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

$$\mathbf{a}\times \mathbf{b}=(a_2{b}_3-a_3{b}_2,a_3{b}_1-a_1{b}_3,a_1{b}_2-a_2{b}_1).$$

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

$$\begin{array}{rcl}\mathbf{a}\times \mathbf{b}& =& \left|\begin{array}{ccc}{\mathbf{e}}_1& {\mathbf{e}}_2& {\mathbf{e}}_3\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|\\ & =& \left|\begin{array}{cc}{a}_2& a_3\\ b_2& b_3\end{array}\right|{\mathbf{e}}_1-\left|\begin{array}{cc}{a}_1& a_3\\ b_1& b_3\end{array}\right|{\mathbf{e}}_2+\left|\begin{array}{cc}{a}_1& a_2\\ b_1& b_2\end{array}\right|{\mathbf{e}}_3\\ & =& (\left|\begin{array}{cc}{a}_2& a_3\\ b_2& b_3\end{array}\right|,-\left|\begin{array}{cc}{a}_1& a_3\\ b_1& b_3\end{array}\right|,\left|\begin{array}{cc}{a}_1& a_2\\ b_1& b_2\end{array}\right|)\\ & =& (a_2{b}_3-a_3{b}_2,a_3{b}_1-a_1{b}_3,a_1{b}_2-a_2{b}_1).\end{array}$$

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 $⟨\mathbf{a},\mathbf{b}⟩$. 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:

$$\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=\left|\begin{array}{c}\mathbf{a}\\ \mathbf{b}\\ \mathbf{c}\end{array}\right|.$$

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,

$$\mathbf{b}\times \mathbf{c}=(b_2{c}_3-b_3{c}_2,b_3{c}_1-b_1{c}_3,b_1{c}_2-b_2{c}_1)$$

so

$$\begin{array}{rcl}\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})& =& a_1(b_2{c}_3-b_3{c}_2)+a_2(b_3{c}_1-b_1{c}_3)+a_3(b_1{c}_2-b_2{c}_1)\\ & =& a_1\left|\begin{array}{cc}{b}_2& b_3\\ c_2& c_3\end{array}\right|-a_2\left|\begin{array}{cc}{b}_1& b_3\\ c_1& c_3\end{array}\right|+a_3\left|\begin{array}{cc}{b}_1& b_2\\ c_1& c_2\end{array}\right|\\ & =& \left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|=\left|\begin{array}{c}\mathbf{a}\\ \mathbf{b}\\ \mathbf{c}\end{array}\right|.\end{array}$$

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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

$$\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=\left|\begin{array}{c}\mathbf{a}\\ \mathbf{b}\\ \mathbf{c}\end{array}\right|=-\left|\begin{array}{c}\mathbf{c}\\ \mathbf{b}\\ \mathbf{a}\end{array}\right|=\left|\begin{array}{c}\mathbf{c}\\ \mathbf{a}\\ \mathbf{b}\end{array}\right|=\mathbf{c}\cdot (\mathbf{a}\times \mathbf{b}).$$

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 $\mathbf{a}\times \mathbf{b}\ne \mathbf{0}$, 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 $\Vert \mathbf{a}\times \mathbf{b}\Vert =\Vert \mathbf{a}\Vert \cdot \Vert \mathbf{b}\Vert$.

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

Suppose $\mathbf{a},\mathbf{b}\ne \mathbf{0}$. Since $\mathbf{a}\cdot \mathbf{b}=0$, we have

$$a_1{b}_1+a_2{b}_2+a_3{b}_3=0.$$

Squaring both sides yields

$$a_1^2{b}_1^2+a_2^2{b}_2^2+a_3^2{b}_3^2+2a_1{a}_2{b}_1{b}_2+2a_2{a}_3{b}_2{b}_3+2a_3{a}_1{b}_3{b}_1=0.$$

so, in particular,

$$-2a_1{a}_2{b}_1{b}_2-2a_2{a}_3{b}_2{b}_3-2a_3{a}_1{b}_3{b}_1=a_1^2{b}_1^2+a_2^2{b}_2^2+a_3^2{b}_3^2.$$

