Constructing the Cross Product

Previously, our discussion on the dot product began with a simple question: What is the angle between two vectors? Our study on the cross product shall also begin with a simple question: What is a vector that is perpendicular to two vectors?

One possible approach is to give the answer, and leave it as it is. However, part of this blog’s goal is to journey from the known to the unknown, and in this context, it means properly constructing the cross product. Just like with determinants, we are going to impose several conditions on the cross product, and the result from these conditions will bring us to one and only one definition for the cross product.

Just like determinants, we are going to build the cross product $\times : \mathbb K^3 \times \mathbb K^3 \to \mathbb K^3$ one step at a time. Here, we denote $\mathbf u \times \mathbf v := \times(\mathbf u, \mathbf v)$ for readability. We will let $(\mathbf i, \mathbf j , \mathbf k) = (\mathbf e_1, \mathbf e_2, \mathbf e_3)$ denote the usual standard ordered basis on $\mathbb K^3$ .

The first observation is that, as a basic normalising step, we want $\mathbf k$ to be the vector perpendicular to $\mathbf i$ and $\mathbf j$ . They are all unit vectors (orthonormal, in fact). Actually, both $\pm \mathbf k$ would be perpendicular to $\mathbf i$ and $\mathbf j$ .

Lemma 1. In the case $\mathbb R = \mathbb K$ , we have $\mathbf i \cdot \mathbf k = 0$ , $\mathbf j \cdot \mathbf k = 0$ . Equivalently, $\mathbf i \perp \mathbf k$ and $\mathbf j \perp \mathbf k$ .

Almost by custom, we will define $\mathbf i \times \mathbf j := \mathbf k$ and $\mathbf j \times \mathbf i := -\mathbf i \times \mathbf j = -\mathbf k$ .

Definition 1. Define $\mathbf i \times \mathbf j := \mathbf k$ , $\mathbf j \times \mathbf k := \mathbf i$ , and $\mathbf k \times \mathbf i := \mathbf j$ . For $\mathbf u, \mathbf v \in \{\mathbf i, \mathbf j, \mathbf k \}$ , define $\mathbf u \times \mathbf v := -\mathbf v \times \mathbf u$ whenever $\mathbf u \neq \mathbf v$ .

By Lemma 1, we obtain the property

$(\mathbf u \times \mathbf v) \cdot \mathbf u = (\mathbf u \times \mathbf v) \cdot \mathbf v = \mathbf 0.$

Now, the stipulation $\mathbf u \neq \mathbf v$ is rather arbitrary. If instead we allow $\mathbf u \times \mathbf v := -\mathbf v \times \mathbf u$ for any $\mathbf u, \mathbf v \in \mathbb K^3$ , we actually obtain an alternating map, which yields $\mathbf u \times \mathbf u = \mathbf 0$ .

Theorem 1. For any $\mathbf u \in \mathbb K^3$ , $\mathbf u \times \mathbf u = \mathbf 0$ . In particular,

$\mathbf i \times \mathbf i = \mathbf j \times \mathbf j = \mathbf k \times \mathbf k = \mathbf 0.$

Next, and rather understandably, we would like $\times$ to be multi-linear (since there are two arguments, we say that $\times$ is bi-linear). It turns out that we are forced into one formula for $\mathbf u \times \mathbf v$ .

Theorem 2. Suppose furthermore that $\times$ is linear in both arguments. Then for any $\mathbf u = \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix} , \mathbf v = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} \in \mathbb K^3$ ,

$\mathbf u \times \mathbf v = \begin{bmatrix} u_2 v_3 - u_3 v_2 \\ -(u_1 v_3 - u_3 v_1) \\ u_1 v_2 - u_2 v_1 \end{bmatrix}.$

Proof. We first exploit the linearity of $\mathbf e_i \times (\cdot)$ . Writing $\mathbf v = v_1 \mathbf e_1 + v_2 \mathbf e_2 + v_3 \mathbf e_3$ ,

$\begin{aligned}\mathbf e_i \times \mathbf v &= \mathbf e_i \times (v_1 \mathbf e_1 + v_2 \mathbf e_2 + v_3 \mathbf e_3) \\ &= v_1 \mathbf e_i \times \mathbf e_1 + v_2 \mathbf e_i \times \mathbf e_2 + v_3 \mathbf e_i \times \mathbf e_3 \\ &= \sum_{n \neq i} v_n \mathbf e_i \times \mathbf e_n.\end{aligned}$

In particular,

$\mathbf e_1 \times \mathbf v = \begin{bmatrix} 0 \\ -v_3 \\ v_2\end{bmatrix}, \quad \mathbf e_2 \times \mathbf v = \begin{bmatrix} v_3 \\ 0 \\ -v_1\end{bmatrix}, \quad \mathbf e_3 \times \mathbf v = \begin{bmatrix} -v_2 \\ v_1 \\ 0\end{bmatrix}.$

