Proof of the cross product formula

Question

I only care if the formula for cross product a x b has a direction consistent with the right-hand rule

$$ \begin{bmatrix} a_2b_3-a_3b_2 \\ a_3b_1-a_1b_3 \\ a_1b_2-a_2b_1 \\ \end{bmatrix} $$

What I understand is that, according to the fact that i × j = k, the cross product follows the right-hand rule — at least in this case. This contradicts the claim that the cross product never satisfies the right-hand rule. We also know that the vector −(a × b) is perpendicular to both a and b, just like a × b, but points in the opposite direction.

My question is: on what basis can we reject the possibility that in some cases, the vector a × b might not satisfy the right-hand rule, while b × a does? The example of i × j = k does not exclude this possibility.

We only know for sure that the formula for a x b always produces a vector perpendicular to both a and b, but we do not know whether it always describes a vector consistent with the right-hand rule or the left-hand rule-this proof would be sufficient to prove the right hand rule for a x b.

And no, the convention is whether we write a x b or b x a, not whether the pattern obeys the right-hand or left-hand rule!

(otherwise the surface integral formula would be incorrect because we assume that a x b has a direction consistent with the right-hand rule, which does not have to be true)

Answer 1

Note that for any rotation matrix $U$, it’s possible to show that $Uv \times Uw=U(v\times w)$. This justifies the intuition that if it follows the right hand rule for the basis set of vectors, then it follows it for any set of orthogonal vectors.

One nice way to prove this rotation property is by noting that you can define the cross product via the box product and use nice properties of the determinant to simplify things. See here for more details Cross product: matrix transformation identity

Answer 2

The right-hand rule arises geometrically, but is algebraically equivalent to the conditions:

$$i \times j = k$$ $$j \times k = i$$ $$k \times i = j$$

Since the cross-product formula follows the right-hand rule for the basis vectors, all other vectors that are a linear combination of these will also follow the right-hand rule.

This follows from the rules for associativity when multiplying with a scalar and the distributivity of the cross product over vector addition.

Answer 3

The concept of orientation is the observation that ordered bases fall under two classes with respect to linear transformations with positive determinant.

More formally, let $$ F = \{(v_1,v_2,v_3) \;:\; v_1,v_2,v_2\in\mathbb R^3\text{ is a basis}\} $$ be the set of frames. If $f\in F$ the define $f_i$ by $f=(f_1,f_2,f_3)$. Define a relation $\sim$ on $f,g\in F$ by $$ f\sim g \equiv \text{there is a matrix }M\text{ with }\det(M)>0\text{ such that }f = Mg. $$ Then $\sim$ is an equivalence relation and the cardinality of $F/{\sim}$ is two; we call the two class the orientations of $\mathbb R^3$, and naturally we say that $f$ and $g$ have the same orientation if $f\sim g$.

We say that $f$ follows the right-hand rule (or is right-handed) if $f_3$ is on the same side of the plane $\mathrm{span}\{f_1,f_2\}$ as indicated by the thumb of your right hand when curling your fingers from $f_1$ to $f_2$ through the (small) angle between them.

Theorem 1. If $f, g$ both follow the right-hand rule, then $f\sim g$.

Proof. Let $e=(e_1,e_2,e_3)$ be the standard basis. If $f$ is orthonormal then the theorem is obvious from geometric intuition: there is a rotation $M$ taking $e_i$ to $f_i$, and rotations have determinant 1. Thus, WLOG we may assume that $g$ is a right-handed orthonormal basis with $\mathrm{span}\{g_1,g_2\}=\mathrm{span}\{f_1,f_2\}$ and $f_1 = \alpha g_1$ for some $\alpha>0$. We can define a linear transformation by $Mg_i = f_i$, and we need to show that $\det(M)>0$. In the basis $g$, the matrix $M$ takes the form $$ M = \begin{pmatrix} \alpha & * & * \\ 0 & g_2\cdot f_2 & * \\ 0 & 0 & g_3\cdot f_3 \end{pmatrix} $$ Thus $$ \det(M) = \alpha(g_2\cdot f_2)(g_3\cdot f_3) $$ and each factor is $>0$, since the angle between $g_2,f_2$ and the angle between $g_3,f_3$ must both be acute (due to both $f$ and $g$ being right-handed). $\quad\square$

Corollary. If $f\sim g$ and $f$ is right-handed, then $g$ is right-handed.

