For vectors one can define the multiplication by a constant $k$ and the addition of vectors in a meaningful way. For the multiplication with a constant this is immediately obvious.

$$\begin{array}{c}k\mathbf{a}=k\left(\begin{array}{c}{a}_{1}\hfill \\ {a}_{2}\hfill \\ {a}_{3}\hfill \\ \hfill \end{array}\right)=\left(\begin{array}{c}k{a}_{1}\hfill \\ k{a}_{2}\hfill \\ k{a}_{3}\hfill \\ \hfill \end{array}\right),\hfill \\ \left|k\mathbf{a}\right|=\sqrt{{\left(k{a}_{1}\right)}^{2}+{\left(k{a}_{2}\right)}^{2}+{\left(k{a}_{3}\right)}^{3}}=k\sqrt{{a}_{1}^{2}+{a}_{2}^{2}+{a}_{3}^{3}}=k\left|\mathbf{a}\right|\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\left\{qed\right\}\hfill \\ \hfill \end{array}$$

For addition and subtraction the following definitions apply:

$$\begin{array}{c}\mathbf{a}+\mathbf{b}=\left(\begin{array}{c}{a}_{1}\hfill \\ {a}_{2}\hfill \\ {a}_{3}\hfill \\ \hfill \end{array}\right)+\left(\begin{array}{c}{b}_{1}\hfill \\ {b}_{2}\hfill \\ {b}_{3}\hfill \\ \hfill \end{array}\right)=\left(\begin{array}{c}{a}_{1}+{b}_{1}\hfill \\ {a}_{2}+{b}_{2}\hfill \\ {a}_{3}+{b}_{3}\hfill \\ \hfill \end{array}\right)\hfill \\ \hfill \\ \mathbf{a}-\mathbf{b}=\left(\begin{array}{c}{a}_{1}\hfill \\ {a}_{2}\hfill \\ {a}_{3}\hfill \\ \hfill \end{array}\right)-\left(\begin{array}{c}{b}_{1}\hfill \\ {b}_{2}\hfill \\ {b}_{3}\hfill \\ \hfill \end{array}\right)=\left(\begin{array}{c}{a}_{1}-{b}_{1}\hfill \\ {a}_{2}-{b}_{2}\hfill \\ {a}_{3}-{b}_{3}\hfill \\ \hfill \end{array}\right)=-\left(\begin{array}{c}{b}_{1}-{a}_{1}\hfill \\ {b}_{2}-{a}_{2}\hfill \\ {b}_{3}-{a}_{3}\hfill \\ \hfill \end{array}\right)=-\left(\mathbf{b}-\mathbf{a}\right)\hfill \\ \hfill \end{array}$$

The rules for multiplication with a constant and for the addition and subtraction are formal as for complex numbers, which one can write in analogy to the vector notation above as matrix with one column/line and two lines/columns. Thus these vector operations are also commutative, associative and distributive, i.e. the sequence of the vectors does not matter.

Vectors however do not constitute an extension of the complex number space to higher dimensions. Vectors are not multiplied with each other according to the rules of complex numbers with each other and the division of one vector by another one cannot be defined.

In the following we will define two different kinds of multiplications between vectors. These are operations that do not have an analogon in the space of the real or complex numbers. They are rather newly introduced for reasons of expediency. It is somewhat unfortunate that one has used the term multiplication. That experts also feels this way, can be seen from the fact, that one referred to the scalar product in earlier times, while it is nowadays preferably called the inner product and what used to be called the vector product is nowadays refereed to as outer product. This is however only a semantic problem, as soon as one understand the specifics.

For the vector addition we assume, that both vectors are similar quantities, i.e. for example that they represent two forces or two distances. It would one make sense to add a force to a distance, although they are both represented by vectors.

However in physics one would like to combine two vectors of different types with each other. Force and distance are suitable examples. We define work = force times distance, where the quantities of force and distance enter with the length of the corresponding vectors. For this easy formula we however assume, that the directions of the force and distance vector coincide. If however the force acts in another direction, for example at a right angle to the direction of movement, the force acting at a right angle does not perform any work. The interplay between force- and distance vector thus depends not only on the absolute value of the two vectors but also on the angle between them.

