FHTW Typst bei Abbildung 4

.gitignore

@@ -0,0 +1,3 @@
+*.log
+*.synctex
+*.aux

FHTW.typ

@@ -0,0 +1,69 @@
+#import "@preview/cetz:0.4.2"
+#import "LineareAlgebraGrafiken.typ": * 
+
+#set page(
+  margin: 2cm,
+  paper: "a4"
+  )
+
+#title("Mathematik FHTW Berlin")
+
+= Lineare Algebra
+== Vektoren
+=== Veranschaulichung von Vektoren in der Ebene
+
+#show math.equation: set text(font: "New Computer Modern Sans Math", size: 12pt)
+
+#show math.equation.where(block: false): it => math.display(it)
+
+
+//#set page(width: auto, height: auto, margin: .5cm)
+
+#table(
+  columns: 3,
+  stroke: none,
+  [#abb1()],[#h(1cm)], [#abb2()]
+)//&#show math.equation: block.with(fill: white, inset: 1pt)
+
+
+
+
+
+$arrow(a)=(a_1, a_2)$, auch üblich $arrow(a)=vec(a_1, a_2) arrow.l$ geordnetes Paar
+
+=== Menge aller Vektoren in der Ebene
+Die Menge aller Vektoren in der Ebene heißt $RR^2$; dabei ist $RR$ die Menge der reellen Zahlen. 
+
+Also: $RR={arrow(a)=(a_1,a_2) | a_1 in RR, a_2 in RR }$
+
+
+=== Addition von Vektoren in der Ebene
+#table(
+  columns: 2,
+  stroke: none,
+  [- liefert wieder einen Vektor
+- rechnerisch: 
+  - $arrow(a)=(a_1, a_2)$, $arrow(b)=(b_1, b_2)$ 
+  - $arrow(a)+arrow(b)=(a_1+b_1, a_2+b_2)$ 
+- Rechenregeln:
+  - $arrow(a)+arrow(b)=arrow(b)+arrow(a)$
+  - $arrow(a)+(arrow(b)+arrow(c)))$; $(arrow(a)+arrow(b))=arrow(c)$],[zeichnerisch:#abb3()]
+)
+
+#pagebreak()
+=== Nullvektor in der Ebene
+
+- $arrow(0)=(0,0)$
+- Rechenregel: $arrow(a)+arrow(0)=arrow(a)$
+
+=== Subtraktion von Vektoren in der Ebene
+
+#table(
+  columns: 2,
+  stroke: none,
+  [- liefert wieder einen Vektor
+- rechnerisch:
+  - $arrow(a)=(a_1,a_2)$, $arrow(b)=(b_1, b_2)$
+  - $arrow(a)-arrow(b)= (a_1-b_1, a_2-b_2$)  ],[#abb4()]
+)
+

LineareAlgebra.tex

@@ -1,3 +1,4 @@
+%!TEX root=MathematikFHTW.tex
 \chapter{Lineare Algebra}
 \section{Vektoren in der Ebene - Übersicht}
 
@@ -28,4 +29,4 @@
 \input{LineareAbhaengigkeit_von_n_dimensionalen_Vektoren.tex}
 \input{Laenge_eines_n_dimensionalen_Vektors.tex}%%04_1%%
 \input{N_Dimensionale_Einheitsvektoren.tex}%%04_1%%
-\input{Skalarprodukt_von_zwei_n_dimensionalen_Vektoren.tex}%%04_1%% 
+\input{Skalarprodukt_von_zwei_n_dimensionalen_Vektoren.tex} \input{Oeffnungswinkel_zwischen_zwei_n_dimensionalen_Vektoren.tex}% 

