normal vector - der Normalenvektor
A normal vector, say $\vec{n}$
, is a vector perpendicular (orthogonal) to two (linearly independent) vectors $\vec{a}$
and $\vec{b}$
.
Note - vectors $\vec{a}$
and $\vec{b}$
also span a plane and the normal vector is perpendicular to the plane.
This means
$$\vec{a}\perp \vec{n} \Rightarrow \vec{a}\cdot \vec{n}=0$$
and
$$\vec{b}\perp \vec{n} \Rightarrow \vec{b}\cdot \vec{n}=0$$
In 3-dimensional space
Let $\vec{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$
and $\vec{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$
, and let the normal we are looking for be $\vec{n} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$
Applying the scalar products above to the vectors, we get two simultaneous equations.
$${\begin{align} a_1x + a_2y + a_3z & = 0 \newline b_1x + b_2y + b_3z & = 0 \end{align}}$$
Writing as a matrix.
$${\begin{vmatrix} a_1 & a_2 & a_3 & \vdots & 0 \\ b_1 & b_2 & b_3 & \vdots & 0 \end{vmatrix}}$$
Multiply the 1st row by $-b_1$
and the second row by $a_1$
$${\begin{vmatrix} -a_1b_1 & -a_2b_1 & -a_3b_1 & \vdots & 0 \\ a_1b_1 & a_1b_2 & a_1b_3 & \vdots & 0 \end{vmatrix}}$$
Adding the equations I and II in the second row
$${\begin{vmatrix} a_1b_1 & a_2b_1 & a_3b_1 & \vdots & 0 \\ 0 & a_1b_2-a_2b_1 & a_1b_3-a_3b_1 & \vdots & 0 \end{vmatrix}}$$
Now, look at the second row as an equation
$${\begin{align} (a_1b_2-a_2b_1)y + (a_1b_3-a_3b_1)z & = 0 \end{align}}$$
Rearranging,
$${\begin{align} (a_1b_2-a_2b_1)y & = -(a_1b_3-a_3b_1)z \end{align}}$$
Writing the RHS without the negative sign (swapping $a_3b_1$
and $a_1b_3$
).
$${\begin{align} (a_1b_2-a_2b_1)y & = (a_3b_1 - a_1b_3)z \end{align}}$$
Next, divide by $(a_1b_2-a_2b_1)$
.
$${\begin{align} y & = \frac{a_3b_1 - a_1b_3}{a_1b_2-a_2b_1} z \end{align}}$$
As we do not have a third equation, we can take any value of z. In this case, for reasons that will become apparent in the process, we choose a random z, say
$$z = a_1b_2-a_2b_1$$
Substituting, to find y we get
$${\Rightarrow \begin{align} y & = a_3b_1 - a_1b_3 \end{align}}$$
Substituting y and z into equation I.
$${\begin{align} a_1x + a_2(a_3b_1 - a_1b_3) + a_3(a_1b_2-a_2b_1) & = 0 \newline a_1x + \color{orange}{ a_2a_3b_1} - a_1a_2b_3 + a_1a_3b_2 \color{orange}{-a_2a_3b_1} & = 0 \newline \color{#b2d232}{a_1}x - \color{#b2d232}{a_1}a_2b_3 + \color{#b2d232}{a_1}a_3b_2 & = 0 \newline x - a_2b_3 + a_3b_2 & = 0 \newline x & = a_2b_3 - a_3b_2 \end{align}}$$
$${\Rightarrow \begin{align} x & = a_2b_3 - a_3b_2 \end{align}}$$
So, one possible normal vector $\vec{n}$
to the vectors $\vec{a}$
and $\vec{b}$
is
$$\vec{n} = \begin{pmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2-a_2b_1 \end{pmatrix}$$
cross product - Das Vektorprodukt
This vector is how we define the cross product $\color{orange}{\vec{a}\times \vec{b}}$
of two vectors $\vec{a}$
and $\vec{b}$
.
$$\begin{align}\vec{a} \times \vec{b} & = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} \times \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} \newline & = \begin{pmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2-a_2b_1 \end{pmatrix}\end{align}$$
How can we remember this?
$${\begin{align} \vec{a} \times \vec{b} & = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} \times \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} \end{align}}$$
video of an example
alternative methods
see rule of Sarrus
using the right hand - ein Rechtssystem
In a “right-handed” system, the axes of the coordinate system are arranged counter-clockwise. This is the mathematically positive sense from x-axis to y-axis to z-axis.
using the left hand instead
As, many write with the right hand, some text books use the left hand to describe the same system. But, here the thumb is the cross product.