Vector product
Given two vectors in 3 dimensions, that is, with three components, we can define a new operation: the vector product. The vector product between two vectors $\vec{a}$ and $\vec{b}$ is another vector $\vec{c}$.
We define the vector product by: $\vec{c}=\vec{a}\times\vec{b}$. Also, it is possible to denote the vector product using the symbol $\land$. So that $\vec{c}=\vec{a}\land\vec{b}$.
The resultant vector $\vec{c}$ to the vector product between two vectors $\vec{b}$ has the following properties:
- The angle is perpendicular to the plane formed by two vectors $\vec{a}$ and $\vec{b}$.
- The direction of the vector $\vec{c}$ is given by applying the "rule of the corkscrew" or the "rule of the right hand":
It is the direction of rotation a corkscrew would move when opening a bottle. With a corkscrew, or a screw "towards the right" (clockwise,) the corkscrew or the screw "goes into" the bottle. Also, it is possible to use the corkscrew or a screw in another sense: when one screws a corkscrew "towards the left" (counterclockwise), the corkscrew or the screw "comes out" of the bottle).
How to determine the vector $\vec{c}$ from the vector product of $\vec{a}$ and $\vec{b}$ in coordinates:
If $\vec{a}=(a_1,a_2,a_3)$ and $\vec{b}=(b_1,b_2,b_3)$. The vector product between $\vec{a}$ and $\vec{b}$ is the vector $\vec{c}$. We need to calculate the following determinant:
$$\vec{c}=(c_1,c_2,c_3)=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}= \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} \vec{i} + (-1)\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} \vec{j} + \begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} \vec{k}$$
Where $\vec{i}$, $\vec{j}$, $\vec{k}$ is the canonical base of $\mathbb{R}^3$. Namely, $\vec{i}=(1,0,0)$, $\vec{j}=(0,1,0)$, $\vec{k}=(0,0,1)$ form an orthonormal base.
If $\vec{a}=(2,0,-1)$, $\vec{b}=(1,1,-2)$. Let's compute $\vec{c}=\vec{a}\times\vec{b}$:
$$\vec{c}=(c_1,c_2,c_3)=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 0 & -1 \\ 1 & 1 & -2 \end{vmatrix}= \begin{vmatrix} 0 & -1 \\ 1 & -2 \end{vmatrix} \vec{i} + (-1)\begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix} \vec{j} + \begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix} \vec{k} = (1,3,2)$$
Another way of determining the vector product of $\vec{a}$ and $\vec{b}$:
$$\vec{c}=\vec{a}\times\vec{b}= |\vec{a}||\vec{b}|\sin(\widehat{ab})\cdot\hat{n}$$
where $\hat{n}$ is a unit vector in the corresponding angle and direction. The angle is the perpendicular to the plane formed by $\vec{a}$ and $\vec{b}$ and the direction (of rotation) given by the rule of the corkscrew.
Properties of the vector product:
- $\vec{a}\times\vec{b}=-\vec{b}\times\vec{a}$
- If $\vec{a}$ and $\vec{b}$ eare in the same straight line; then the vector product is zero.
If $\vec{a}=(1,0,0)$ and $\vec{b}=(-2,0,0)$ then:
$$\vec{c}=(c_1,c_2,c_3)=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & 0 \\ -2 & 0 & 0 \end{vmatrix}= \begin{vmatrix} 0 & 0 \\ 0 & 0 \end{vmatrix} \vec{i} + (-1)\begin{vmatrix} 1 & -2 \\ 0 & 0 \end{vmatrix} \vec{j} + \begin{vmatrix} 1 & -2 \\ 0 & 0 \end{vmatrix} \vec{k} = (0,0,0)=\vec{0}$$