Vector product

The scalar product maps a pair of vectors to a scalar: \(\mathbb{R}^n\times \mathbb{R}^n \to \mathbb{R}\). The scalar product can be defined in the vector space \(\mathbb{R}^n\) with any \(n\). In contrast, the vector product (also known as the cross product), which maps a pair of vectors to another vector (\(\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n\)) in the same vector space, can be defined only in the 3-dimensional space, \(\mathbb{R}^3\).

Definition (Vector product)

Let \(\mathbf{a}, \mathbf{b}\in\mathbb{R}^3\) be row vectors \(\mathbf{a}= (a_1, a_2, a_3)\) and \(\mathbf{b} = (b_1, b_2, b_3)\). Then their vector product (also called the cross product or outer product), denoted \(\mathbf{a}\times\mathbf{b}\), is the vector defined as

\[\mathbf{a}\times\mathbf{b} = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1).\]

Remark. Some authors prefer to denote the vector product by \([\mathbf{a},\mathbf{b}]\) rather than \(\mathbf{a}\times \mathbf{b}\). □

Let \(\mathbf{e}_1 = (1, 0, 0), \mathbf{e}_2 = (0, 1, 0), \mathbf{e}_3 = (0, 0, 1)\), then the cross product may be formally expressed as a determinant as in

\[ \begin{eqnarray*} \mathbf{a}\times\mathbf{b} &=& \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ \end{vmatrix}\\ &=& \begin{vmatrix} a_2 & a_3\\ b_2 & b_3 \end{vmatrix}\mathbf{e}_1 -\begin{vmatrix} a_1 & a_3\\ b_1 & b_3 \end{vmatrix}\mathbf{e}_2 +\begin{vmatrix} a_1 & a_2\\ b_1 & b_2 \end{vmatrix}\mathbf{e}_3\\ &=& \left(\begin{vmatrix} a_2 & a_3\\ b_2 & b_3 \end{vmatrix}, -\begin{vmatrix} a_1 & a_3\\ b_1 & b_3 \end{vmatrix}, \begin{vmatrix} a_1 & a_2\\ b_1 & b_2 \end{vmatrix}\right)\\ &=& (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1). \end{eqnarray*} \]

Theorem 

Let \(\mathbf{a}, \mathbf{b} \in \mathbb{R}^3\). Then we have the following.

1. If \(\mathbf{a} = \mathbf{0}\) or \(\mathbf{b} = \mathbf{0}\), then \(\mathbf{a}\times\mathbf{b} = \mathbf{0}\).
2. If \(\mathbf{a}\) and \(\mathbf{b}\) are parallel, that is, \(\mathbf{a} = \lambda\mathbf{b}\) or \(\mathbf{b} = \mu\mathbf{a}\) for some non-zero \(\lambda,\mu\in\mathbb{R}\), then \(\mathbf{a}\times\mathbf{b} = \mathbf{0}\).
3. \(\mathbf{a}\times \mathbf{b} = -(\mathbf{b}\times\mathbf{a})\). (Thus, the vector product is not commutative.)
4. For any \(\lambda\in\mathbb{R}\), \(\lambda(\mathbf{a}\times\mathbf{b}) = (\lambda\mathbf{a})\times\mathbf{b} = \mathbf{a}\times(\lambda\mathbf{b})\).
5.  \(\mathbf{e}_1\times\mathbf{e}_2 = \mathbf{e}_3\), \(\mathbf{e}_2\times\mathbf{e}_3 = \mathbf{e}_1\), \(\mathbf{e}_3\times\mathbf{e}_1 = \mathbf{e}_2\).

Proof. (Exercise.) ■

In the following, the scalar product between \(\mathbf{a}\) and \(\mathbf{b}\) is denoted by \(\mathbf{a}\cdot\mathbf{b}\) rather than \(\langle\mathbf{a},\mathbf{b}\rangle\). Recall that the scalar product is also called the dot product.

Definition (Triple product)

The triple product of three vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c} \in\mathbb{R}^3\) is defined by

\[\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}).\]

Theorem

Let \(\mathbf{a}, \mathbf{b}, \mathbf{c} \in \mathbb{R}^3\) be row vectors. Their triple product is given by the determinant:

\[ \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = \begin{vmatrix} \mathbf{a}\\ \mathbf{b}\\ \mathbf{c} \end{vmatrix}. \]