LineareAlgebraGrafiken.typ

@@ -0,0 +1,205 @@
+#import "@preview/cetz:0.3.3" //Grafiken
+#set text(font: "New Computer Modern Sans Math")
+
+#let abb1() = {
+
+cetz.canvas(length: 3cm, {
+  import cetz.draw: *
+
+  set-style(
+    mark: (fill: black, scale: 2),
+    stroke: (thickness: 0.4pt, cap: "round"),
+    angle: (
+      radius: 0.3,
+      label-radius: .22,
+      fill: green.lighten(80%),
+      stroke: (paint: green.darken(50%))
+    ),
+    content: (padding: 1pt)
+  )
+
+  //grid((-.25, -.25), (2, 2 ), step: 0.5, stroke: gray + 0.2pt)
+
+  line((-.05, 0), (1.85, 0), mark: (end: "stealth", scale: 0.8))
+  content((), $ x $, anchor: "west")
+  line((0, -.05), (0, 1.85), mark: (end: "stealth", scale: 0.8))
+  content((), $ y $, anchor: "south")
+
+  for (x, ct) in ((0, $ 0 $), (0.5, $ 1 $), (1, $ 2 $), (1.5, $ 3 $), (1.25, $a_1$)) {
+    line((x, 3pt), (x, -3pt))
+    content((), anchor: "north", ct)
+  }
+
+  for (y, ct) in ((0.5, $ 1 $), (1, $ 2 $), (1.5, $ 3 $), (1.25, $a_2$)) {
+    line((3pt, y), (-3pt, y))
+    content((), anchor: "east", ct)
+  }
+
+  
+
+  //line((0,0), (1.25, 1.25),name: "vec",mark: (end: "stealth", fill: green.darken(30%)),
+  //stroke: (paint: green.darken(30%), thickness: 1.25pt))
+
+line((0,0), (1.25, 1.25),name: "vec",stroke: (paint: red.darken(30%), thickness: 1.25pt))
+
+
+  line((1.25,0), (1.25, 1.25), stroke: (dash: "dashed", paint: blue.darken(30%), thickness: 0.5pt))
+
+  line((0,1.25), (1.25, 1.25), stroke: (dash: "dashed", paint: blue.darken(30%), thickness: 0.5pt))
+
+circle((1.25,1.25), radius: 0.025, fill: green.darken(30%), stroke: (paint: green.darken(30%), thickness: 1.25pt), name: "punkt")
+  content("punkt.end", anchor: "south-west", padding: (left: 3mm, bottom: 2mm),  text(green.darken(30%))[Punkt $(a_1, a_2)$])})
+}
+
+#let abb2() = {
+  cetz.canvas(length: 3cm, {
+  import cetz.draw: *
+
+  set-style(
+    mark: (fill: black, scale: 2),
+    stroke: (thickness: 0.4pt, cap: "round"),
+    angle: (
+      radius: 0.3,
+      label-radius: .22,
+      fill: green.lighten(80%),
+      stroke: (paint: green.darken(50%))
+    ),
+    content: (padding: 1pt)
+  )
+
+//  grid((-.25, -.25), (2, 2 ), step: 0.5, stroke: gray + 0.2pt)
+
+  line((-.05, 0), (1.85, 0), mark: (end: "stealth", scale: 0.8))
+  content((), $ x $, anchor: "west")
+  line((0, -.05), (0, 1.85), mark: (end: "stealth", scale: 0.8))
+  content((), $ y $, anchor: "south")
+
+  for (x, ct) in ((0, $ 0 $), (0.5, $ 1 $), (1, $ 2 $), (1.5, $ 3 $), (1.25, $a_1$)) {
+    line((x, 3pt), (x, -3pt))
+    content((), anchor: "north", ct)
+  }
+
+  for (y, ct) in ((0.5, $ 1 $), (1, $ 2 $), (1.5, $ 3 $), (1.25, $a_2$)) {
+    line((3pt, y), (-3pt, y))
+    content((), anchor: "east", ct)
+  }
+
+
+line((0,0), (1.25, 1.25),name: "vec",stroke: (paint: red.darken(30%), thickness: 1.25pt))
+
+
+  line((1.25,0), (1.25, 1.25), stroke: (dash: "dashed", paint: blue.darken(30%), thickness: 0.5pt))
+
+  line((0,1.25), (1.25, 1.25), stroke: (dash: "dashed", paint: blue.darken(30%), thickness: 0.5pt))
+
+  line((0,0), (1.25, 1.25),name: "vec",mark: (end: "stealth", fill: green.darken(30%),scale: 0.8), stroke: (paint: green.darken(30%), thickness: 1.25pt))
+
+  content("vec.end", anchor: "north-west", padding: (left: 3mm, top: -2mm), text(green.darken(30%))[Vektor $arrow(a)$])
+})
+
+}
+
+#let abb3() = {
+  cetz.canvas(length: 40mm,
+   {
+  import cetz.draw: *
+
+  set-style(
+    mark: (fill: black, scale: 2),
