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":

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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:

$\vec{c} = (c_{1}, c_{2}, c_{3}) = | \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix} | = | \begin{matrix} a_{2} & a_{3} \\ b_{2} & b_{3} \end{matrix} | \vec{i} + (- 1) | \begin{matrix} a_{1} & a_{3} \\ b_{1} & b_{3} \end{matrix} | \vec{j} + | \begin{matrix} a_{1} & a_{2} \\ b_{1} & b_{2} \end{matrix} | \vec{k}$

Where $\vec{i}$ , $\vec{j}$ , $\vec{k}$ is the canonical base of $R^{3}$ . Namely, $\vec{i} = (1, 0, 0)$ , $\vec{j} = (0, 1, 0)$ , $\vec{k} = (0, 0, 1)$ form an orthonormal base.

Example

If $\vec{a} = (2, 0, - 1)$ , $\vec{b} = (1, 1, - 2)$ . Let's compute $\vec{c} = \vec{a} \times \vec{b}$ :

$\vec{c} = (c_{1}, c_{2}, c_{3}) = | \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 0 & - 1 \\ 1 & 1 & - 2 \end{matrix} | = | \begin{matrix} 0 & - 1 \\ 1 & - 2 \end{matrix} | \vec{i} + (- 1) | \begin{matrix} 2 & - 1 \\ 1 & - 2 \end{matrix} | \vec{j} + | \begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix} | \vec{k} = (1, 3, 2)$

Another way of determining the vector product of $\vec{a}$ and $\vec{b}$ :

$\vec{c} = \vec{a} \times \vec{b} = | \vec{a} | | \vec{b} | \sin (\hat{a b}) \cdot \hat{n}$

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.

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Properties of the vector product:

$\vec{a} \times \vec{b} = - \vec{b} \times \vec{a}$

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If $\vec{a}$ and $\vec{b}$ eare in the same straight line; then the vector product is zero.

Example

If $\vec{a} = (1, 0, 0)$ and $\vec{b} = (- 2, 0, 0)$ then:

$\vec{c} = (c_{1}, c_{2}, c_{3}) = | \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & 0 \\ - 2 & 0 & 0 \end{matrix} | = | \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} | \vec{i} + (- 1) | \begin{matrix} 1 & - 2 \\ 0 & 0 \end{matrix} | \vec{j} + | \begin{matrix} 1 & - 2 \\ 0 & 0 \end{matrix} | \vec{k} = (0, 0, 0) = \vec{0}$