Proof. Let \(\mathbf{a} = (a_1, a_2, a_3)\), \(\mathbf{b} = (b_1, b_2, b_3)\), \(\mathbf{c} = (c_1, c_2, c_3)\). By definition,

\[\mathbf{b}\times\mathbf{c} = (b_2c_3 - b_3c_2, b_3c_1 - b_1c_3, b_1c_2 - b_2c_1)\]

so

\[ \begin{eqnarray} \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) &=& a_1(b_2c_3 - b_3c_2) + a_2(b_3c_1 - b_1c_3) + a_3(b_1c_2 - b_2c_1)\\ &=& a_1 \begin{vmatrix} b_2 & b_3\\ c_2 & c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1 & b_3\\ c_1 & c_3 \end{vmatrix} + a_3 \begin{vmatrix} b_1 & b_2\\ c_1 & c_2 \end{vmatrix}\\ &=& \begin{vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3 \end{vmatrix} = \begin{vmatrix} \mathbf{a}\\ \mathbf{b}\\ \mathbf{c} \end{vmatrix}. \end{eqnarray} \]

■

Corollary 

\[ \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = \mathbf{c}\cdot(\mathbf{a}\times\mathbf{b}) = \mathbf{b}\cdot(\mathbf{c}\times\mathbf{a}). \]

Proof. By the property of determinants, we have

\[ \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = \begin{vmatrix} \mathbf{a}\\ \mathbf{b}\\ \mathbf{c} \end{vmatrix} = -\begin{vmatrix} \mathbf{c}\\ \mathbf{b}\\ \mathbf{a} \end{vmatrix} = \begin{vmatrix} \mathbf{c}\\ \mathbf{a}\\ \mathbf{b} \end{vmatrix} = \mathbf{c}\cdot(\mathbf{a}\times\mathbf{b}). \]

The rest is similar. ■

Corollary 

\[\mathbf{a}\cdot(\mathbf{a}\times\mathbf{b}) = 0.\]

Proof. (Exercise.) ■

Remark. Similarly, we have \(\mathbf{b}\cdot(\mathbf{a}\times\mathbf{b}) = 0\).

This result indicates that the vector product \(\mathbf{a}\times\mathbf{b}\) is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\). If \(\mathbf{a}\times\mathbf{b}\neq \mathbf{0}\), the three vectors \(\mathbf{a}\), \(\mathbf{b}\) and their vector product form a right-handed set. This means if your right index and middle fingers to directions of \(\mathbf{a}\) and \(\mathbf{b}\), respectively, then your right thumb points to the direction of \(\mathbf{a}\times\mathbf{b}\). □

Corollary

\[\mathbf{a}\times(\mathbf{b} + \mathbf{c}) = \mathbf{a}\times\mathbf{b} + \mathbf{a}\times\mathbf{c}.\]

Proof. (Exercise.) ■

Lemma

Let \(\mathbf{a}, \mathbf{b}\in\mathbb{R}^3\). If \(\mathbf{a}\cdot\mathbf{b} = 0\), then \(\|\mathbf{a}\times\mathbf{b} \| = \|\mathbf{a}\|\cdot\|\mathbf{b}\|\).

Proof. If \(\mathbf{a} = \mathbf{0}\) or \(\mathbf{b} = \mathbf{0}\), the result is trivial.

Suppose \(\mathbf{a},\mathbf{b}\neq\mathbf{0}\). Since \(\mathbf{a}\cdot\mathbf{b} = 0\), we have

\[a_1b_1 + a_2b_2 + a_3b_3 = 0.\]

Squaring both sides yields

\[a_1^2b_1^2 + a_2^2b_2^2 + a_3^2b_3^2 + 2a_1a_2b_1b_2 + 2a_2a_3b_2b_3 + 2a_3a_1b_3b_1 = 0.\]

so, in particular,

\[-2a_1a_2b_1b_2 -2a_2a_3b_2b_3 - 2a_3a_1b_3b_1 = a_1^2b_1^2 + a_2^2b_2^2 + a_3^2b_3^2.\]

Using this, we compute 

\[ \begin{eqnarray*} \|\mathbf{a}\times\mathbf{b}\|^2 &=& (a_2b_3 - a_3b_2)^2 + (a_3b_1 - a_1b_3)^2 + (a_1b_2 - a_2b_1)^2\\ &=& a_2^2b_3^2 + a_3^2b_2^2 -2a_2a_3b_2b_3\\ && + a_3^2b_1^2 + a_1^2b_3^2 -2a_1a_3b_1b_3\\ && + a_1^2b_2^2 + a_2^2b_1^2 -2a_1a_2b_1b_2\\ &=& a_2^2b_3^2 + a_3^2b_2^2+ a_3^2b_1^2 + a_1^2b_3^2+ a_1^2b_2^2 + a_2^2b_1^2\\ && + (a_1^2b_1^2 + a_2^2b_2^2 + a_3^2b_3^2)\\ &=& (a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2)\\ &=& \|\mathbf{a}\|^2\cdot\|\mathbf{b}\|^2. \end{eqnarray*} \]