Using this, we compute 

$$\begin{array}{rcl}\Vert \mathbf{a}\times \mathbf{b}{\Vert }^2& =& (a_2{b}_3-a_3{b}_2{)}^2+(a_3{b}_1-a_1{b}_3{)}^2+(a_1{b}_2-a_2{b}_1{)}^2\\ & =& a_2^2{b}_3^2+a_3^2{b}_2^2-2a_2{a}_3{b}_2{b}_3\\ & & +a_3^2{b}_1^2+a_1^2{b}_3^2-2a_1{a}_3{b}_1{b}_3\\ & & +a_1^2{b}_2^2+a_2^2{b}_1^2-2a_1{a}_2{b}_1{b}_2\\ & =& a_2^2{b}_3^2+a_3^2{b}_2^2+a_3^2{b}_1^2+a_1^2{b}_3^2+a_1^2{b}_2^2+a_2^2{b}_1^2\\ & & +(a_1^2{b}_1^2+a_2^2{b}_2^2+a_3^2{b}_3^2)\\ & =& (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)\\ & =& \Vert \mathbf{a}{\Vert }^2\cdot \Vert \mathbf{b}{\Vert }^2.\end{array}$$

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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

$$\Vert \mathbf{a}\times \mathbf{b}\Vert =\Vert \mathbf{a}\Vert \cdot \Vert \mathbf{b}\Vert \cdot |\sin \theta |.$$

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

$$\mathbf{a}\cdot \mathbf{b}=\Vert \mathbf{a}\Vert \cdot \Vert \mathbf{b}\Vert \cos \theta .$$

Using the unit vector defined by

$$\hat{\mathbf{a}}=\frac{1}{\Vert \mathbf{a}\Vert }\mathbf{a},$$

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

$$\begin{array}{}\text{(eq:bbpp)}& \mathbf{b}={\mathbf{b}}_{\parallel }+{\mathbf{b}}_{\perp }.\end{array}$$

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

$$\begin{array}{}\text{(eq:paracos)}& \Vert {\mathbf{b}}_{\parallel }{\Vert }^2=(\hat{\mathbf{a}}\cdot \mathbf{b}{)}^2=\Vert \mathbf{b}{\Vert }^2{\cos }^2\theta .\end{array}$$

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

$${\mathbf{b}}_{\perp }\cdot \hat{\mathbf{a}}=\mathbf{b}\cdot \hat{\mathbf{a}}-(\hat{\mathbf{a}}\cdot \mathbf{b})\Vert \hat{\mathbf{a}}{\Vert }^2=0.$$

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

$$\Vert \mathbf{b}{\Vert }^2=\Vert {\mathbf{b}}_{\parallel }{\Vert }^2+\Vert {\mathbf{b}}_{\perp }{\Vert }^2.$$

Substituting (eq:paracos), 

$$\Vert {\mathbf{b}}_{\perp }{\Vert }^2=\Vert \mathbf{b}{\Vert }^2-\Vert \mathbf{b}{\Vert }^2{\cos }^2\theta =\Vert \mathbf{b}{\Vert }^2{\sin }^2\theta .$$

Hence

$$\Vert {\mathbf{b}}_{\perp }\Vert =\Vert \mathbf{b}\Vert |\sin \theta |.$$

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,

$$\Vert \mathbf{a}\times \mathbf{b}\Vert =\Vert \mathbf{a}\Vert \cdot \Vert {\mathbf{b}}_{\perp }\Vert =\Vert \mathbf{a}\Vert \cdot \Vert \mathbf{b}\Vert \cdot |\sin \theta |.$$

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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 $\Vert \mathbf{a}\Vert \cdot \Vert \mathbf{b}\Vert \cdot |\sin \theta |$ 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. □