# 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: 1. $$\vec{a}\times\vec{b}=-\vec{b}\times\vec{a}$$ 1. 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: $$\vec{c}=(c_1,c_2,c_3)=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & 0 \\ -2 & 0 & 0 \end{vmatrix}= \begin{vmatrix} 0 & 0 \\ 0 & 0 \end{vmatrix} \vec{i} + (-1)\begin{vmatrix} 1 & -2 \\ 0 & 0 \end{vmatrix} \vec{j} + \begin{vmatrix} 1 & -2 \\ 0 & 0 \end{vmatrix} \vec{k} = (0,0,0)=\vec{0}$$$