■

Remark. This lemma says that, if the vectors \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular to each other (including the case one (or both) of them is zero), then the length of their vector product is the product of their lengths. □

Theorem

Let \(\mathbf{a}, \mathbf{b}\in\mathbb{R}^3\) such that the angle between them is \(\theta\). Then

\[\|\mathbf{a}\times\mathbf{b}\| = \|\mathbf{a}\|\cdot\|\mathbf{b}\|\cdot|\sin\theta|.\]

Proof. Let \(\mathbf{a}, \mathbf{b}\in\mathbb{R}^3\) be non-zero vectors, and let the angle between them be \(\theta\). So we have

\[\mathbf{a}\cdot\mathbf{b} = \|\mathbf{a}\|\cdot\|\mathbf{b}\|\cos\theta.\]

Using the unit vector defined by

\[\hat{\mathbf{a}} = \frac{1}{\|\mathbf{a}\|}\mathbf{a},\]

let us define

\[\mathbf{b}_{\parallel} = (\hat{\mathbf{a}}\cdot\mathbf{b}) \hat{\mathbf{a}}\]

and

\[\mathbf{b}_{\perp} = \mathbf{b} - \mathbf{b}_{\parallel}.\]

By this definition, we have

\[\mathbf{b} = \mathbf{b}_{\parallel} + \mathbf{b}_{\perp}.\tag{eq:bbpp}\]

Clearly, \(\mathbf{b}_{\parallel}\) is parallel to \(\mathbf{a}\) and

\[\|\mathbf{b}_{\parallel}\|^2 =(\hat{\mathbf{a}}\cdot\mathbf{b})^2 = \|\mathbf{b}\|^2\cos^2\theta.\tag{eq:paracos}\]

On the other hand, \(\mathbf{b}_{\perp}\) is perpendicular to \(\mathbf{a}\) because

\[\mathbf{b}_{\perp}\cdot\hat{\mathbf{a}} = \mathbf{b}\cdot\hat{\mathbf{a}} - (\hat{\mathbf{a}}\cdot\mathbf{b})\|\hat{\mathbf{a}}\|^2 = 0.\]

It follows that \(\mathbf{b}_{\perp}\cdot\mathbf{b}_{\parallel} = 0\). Therefore, from (eq:bbpp), we have

\[\|\mathbf{b}\|^2 = \|\mathbf{b}_{\parallel}\|^2 + \|\mathbf{b}_{\perp}\|^2.\]

Substituting (eq:paracos), 

\[\|\mathbf{b}_{\perp}\|^2 = \|\mathbf{b}\|^2 - \|\mathbf{b}\|^2\cos^2\theta = \|\mathbf{b}\|^2\sin^2\theta.\]

Hence

\[\|\mathbf{b}_{\perp}\| = \|\mathbf{b}\||\sin\theta|.\]

Now consider the vector product \(\mathbf{a}\times\mathbf{b}\). Since \(\mathbf{a}\) and \(\mathbf{b}_{\parallel}\) are parallel to each other, \(\mathbf{a}\times\mathbf{b}_{\parallel} = \mathbf{0}\). Therefore,

\[\mathbf{a}\times\mathbf{b} = \mathbf{a}\times(\mathbf{b}_{\parallel} + \mathbf{b}_{\perp}) = \mathbf{a}\times\mathbf{b}_{\parallel} + \mathbf{a}\times\mathbf{b}_{\perp} = \mathbf{a}\times\mathbf{b}_{\perp}.\]

Since \(\mathbf{a}\cdot\mathbf{b}_{\perp} = 0\) (i.e., they are perpendicular), using the above lemma,

\[\|\mathbf{a}\times\mathbf{b}\| = \|\mathbf{a}\|\cdot\|\mathbf{b}_{\perp}\| = \|\mathbf{a}\|\cdot\|\mathbf{b}\|\cdot|\sin\theta|.\]

■

Remark. Note that any two vectors \(\mathbf{a}\) and \(\mathbf{b}\) define a parallelogram. That is, the origin and the position vectors \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{a} + \mathbf{b}\) make the vertices of a parallelogram. You should prove that the area of this parallelogram is given by \(\|\mathbf{a}\|\cdot\|\mathbf{b}\|\cdot|\sin\theta|\) where \(\theta\) is the angle between \(\mathbf{a}\) and \(\mathbf{b}\). □

Remark. Similarly, any three vectors \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\) can define a parallelepiped (a solid body of which each face is a parallelogram). We can show that the triple product \(\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})\) corresponds to the signed volume of that parallelepiped. Here, the term ``signed'' indicates that this ``volume'' can be negative. More specifically, the sign of this volume is positive if the triple \((\mathbf{a}, \mathbf{b}, \mathbf{c})\), in this order, makes a right-hand set, or negative if it makes a left-hand set. □