Proof. "$f$ and $g$ have the same handedness" is an equivalence relation with equivalence classes $R,L$ for right- and left-handed frames. Denote $F/{\sim} = \{O_1,O_2\}$; WLOG, the Theorem allows us to assume $R\subseteq O_1$. But an identical proof shows that $L\subseteq O_1$ or $L\subseteq O_2$; it must be the latter since $R\sqcup L = F = O_1\sqcup O_2$, and so in fact $R = O_1$ and $L = O_2$. $\quad\square$

Theorem 2. For all linearly independent $a,b$, $$ (a,b,a\times b) \sim (e_1,e_2,e_3). $$

Proof. Using the well-known fact that $$ R(a\times b) = (Ra)\times(Rb) $$ for any rotation $R$, $$ (a,b,a\times b) \sim (Ra,Rb,R(a\times b)) = (Ra,Rb,(Ra)\times(Rb)) $$ so WLOG we may assume that $a = \alpha e_1$ for some $\alpha>0$ and that $\mathrm{span}\{a,b\} = \mathrm{span}\{e_1,e_2\}$. Thus $$ a\times b = \alpha(b\cdot e_2)e_3. $$ There is a matrix $M$ with $Me_1 = a$, $Me_2 = b$, and $Me_3 = a\times b$, namely $$ M = \begin{pmatrix} \alpha & * & 0 \\ 0 & b\cdot e_2 & 0 \\ 0 & 0 & \alpha(b\cdot e_2) \end{pmatrix} $$ and $$ \det(M) = \alpha^2(b\cdot e_2)^2 > 0.\quad\square $$

Corollary. $(a,b,a\times b)$ is right-handed.

Answer 4

You have to define what the righthand rule is. Let's call an ordered triple of vectors $(a,b,c)$ a frame if $$(a\times b)\cdot c \ne 0.$$ It is straightforward to show that $(a,b,c)$ is not a frame if and only if the three vectors lie in a plane.

The standard definition of a righthanded frame uses the cross product. $a,b,c$ satisfy the righthand rule if $$(a\times b)\cdot c > 0.$$ A little bit of analysis is needed next. The left side of the inequality is a continuous function of $a, b, c$. So if a frame $(a_1,b_1,c_1)$ can be changed continuously into another frame $(a_2,b_2,c_2)$, then the sign of the continuously changing frames cannot change.

What's still unclear is whether any two righthanded frames can be connected by a continuously changing family of righthanded frames This takes more work to show, but it is indeed the case.

Answer 5

What our hands do in 3-dimensional space is pick an orientation.

Left hands pick one orientation, right hands pick the other, and this covers all possibilities, except for the degenerate case when all fingers lie on the same plane (then the hand is flat and we cannot unambiguously choose between left and right - this is the zero orientation).

---

An orientation in 3 dimensions can be determined by an ordered triplet $(u,v,w)$. The order is important - interchanging any two vectors switches the orientation (as if you changed your thumb for your pointer).

The (sized) orientation determined by the 3 vectors above is usually denoted by $u\wedge v\wedge w$. For example, the interchanging property reads $u\wedge v\wedge w = -\boldsymbol v\wedge \boldsymbol u\wedge w$. This object is linear in each argument, e.g. $(u+u')\wedge v \wedge w=u\wedge v \wedge w+u'\wedge v \wedge w$.

The sized orientations (called oriented volumes) determine the same orientation if one is a positive multiple of the other.

---

Your question is whether the cross product completes $u$ and $v$ to a triplet having the same orientation as $(i,j,k)$, i.e. whether $u \wedge v \wedge (u\times v)$ and $i\wedge j\wedge k$ differ by a positive constant (in fact, they should turn out to be equal).

And indeed you can verify this by decomposing $u,v$ and $u\times v$ in the $i,j,k$ basis, expanding braces (in $u \wedge v \wedge (u\times v)$) and simplifying the expression.