The corresponding combination of two vectors $\mathbf{a}$ and $\mathbf{b}$ is denoted as scalar product or inner product and is defined by

$$\mathbf{a}\cdot \mathbf{b}=\left|\mathbf{a}\right|\left|\mathbf{b}\right|cos\left(\mathbf{a},\mathbf{b}\right)=\left(\begin{array}{c}{a}_{1}\hfill \\ {a}_{2}\hfill \\ {a}_{3}\hfill \\ \hfill \end{array}\right)\cdot \left(\begin{array}{c}{b}_{1}\hfill \\ {b}_{2}\hfill \\ {b}_{3}\hfill \\ \hfill \end{array}\right)={a}_{1}{b}_{1}+{a}_{2}{b}_{2}+{a}_{3}{b}_{3}={b}_{1}{a}_{1}+{b}_{2}{a}_{2}+{b}_{3}{a}_{3}=\mathbf{b}\cdot \mathbf{a}\phantom{\rule{0em}{0ex}},$$

where $\left(\mathbf{a},\mathbf{b}\right)$ is used as a sign for the angle between $\mathbf{a}$ and $\mathbf{b}$. For the combination sign a dot is used and the combination is read as a dot b.

The inner product is a maximum, if both vectors are parallel
($cos\left(0\right)=cos\left(\pi \right)=1$)
and is equal to zero if they are orthogonal to each other
($cos\left(\pi \u22152\right)=cos(\frac{3\pi}{2}=0$). It is a number,
a scalar ^{1} ,
not a vector. This product is commutative, i.e. it does not matter which vector
appears first in the product. The resulting number is equal to the length
of one of the vectors multiplied with the projection of the other vector on
it.

A second well defined way to combine two vectors of different types defines a vector as result of a multiplication. Its direction is orthogonal to both input vectors and therefore also on the plane defined by the two vectors. It s absolute value is $\left|\mathbf{a}\right|\times \left|\mathbf{b}\right|sin\left(\mathbf{a},\mathbf{b}\right)$. The product is a maximum, if both vectors are orthogonal to each other. An example from physics is the deflecting force on a moving charge in a magnetic field.

For this outer product or vector product the following definitions apply:

$$\begin{array}{c}\mathbf{c}=\mathbf{a}\times \mathbf{b}=\left(\begin{array}{c}{a}_{1}\hfill \\ {a}_{2}\hfill \\ {a}_{3}\hfill \\ \hfill \end{array}\right)\times \left(\begin{array}{c}{b}_{1}\hfill \\ {b}_{2}\hfill \\ {b}_{3}\hfill \\ \hfill \end{array}\right)=\left(\begin{array}{c}{a}_{2}{b}_{3}-{a}_{3}{b}_{2}\hfill \\ {a}_{3}{b}_{1}-{a}_{1}{b}_{3}\hfill \\ {a}_{1}{b}_{2}-{a}_{2}{b}_{1}\hfill \\ \hfill \end{array}\right)=-\left(\begin{array}{c}{a}_{3}{b}_{2}-{a}_{2}{b}_{3}\hfill \\ {a}_{1}{b}_{3}-{a}_{3}{b}_{1}\hfill \\ {a}_{2}{b}_{1}-{a}_{1}{b}_{2}\hfill \\ \hfill \end{array}\right)=-\mathbf{b}\times \mathbf{a},\hfill \\ \left|\mathbf{c}\right|=\left|\mathbf{a}\right|\left|\mathbf{b}\right|sin\left(\mathbf{a},\mathbf{b}\right).\hfill \\ \hfill \end{array}$$

This initially slightly confusing formula for the resulting vector can easily be analyzed mnemonically: In the first component the first coordinate of the input vector does not appear, and in its negative term the indices are simply exchanged. For the second and third component the indices are cyclically changed.

$$\mathbf{a}\times \mathbf{b}\text{isreadas}a\text{}cross\text{}b\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\text{versus}\mathbf{a}\cdot \mathbf{b}\text{as}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}a\text{}dot\text{}b\phantom{\rule{0em}{0ex}}$$

The vector product is not commutative; it does indeed depend on the sequence of the vectors. Swapping the sequence changes the sign.

Since $\mathbf{a}\times \mathbf{b}$ is a vector one can multiply this resulting vector with a third vector $\mathbf{c}$ both in the inner as well as the outer sense. Then we have:

$$\left(\mathbf{a}\times \mathbf{b}\right)\cdot \mathbf{c}\text{isascalar,}\left(\mathbf{a}\times \mathbf{b}\right)\times \mathbf{c}\text{isavector.}$$