+    stroke: (thickness: 0.4pt, cap: "round"),
+    angle: (
+      radius: 0.3,
+      label-radius: .22,
+      fill: green.lighten(80%),
+      stroke: (paint: green.darken(50%))
+    ),
+    content: (padding: 1pt)
+  )
+
+  //grid((-.25, -.25), (2.5 , 2 ), step: 0.5, stroke: gray + 0.2pt)
+
+  line((-.05, 0), (80mm, 0), mark: (end: "stealth",scale: 0.9))
+  content((), $ x $, anchor: "west")
+  line((0, -.05), (0, 70mm), mark: (end: "stealth",scale: 0.9))
+  content((), $ y $, anchor: "south")
+
+  for (x, ct) in ((10mm, $ 1 $), (20mm, $ b_1$), (30mm, $ 3 $),(40mm, $a_1$), (50mm,$5$), (60mm, $a_1 + b_1$),(70mm, $7$)) {
+    line((x, 3pt), (x, -3pt))
+    content((), anchor: "north", ct)
+  }
+
+  for (y, ct) in ((10mm, $ a_2 $), (20mm, $ 2 $), (30mm, $ 3 $), (40mm, $b_2$),(1.25,$a_2+b_2$), (1.5, $6$)) {
+    line((3pt, y), (-3pt, y))
+    content((), anchor: "east", ct)
+  }
+
+  line((0,0), (40mm, 10mm),name: "veca",mark: (end: "stealth", fill: blue.darken(30%),scale:0.9), stroke: (paint: blue.darken(30%), thickness: 0.75pt))
+  
+ content("veca.100%", anchor: "west", padding: (left:1mm, bottom:-5mm),text(blue.darken(30%))[$arrow(a)$])
+
+ line((40mm,0), (40mm, 10mm ), stroke: (dash: "dotted",paint: blue.darken(30%), thickness: 0.5pt))
+
+ line((0,10mm), (40mm, 10mm), stroke: (dash: "dotted",paint: blue.darken(30%), thickness: 0.5pt))
+ 
+ line((0,0), (20mm, 40mm),name: "vecb",mark: (end: "stealth", fill: red.darken(30%),scale: 0.9), stroke: (paint: red.darken(30%), thickness: 0.75pt))
+
+ content("vecb.80%", anchor: "east", padding: (left:0mm, bottom:5mm),text(red.darken(30%))[$arrow(b)$])
+
+ line((20mm,0mm), (20mm, 40mm), stroke: (dash: "dotted",paint: red.darken(30%), thickness: 0.5pt))
+
+ line((0mm,40mm), (20mm, 40mm), stroke: (dash: "dotted",paint: red.darken(30%), thickness: 0.5pt))
+
+ line((0,0), (60mm, 50mm),name: "vec_ab",mark: (end: "stealth", fill: orange ,scale: 0.9), stroke: (paint: orange, thickness: 0.75pt))
+
+ content("vec_ab.100%", anchor: "east", padding: (left:0mm, bottom:5mm),text(green.darken(60%))[$arrow(a) + arrow(b)$])
+
+ line((20mm,40mm), (60mm, 50mm),name: "veca_dotted",mark: (end: "stealth", scale: 0.9), stroke: (dash: "dotted", paint: blue, thickness: 0.75pt))
+ 
+ line((40mm,10mm), (60mm, 50mm),name: "vecb_dotted",mark: (end: "stealth", scale: 0.9), stroke: (dash: "dotted", paint: red, thickness: 0.75pt))
+
+ content((50mm, 20mm), [Parallelogramm])
+
+  arc(
+    (48mm, 18mm),
+    radius: 0.25,
+    start: -45deg,
+    delta: -90deg,
+    mark: (
+      end: "stealth",
+    ))
+
+  line((60mm,0mm), (60mm, 50mm),name: "vecab_dotted", stroke: (dash: "dotted", paint: orange, thickness: 0.75pt))
+
+  line((0mm,50mm), (60mm, 50mm),name: "vecab_dotted",stroke: (dash: "dotted", paint: orange, thickness: 0.75pt))
+    
+  })}
+
+  #let abb4() = {
+
+      cetz.canvas(length: 40mm,
+   {
+  import cetz.draw: *
+
+  set-style(
+    mark: (fill: black, scale: 2),
+    stroke: (thickness: 0.4pt, cap: "round"),
+    angle: (
+      radius: 0.3,
+      label-radius: .22,
+      fill: green.lighten(80%),
+      stroke: (paint: green.darken(50%))
+    ),
+    content: (padding: 1pt)
+  )
+
+  //grid((-.25, -.25), (2.5 , 2 ), step: 0.5, stroke: gray + 0.2pt)
+
+  line((-30mm, 0), (50mm, 0), mark: (end: "stealth",scale: 0.9))
+  content((), $ x $, anchor: "west")
+  line((0, -.05), (0, 70mm), mark: (end: "stealth",scale: 0.9))
+  content((), $ y $, anchor: "south")
+
+  
+})
+    
+  }