Hence, exploiting the linearity of $(\cdot)\times \mathbf v$ , writing $\mathbf u = u_1 \mathbf e_1 + u_2 \mathbf e_2 + u_3 \mathbf e_3$ ,

$\begin{aligned} \mathbf u \times \mathbf v &= (u_1 \mathbf e_1 + u_2 \mathbf e_2 + u_3 \mathbf e_3) \times \mathbf v \\ &= u_1 \mathbf e_1 \times \mathbf v + u_2 \mathbf e_2 \times \mathbf v + u_3 \mathbf e_3 \times \mathbf v \\ &= u_1 \begin{bmatrix} 0 \\ -v_3 \\ v_2\end{bmatrix} + u_2 \begin{bmatrix} v_3 \\ 0 \\ -v_1\end{bmatrix} + u_3 \begin{bmatrix} -v_2 \\ v_1 \\ 0\end{bmatrix} \\ &= \begin{bmatrix} u_2 v_3 - u_3 v_2 \\ -(u_1 v_3 - u_3 v_1) \\ u_1 v_2 - u_2 v_1 \end{bmatrix}. \end{aligned}$

Corollary 1. For vectors $\mathbf u, \mathbf v, \mathbf w \in \mathbb K^3$ , $(\mathbf u \times \mathbf v) \cdot \mathbf w = \det(\mathbf u, \mathbf v, \mathbf w)$ and identifying

$\mathbf u \times \mathbf v \equiv ( \mathbf u \times \mathbf v )^{\mathrm T} = (\mathbf u \times \mathbf v) \cdot (\cdot ) = \det(\mathbf u, \mathbf v, \cdot)$ ,

we have

$\mathbf u \times \mathbf v \equiv \det(\mathbf u, \mathbf v, \cdot) =: \left|\begin{matrix} \mathbf i & \mathbf j & \mathbf k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{matrix}\right|.$

Proof. By expanding each component,

$\mathbf u \times \mathbf v = \left| \begin{matrix} u_2 & u_3 \\ v_2 & v_3 \end{matrix} \right| \mathbf i - \left| \begin{matrix} u_1 & u_3 \\ v_1 & v_3 \end{matrix} \right| \mathbf j + \left| \begin{matrix} u_1 & u_2 \\ v_1 & v_2 \end{matrix} \right| \mathbf k.$

Hence, taking the dot product with $\mathbf w = w_1 \mathbf e_1 + w_2 \mathbf e_2 + w_3 \mathbf w_3$ ,

$\begin{aligned} (\mathbf u \times \mathbf v) \cdot \mathbf w &= \left| \begin{matrix} u_2 & u_3 \\ v_2 & v_3 \end{matrix} \right| w_1 - \left| \begin{matrix} u_1 & u_3 \\ v_1 & v_3 \end{matrix} \right| w_2 + \left| \begin{matrix} u_1 & u_2 \\ v_1 & v_2 \end{matrix} \right| w_3 \\ &= \left|\begin{matrix} w_1 & w_2 & w_3 \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{matrix}\right| = \det(\mathbf w, \mathbf u, \mathbf v) = -\det(\mathbf u, \mathbf w, \mathbf v) \\ &= -(-\det(\mathbf u, \mathbf v, \mathbf w)) = \det(\mathbf u, \mathbf v, \mathbf w). \end{aligned}$

Interestingly however, if we began with this definition of the cross product, then we will recover all of our previous properties.

Corollary 2. For any $\mathbf u,\mathbf v \in \mathbb K^3$ , define

$\mathbf u \times \mathbf v = \begin{bmatrix} u_2 v_3 - u_3 v_2 \\ -(u_1 v_3 - u_3 v_1) \\ u_1 v_2 - u_2 v_1 \end{bmatrix} \equiv \det(\mathbf u, \mathbf v, \cdot).$

Then we recover the hypotheses in Definition 1, Theorem 1 and Theorem 2.

Proof. Since $\det$ is alternative, $(\mathbf u, \mathbf v) \in \{\mathbf i, \mathbf j , \mathbf k\}^2$ recovers our hypotheses in Definition 1 and Theorem 1. Since the determinant is multilinear, we recover Theorem 2.

Using the usual dot product (which still works as a computation in $\mathbb K^3$ , barring some technical interpretive caveats), we have that $\mathbf u \times \mathbf v$ is orthogonal to both $\mathbf u$ and $\mathbf v$ .

Corollary 3. For vectors $\mathbf u, \mathbf v, \mathbf w \in \mathbb K^3$ ,

$(\mathbf u \times \mathbf v) \cdot \mathbf u = (\mathbf u \times \mathbf v) \cdot \mathbf v = \mathbf 0.$

Proof. The results follow since the determinant is alternative.

