbf u$ as the vector product and call this "right hand".

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A bilinear operator on $\mathbb{R}^3$ that returns a vector in $\mathbb{R}^3$ orthogonal to both its arguments must have $i$th component of the form $\sum_{jk}\epsilon_{ijk}u_jv_k$ for some symbol $\epsilon_{ijk}$ to determine. Since $u_i u_j v_k$ is $i\leftrightarrow j$=symmetric, orthogonality to $\mathbf{u}$ us equivalent to $\epsilon_{ijk}=-\epsilon_{jik}$. Similarly, we require $\epsilon_{ijk}=-\epsilon_{ikj}$. Thus $\epsilon_{ijk}=-\epsilon_{jik}=\epsilon_{jki}=-\epsilon_{kji}$, and exchanging any two indices changes the symbol's sign. Since all permutations of $123$ have a definite parity, to completely specify $\epsilon_{ijk}$ requires only the value of $\epsilon_{123}$. The choice $\epsilon_{123}=1$ gets the usual cross product, and since $\mathbf{i}\times\mathbf{j}=\mathbf{k}$ it's right-handed. Similarly, $\epsilon_{123}=-1$ would send $\mathbf{i},\,\mathbf{j}$ to $-\mathbf{k}$ and give us a left-handed rule.

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I'm partial to a proof published in The College Mathematics Journal, Vol. 46, No. 3, May 2015, p. 215 - 217, perhaps because I wrote it, although I'm sure that thousands of others came up with the same proof. You can read and download the complete paper for free at https://www.academia.edu/18549734/An_Even_Simpler_Proof_of_the_Right-Hand_Rule However, you will have to sign up on Academia.edu, so if you don't want to do that here's a proof of the main case:

Imagine you're way up on the positive z-axis, looking down on the x-y plane, and a and b both have initial points at the origin. Let $\textbf{a}_\textbf{p} = (a{_1}, a{_2}, 0 )$ and $\textbf{b}_\textbf{p} = (b{_1}, b{_2}, 0)$ be respectively the projections of a and b onto the x-y plane. Assume that $\textbf{a}_\textbf{p}$ and $\textbf{b}_\textbf{p}$ are not collinear. Look at the smaller of the two angles formed by $\textbf{a}_\textbf{p}$ and $\textbf{b}_\textbf{p}$.

The right-hand rule says that from this perspective, if a is to the right of b then the z-component of a x b will be positive, since your thumb will point up, and if a is to the left of b then the z-component of a x b will be negative, since your thumb will point down. Note that the left-right orientation of $\textbf{a}_\textbf{p}$ and $\textbf{b}_\textbf{p}$ is the same as the left-right orientation of a and b, i.e., if a is to the right of b then $\textbf{a}_\textbf{p}$ is to the right of $\textbf{b}_\textbf{p}$, and if a is to the left of b then $\textbf{a}_\textbf{p}$ is to the left of $\textbf{b}_\textbf{p}$.

Let $\alpha$ and $\beta$ be the measures of the angles which $\textbf{a}_\textbf{p}$ and $\textbf{b}_\textbf{p}$ respectively form with the positive x-axis, where $0 \leq \alpha, \beta < 2\pi$, and as usual, positive angle measurements correspond to counterclockwise rotations. Note that a x b and $\textbf{a}_\textbf{p}$ x $\textbf{b}_\textbf{p}$ have the same z-component,

$$a{_1}b{_2} - a{_2}b{_1} = (\left|\textbf{a}_\textbf{p}\right|\cos\alpha)(\left|\textbf{b}_\textbf{p}\right|\sin\beta) - (\left|\textbf{a}_\textbf{p}\right|\sin\alpha)(\left|\textbf{b}_\textbf{p}\right|\cos\beta )\\ = \left|\textbf{a}_\textbf{p}\right|\left|\textbf{b}_\textbf{p}\right|[(\sin\beta)(\cos\alpha) - (\sin\alpha)(\cos\beta)]\\ = \left |\textbf{a}_\textbf{p}\right|\left|\textbf{b}_\textbf{p}\right|\sin(\beta - \alpha)$$

Suppose a is to the right of b. Then $\textbf{a}_\textbf{p}$ is to the right of $\textbf{b}_\textbf{p}$, so $0 < \beta - \alpha < \pi$ (e.g., if $\textbf{a}_\textbf{p}$ is in Quadrant I and $\textbf{b}_\textbf{p}$ is in Quadrant II) or $-2\pi < \beta - \alpha < -\pi$ (e.g., if $\textbf{a}_\textbf{p}$ is in Quadrant IV and $\textbf{b}_\textbf{p}$ is in Quadrant I). In each of these cases, $\sin(\beta - \alpha) > 0$, so the z-component of a x b is positive, as stated by the right-hand rule. Similarly, if a is to the left of b then $\textbf{a}_\textbf{p}$ is to the left of $\textbf{b}_\textbf{p}$, so $-\pi < \beta - \alpha < 0$ or $\pi < \beta - \alpha < 2\pi$, so $\sin(\beta - \alpha) < 0$, thus the z-component is negative.

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  • $\begingroup$ Yes I did read your paper, but I'm not sure if I'm comfortable with phrases like "imagine you're way up", "looking down", "to the right/left", "your thumb will point up/down". This doesn't seem like a fully formal/rigorous proof. I think we'd need to formally define what all of these phrases mean. (I didn't downvote your answer though.) $\endgroup$ – dtcm840 Jun 28 '18 at 9:54

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