Loesung120.tex

@@ -3,19 +3,19 @@
 Sei $x=\sum\limits_{k=0}^{\infty}\left(  -1\right)  ^{k}\frac{1}{k};$ \ $\widetilde{x}= \sum\limits_{k=0}^{4}\left(  -1\right)  ^{k}\frac{1}{k!}$
 
 \begin{description}
-\item[a.] Mit Hilfe von welcher speziellen Funktion l\"{a}\ss t sich $x$ genau beschreiben? Wie? (Tip: 3.3.5)
+\item[a.] Mit Hilfe von welcher speziellen Funktion läßt sich $x$ genau beschreiben? Wie? (Tip: 3.3.5)
 
-\item[b.] Berechnen Sie $\widetilde{x}$.
+\item[b.] Berechnen Sie $\displaystyle \widetilde{x}$.
 
-\item[c.] Geben Sie einen absoluten H\"{o}chstfehler von $\widetilde{x}$ an.
+\item[c.] Geben Sie einen absoluten Höchstfehler von $\widetilde{x}$ an.
 (Tip: 3.2.7)
 \end{description}
 
 \subsection{Lösung}
 \begin{description}
-\item[a.] $\exp(z)=\sum\limits_{k=0}^{\infty}\frac{1}{k!}z^{k}$ \ $\Longrightarrow$ \ \ $x=\sum\limits_{k=0}^{\infty}\frac{1}{k!}(-1)^{k}=\exp(-1)=\underline{\underline{\frac{1}{e}}}$
+\item[a.] $\displaystyle \exp(z)=\sum\limits_{k=0}^{\infty}\frac{1}{k!}z^{k}$ \ $\Longrightarrow$ \ \ $x=\sum\limits_{k=0}^{\infty}\frac{1}{k!}(-1)^{k}=\exp(-1)=\underline{\underline{\frac{1}{e}}}$
 
-\item[b.] $\widetilde{x}=\sum\limits_{k=0}^{4}\frac{1}{k!}(-1)^{k}=\allowbreak 1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}=\frac{12-4+1}{24}=\allowbreak\frac{9}{24}=\frac{3}{8}=\allowbreak\underline{\underline{0.375}}\,$
+\item[b.] $\displaystyle \widetilde{x}=\sum\limits_{k=0}^{4}\frac{1}{k!}(-1)^{k}=\allowbreak 1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}=\frac{12-4+1}{24}=\allowbreak\frac{9}{24}=\frac{3}{8}=\allowbreak\underline{\underline{0.375}}\,$
 
-\item[c.] Da die vorliegende Reihe eine alternierende Reihe ist, gilt $|x-\widetilde{x}|\leq\frac{1}{5!}=\frac{1}{120}=8.\overline{3}\cdot 10^{-3}$.  Damit ist $\underline{\underline{\alpha_x=8.\overline{3}\cdot 10^{-3}}}$ ein absoluter Höchstfehler von $\widetilde{x}$.
+\item[c.] Da die vorliegende Reihe eine alternierende Reihe ist, gilt $\displaystyle |x-\widetilde{x}|\leq\frac{1}{5!}=\frac{1}{120}=8.\overline{3}\cdot 10^{-3}$.  Damit ist $\underline{\underline{\alpha_x=8.\overline{3}\cdot 10^{-3}}}$ ein absoluter Höchstfehler von $\widetilde{x}$.
 \end{description} 

Loesung125.tex

@@ -1,9 +1,9 @@
 \section{Aufgabe 125}
 
-Gegeben sei das eindeutig lösbare lineare Gleichungssystem $\ A\cdot
+Gegeben sei das eindeutig lösbare lineare Gleichungssystem $\displaystyle A\cdot
 \overrightarrow{x}=\overrightarrow{b}$ mit
 
-$A=\left(
+$\displaystyle A=\left(
 \begin{array}
 [c]{cccccc}%
 4 & -1 & 0 & -1 & 0 & 0\\
@@ -26,48 +26,48 @@ $A=\left(
 \right)  $
 
 \begin{itemize}
-\item[a.] Sei $\overrightarrow{x}^{\left(  0\right)  }=\overrightarrow{0}$.
-Berechnen Sie die Näherungslösung $\overrightarrow{x}^{\left(
+\item[a.] Sei $\displaystyle \overrightarrow{x}^{\left(  0\right)  }=\overrightarrow{0}$.
+Berechnen Sie die Näherungslösung $\displaystyle \overrightarrow{x}^{\left(
 3\right)  }$ des Systems, die man nach 3 Schritten des
 Gesamtschrittverfahrens erhält.
 
