definition
Consider, the cosine rule.
From the cosine rule, we know that
$$|\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2 \cdot |\vec{a}| \cdot |\vec{b}| \cdot \cos{\gamma}$$
We know, by definition, that $|\vec{a}|^2 = a_1^2 + a_2^2$
and $|\vec{b}|^2 = b_1^2 + b_2^2$
Also, $$\begin{align}|\vec{c}|^2 & = |\vec{b} - \vec{a}|^2 \\ & = (b_1-a_1)^2 + (b_2 - a_2)^2 \\ & = b_1^2 - 2a_1b_1 + a_1^2 + b_2^2 - 2a_2b_2 + a_2^2 \end{align}$$
Combining this with the cosine rule.
$$\color{orange}{b_1^2} - 2a_1b_1 + \color{orange}{a_1^2 + b_2^2} - 2a_2b_2 + \color{orange}{a_2^2} = \color{orange}{a_1^2 + a_2^2 + b_1^2 + b_2^2} - 2 \cdot |\vec{a}| \cdot |\vec{b}| \cdot \cos{\gamma}$$
$$ \color{orange}{- 2}(a_1b_1 + a_2b_2) = \color{orange}{- 2} \cdot |\vec{a}| \cdot |\vec{b}| \cdot \cos{\gamma}$$
$$ a_1b_1 + a_2b_2 =|\vec{a}| \cdot |\vec{b}| \cdot \cos{\gamma}$$
The scalar product
$\vec{a}\cdot\vec{b}$
is defined as$ a_1b_1 + a_2b_2 = |\vec{a}| \cdot | \vec{b}| \cdot \cos{\gamma}$
for vectors$\vec{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$
and$\vec{b} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$
, where$\gamma$
is the angle enclosed by the vectors pointing away from each other and a and b are the magnitudes, respectively.
Similarly, in 3-dimensional space (or even n-dimensional vector spaces)
$$\vec{a}\cdot\vec{b}= a_1b_1 + a_2b_2 + a_3b_3 = |\vec{a}| \cdot | \vec{b}| \cdot \cos{\gamma}$$
calculating the angle between two vectors
It follows that
$$\cos{\gamma} = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}| \cdot | \vec{b}|}$$
$$\gamma = \arccos{\left(\frac{\vec{a}\cdot\vec{b}}{|\vec{a}| \cdot | \vec{b}|}\right)}$$
scalar product of the vector with itself
$$\vec{a}\cdot\vec{a}= a_1a_1 + a_2a_2 + a_3a_3 = a_1^2 + a_2^2 + a_3^2 = | \vec{a}|^2$$
as $\gamma = 0°$
and therefore $\cos \gamma = 1$
.
perpendicular (orthogonal) vectors - orthogonale Vektoren
As $\cos(90°) = 0$
it follows, by definition that
$$\vec{a}\cdot\vec{b} = 0 \Leftrightarrow \vec{a}\perp \vec{b}$$
area of triangle
$$\begin{align}A & = \frac{1}{2}ab\sin{\gamma} \newline & = \frac{1}{2}| \vec{a}|\cdot | \vec{b}|\sqrt{1 - \cos^2{\gamma}} \newline & = \frac{1}{2}\sqrt{| \vec{a}|^2\cdot | \vec{b}|^2 - | \vec{a}|^2\cdot | \vec{b}|^2\cos^2{\gamma}} \\ & = \frac{1}{2}\sqrt{ \vec{a}^2\cdot \vec{b}^2 - (\vec{a}\cdot \vec{b})^2} \end{align} $$