Now, the discussion on angles only really makes sense when $\mathbb K = \mathbb R$ , in that

$\mathbf u \cdot \mathbf v = 0 \quad \iff \quad \mathbf u \perp \mathbf v,$

so let’s zoom in on this situation. We observe that for any $k \in \mathbb R$ , $k(\mathbf u \times \mathbf v) \perp \mathbf u$ . So what is it about $\mathbf u \times \mathbf v$ that makes it special? Perhaps $\|\mathbf u \times \mathbf v \|$ is intimately connected with the “degree” of separation, no pun intended, between $\mathbf u$ and $\mathbf v$ . Obviously, due to linearity, $\| \mathbf u \times \mathbf v\| = 0$ whenever either $\mathbf u$ or $\mathbf v$ are the zero vector $\mathbf 0$ .

Assume therefore that $\mathbf u, \mathbf v$ are nonzero vectors. Due to bilinearity, if we can decode the case for $\| \mathbf u \| = \| \mathbf v \| = 1$ , then using the property that $\mathbf u = \|\mathbf u \| \hat{\mathbf u}$ , we obtain

$\| \mathbf u \times \mathbf v\| = \| (\|\mathbf u \| \hat{\mathbf u}) \times (\|\mathbf v \| \hat{\mathbf v})\| = \| \mathbf u \| \| \mathbf v \| \| \hat{\mathbf u} \times \hat{\mathbf v}\|.$

Therefore, let’s work with the simplified case $\| \mathbf u \| = \| \mathbf v \| = 1$ . At this point, we could bash the algebra out. However, taking inspiration from this thread, we will adopt a much cleaner approach. Define the matrix

$\mathbf A = \begin{bmatrix} \mathbf u & \mathbf v & \mathbf u \times \mathbf v \end{bmatrix}.$

Then

$\begin{aligned} \det(\mathbf A^{\mathrm T} \mathbf A) &= \det(\mathbf A^{\mathrm T}) \det(\mathbf A) = (\det \mathbf A)^2. \end{aligned}$

On the left-hand side,

$\begin{aligned} \det(\mathbf A^{\mathrm T} \mathbf A) &= \left| \begin{matrix} \mathbf u^{\mathrm T} \mathbf u & \mathbf u^{\mathrm T} \mathbf v & \mathbf u^{\mathrm T} (\mathbf u \times \mathbf v) \\ \mathbf v^{\mathrm T} \mathbf u & \mathbf v^{\mathrm T} \mathbf v & \mathbf v^{\mathrm T} (\mathbf u \times \mathbf v) \\ (\mathbf u \times \mathbf v)^{\mathrm T} \mathbf u & (\mathbf u \times \mathbf v)^{\mathrm T} \mathbf v & (\mathbf u \times \mathbf v)^{\mathrm T} (\mathbf u \times \mathbf v) \end{matrix} \right| \\ &= \left| \begin{matrix} \| \mathbf u \|^2 & \mathbf u \cdot \mathbf v & 0 \\ \mathbf u \cdot \mathbf v & \| \mathbf v \|^2 & 0 \\ 0 & 0 & \| \mathbf u \times \mathbf v \|^2 \end{matrix} \right| \\ &= \| \mathbf u \times \mathbf v \|^2\left| \begin{matrix} \| \mathbf u \|^2 & \mathbf u \cdot \mathbf v \\ \mathbf u \cdot \mathbf v & \| \mathbf v \|^2 \end{matrix} \right| \\ &= \| \mathbf u \times \mathbf v \|^2 (\|\mathbf u\|^2 \|\mathbf v\|^2 - (\mathbf u \cdot \mathbf v)^2).\end{aligned}$

On the right-hand side,

$(\det \mathbf A)^2 = (\det(\mathbf u, \mathbf v, \mathbf u \times \mathbf v))^2 = ((\mathbf u \times \mathbf v) \cdot (\mathbf u \times \mathbf v))^2 = \|\mathbf u \times \mathbf v\|^4.$

Therefore, dividing by $\|\mathbf u \times \mathbf v\|^2$ on both sides,

$\begin{aligned} \|\mathbf u \times \mathbf v\|^2 &= \|\mathbf u\|^2 \|\mathbf v\|^2 - (\mathbf u \cdot \mathbf v)^2 \\ &= 1^2 \cdot 1^2 - (1 \cdot 1 \cdot \cos \theta)^2 \\ &= 1 - \cos^2 (\theta) = \sin^2(\theta). \end{aligned}$

This equation can only mean one thing: $\| \mathbf u \times \mathbf v \| = \sin(\theta)$ , where we don’t even need to worry about $| \cdot |$ on the right-hand side because $\sin(\theta) \geq 0$ whenever $\theta \in [0, \pi]$ . This yields a rather elegant geometric interpretation for the cross product:

Corollary 4. For $\mathbf u, \mathbf v \in \mathbb R^3$ with angle $\theta$ between them as computed by the dot product (and consider $\theta$ undefined when either $\mathbf u$ or $\mathbf v$ are the zero vector $\mathbf 0$ ),