 \item[b.] Zeigen Sie, daß das Gesamtschrittverfahren konvergiert.
 
 \item[c.] Führen Sie eine Apeoteriori-Fehlerabschätzung für
-$\overrightarrow{x}^{\left(  3\right)  }$\ durch.
+$\displaystyle \overrightarrow{x}^{\left(  3\right)  }$\ durch.
 
 \item[d.] Führen Sie eine Apriori-Fehlerabschätzung für
-$\overrightarrow{x}^{\left(  10\right)  }$ durch.
+$\displaystyle \overrightarrow{x}^{\left(  10\right)  }$ durch.
 \end{itemize}
 
 \subsection{Lösung}
 
 \begin{itemize}
-\item[a.] Rechenvorschriften:\newline$x_{1}^{\left(  Z\right)  }=\frac{1}%
+\item[a.] Rechenvorschriften:\newline$\displaystyle x_{1}^{\left(  Z\right)  }=\frac{1}%
 {4}\left(  2+x_{2}^{\left(  Z-1\right)  }+x_{4}^{\left(  Z-1\right)  }\right)
 =\frac{1}{2}+\frac{1}{4}x_{2}^{\left(  Z-1\right)  }+\frac{1}{4}x_{4}^{\left(
-Z-1\right)  }$\newline$x_{2}^{\left(  Z\right)  }=\frac{1}{4}\left(
+Z-1\right)  }$\newline$\displaystyle x_{2}^{\left(  Z\right)  }=\frac{1}{4}\left(
 1+x_{1}^{\left(  Z-1\right)  }+x_{3}^{\left(  Z-1\right)  }+x_{5}^{\left(
 Z-1\right)  }\right)  =\frac{1}{4}+\frac{1}{4}x_{1}^{\left(  Z-1\right)
 }+\frac{1}{4}x_{3}^{\left(  Z-1\right)  }+\frac{1}{4}x_{5}^{\left(
-Z-1\right)  }$\newline$x_{3}^{\left(  Z\right)  }=\frac{1}{4}\left(
+Z-1\right)  }$\newline$\displaystyle x_{3}^{\left(  Z\right)  }=\frac{1}{4}\left(
 2+x_{2}^{\left(  Z-1\right)  }+x_{6}^{\left(  Z-1\right)  }\right)  =\frac
 {1}{2}+\frac{1}{4}x_{2}^{\left(  Z-1\right)  }+\frac{1}{4}x_{6}^{\left(
-Z-1\right)  }$\newline$x_{4}^{\left(  Z\right)  }=\frac{1}{4}\left(
+Z-1\right)  }$\newline$\displaystyle x_{4}^{\left(  Z\right)  }=\frac{1}{4}\left(
 2+x_{1}^{\left(  Z-1\right)  }+x_{5}^{\left(  Z-1\right)  }\right)  =\frac
 {1}{2}+\frac{1}{4}x_{1}^{\left(  Z-1\right)  }+\frac{1}{4}x_{5}^{\left(
-Z-1\right)  }$\newline$x_{5}^{\left(  Z\right)  }=\frac{1}{4}\left(
+Z-1\right)  }$\newline$\displaystyle x_{5}^{\left(  Z\right)  }=\frac{1}{4}\left(
 1+x_{2}^{\left(  Z-1\right)  }+x_{4}^{\left(  Z-1\right)  }+x_{6}^{\left(
 Z-1\right)  }\right)  =\frac{1}{4}+\frac{1}{4}x_{2}^{\left(  Z-1\right)
 }+\frac{1}{4}x_{4}^{\left(  Z-1\right)  }+\frac{1}{4}x_{6}^{\left(
-Z-1\right)  }$\newline$x_{6}^{\left(  Z\right)  }=\frac{1}{4}\left(
+Z-1\right)  }$\newline$\displaystyle x_{6}^{\left(  Z\right)  }=\frac{1}{4}\left(
 2+x_{3}^{\left(  Z-1\right)  }+x_{5}^{\left(  Z-1\right)  }\right)  =\frac
 {1}{2}+\frac{1}{4}x_{3}^{\left(  Z-1\right)  }+\frac{1}{4}x_{5}^{\left(
 Z-1\right)  }$\newline\newline%
 \begin{tabular}
 [c]{l||llllll}%
-z & $x_{1}^{\left(  Z\right)  }$ & $x_{2}^{\left(  Z\right)  }$ &
-$x_{3}^{\left(  Z\right)  }$ & $x_{4}^{\left(  Z\right)  }$ & $x_{5}^{\left(
+z & $\displaystyle x_{1}^{\left(  Z\right)  }$ & $\displaystyle x_{2}^{\left(  Z\right)  }$ &
+$x_{3}^{\left(  Z\right)  }$ & $\displaystyle x_{4}^{\left(  Z\right)  }$ & $\displaystyle x_{5}^{\left(
 Z\right)  }$ & $x_{6}^{\left(  Z\right)  }$\\\hline\hline
 1 & $0.5$ & $0.25$ & $0.5$ & $0.5$ & $0.25$ & $0.5$\\
 2 & $0.6875$ & $0.5625$ & $0.6875$ & $0.6875$ & $0.5625$ & $0.6875$\\

