Why is the vector chosen by the right hand rule?

Question

I'm reading first year Physics and the Young & Freedman (13e) text describes how to find the vector (cross) product. Notably, the authors simplify the description of finding the product direction with the right hand rule which I've previously seen used to explain the direction of EM force from a wire with current flowing through it (I think?)

The decision here sounds almost arbitrary

There are always two directions perpendicular to a given plane, one on each side of the plane. We choose which of these the direction of ${\mathrm{\overrightarrow{A}}}$ ${\mathrm{\times}}$ ${\mathrm{\overrightarrow{B}}}$ as follows. Imagine rotating vector ${\mathrm{\overrightarrow{A}}}$ about the perpendicular line until it is aligned with ${\mathrm{\overrightarrow{B}}}$, choosing the smaller of the two possible possible angles between ${\mathrm{\overrightarrow{A}}}$ and ${\mathrm{\overrightarrow{B}}}$. Curl the fingers of your right hand around the perpendicular line so that the fingertips point in the direction of rotation; your thumb will then point in the direction of ${\mathrm{\overrightarrow{A}}}$ ${\mathrm{\times}}$ ${\mathrm{\overrightarrow{B}}}$.

Then on the next page,

We see that there are two kinds of coordinate systems, differing in the signs of the vector products of unit vectors. An axis system in which ${\mathcal{\hat{i}}}$ ${\mathrm{\times}}$ ${\mathcal{\hat{j}}}$ = ${\mathcal{\hat{k}}}$, as in [the example] is called a right-handed system. The usual practice is to use only right-handed systems, and we will follow that practice throughout this book.

Now I'm wondering what I'm not being told about and what the significance is? I understand that most of the students reading the book probably don't know enough linear algebra to understand complex derivations of this choice but it's also kind of frustrating not to know any more than the simplified version... Can anyone enlighten me?

Answer 1

There's not exactly a derivation here, just a convention that identifies an algebraic condition (the sign of a triple product) with a geometric condition (the right-hand rule). In this sense you're absolutely right: The definition is arbitrary. :)

Suppose $A = (a_1, a_2, a_3)$, $B = (b_1, b_2, b_3)$, and $C = (c_1, c_2, c_3)$ are vectors in $\mathbf{R}^3$. Their triple product is defined to be the scalar \begin{align*} \langle A, B, C\rangle &= \left|\begin{matrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{matrix}\right| \\ &= a_1 b_2 c_3 + a_2 b_3 c_1 + a_3 b_1 c_2 - a_1 b_3 c_2 - a_2 b_1 c_3 - a_3 b_2 c_1. \end{align*}

In linear algebra (possibly at the high school level), one shows that if three vectors $A$, $B$, and $C$ form a "basis" of $\mathbf{R}^3$ (the vectors are non-coplanar, i.e. none can be written as a linear combination of the other two), then their triple product is non-zero, hence is either positive or negative.

Call an ordered basis $\{A, B, C\}$ positively oriented if its triple product $\langle A, B, C\rangle$ is positive. (This is a purely algebraic condition.) For example, the standard basis $\{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\}$ is positively-oriented.

We'd like a geometric criterion for distinguishing positively-oriented bases from negatively-oriented bases. The convention is to use the right-hand rule: Point the fingers of your right hand along $A$ and curl them in the direction of the small angle (i.e., the angle less than $\pi$) toward $B$. If your right thumb makes an acute angle with $C$, the ordered triple $\{A, B, C\}$ is said to be right-handed. (Otherwise the ordered triple $\{A, B, C\}$ is left-handed.)

Right-handedness is a geometric condition, not an algebraic one. (This is a subtle point!)

Algebraically, the cross product of $A$ and $B$ is defined by the formula $$ A \times B = (a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1), $$ and consequently $(A \times B) \cdot C = \langle A, B, C\rangle$.

Thanks to the convention about right-handed bases (and properties of determinants/the triple product), the cross product $A\times B$ is (equivalently) specified by the geometric conditions

(i) $A \times B$ is orthogonal to $A$ and $B$,

(ii) the length of $A \times B$ is the area of the parallelogram spanned by $A$ and $B$, and

(iii) the ordered set $\{A, B, A \times B\}$ is right-handed (provided $A\times B$ is not zero).

As your textbook notes, there are two vectors satisfying conditions (i) and (ii), each the negative of the other; the right-hand rule tells us how to distinguish $A \times B$ geometrically from $-A \times B$.

Answer 2

This has to do with a basic fact of $n$-dimensional euclidean geometry. I'll explain it for $n=3$.

Two bases $({\bf e}_1,{\bf e}_2,{\bf e}_3)$ and $({\bf f}\,_1,{\bf f}\,_2,{\bf f}\,_3)$ of ${\mathbb R}^3$ have the same orientation if one of them can be deformed continuously into the other, so that at no time during this deformation the three vectors become linearly dependent. This is an equivalence relation on the set of bases of ${\mathbb R}^3$. It turns out that there are exactly two equivalence classes. Two bases belong to the same class iff the matrix expressing one in terms of the other has positive determinant.

There is no algebraic way to distinguish one of these two classes as "positive". If the little green men on Mars were blind we couldn't tell them what a right handed system is. But we humans on earth carry with us a common feeling about "left" and "right" (derived from the common direction of gravity), and this allows us to define one of the two equivalence classes of bases as "positive" when ${\mathbb R}^3$ is modeling our ambient space. The cross product ${\bf a}\times{\bf b}$ of two noncollinear vectors ${\bf a}$ and ${\bf b}$ is then defined such that the triple $({\bf a},{\bf b},{\bf a}\times{\bf b})$ is positively oriented.