Oeffnungswinkel_zwischen_zwei_n_dimensionalen_Vektoren.tex

@@ -0,0 +1,6 @@
+
+%!TEX root=MathematikFHTW.tex
+\subsection{Öffnungswinkel zwischen zwei n-dimensionalen Vektoren $\vec{a}\neq \vec{0}$ und $\vec{b}\neq \vec{0}$}
+\begin{itemize}
+\item $\cos \left(\alpha\right)$	
+\end{itemize}

Skalarprodukt_von_zwei_n_dimensionalen_Vektoren.tex

@@ -1,5 +1,9 @@
+%!TEX root=MathematikFHTW.tex
 \subsection{Skalarprodukt von zwei n-dimensionalen Vektoren}
 \begin{itemize}
 	\item liefert einen Skalar
 	\item Berechnung: $\vec{a}=\left(a_1,\ldots,a_n\right),\;\vec{b}=\left(b_1,\ldots,b_n\right)$
+
+$\displaystyle<\vec{a},\vec{b}>=a_1b_1+ \ldots +a_nb_n= \sum_{i=1}^n a_1b_i $
+\item Rechenregel wie für $n=2$
 \end{itemize}

tikzgrafiken/Grafiken.tex

@@ -0,0 +1,25 @@
+%\documentclass[convert={density=400,outext=.png}]{standalone}
+
+\documentclass[11pt]{book}
+\usepackage{amssymb}
+\usepackage{pgf,tikz,pgfplots}
+\usetikzlibrary{calc}
+
+\usetikzlibrary{arrows,shapes,calc,decorations.pathreplacing,fit,positioning,backgrounds}
+
+\usetikzlibrary{shadings,patterns}
+
+\usepackage{pst-fractal}
+
+\definecolor{azure}{rgb}{0.0, 0.5, 1.0}
+
+\begin{document}
+
+
+\input{tikz_3_2.tex}
+
+\newpage
+
+\input{tikz_3_3.tex}
+
+\end{document}

tikzgrafiken/tikz_3_2.tex

@@ -0,0 +1,29 @@
+%!TEX root=Grafiken.tex
+
+\usetikzlibrary{arrows}
+\usetikzlibrary{snakes}
+
+\begin{tikzpicture}[scale=1.5]
+
+% https://tex.stackexchange.com/questions/175016/how-is-arc-defined-in-tikz
+
+%\draw (x,y) arc (start:stop:radius); draws an arc
+
+%with radius radius
+%starts from (x,y)
+%with center (x-r*cos(start), y-r*sin(start)) and
+%ends at (x-r*cos(start)+r*cos(stop), y-r*sin(start)+r*sin(stop)).
+
+
+
+
+
+    \draw[help lines] (-5,-5) grid (5,5);
+  
+	\draw[red] (1,0) arc (0:56:1) node[midway,xshift=-15, yshift=-6]{$\alpha$};
+	
+        \draw [thick,-latex] (0,0) -- (3,0);
+        \draw [thick,dashed] (1,1.5) -- (4,1.5);
+        \draw [thick, -latex] (0,0) -- (1,1.5);
+	    \draw [thick,dashed] (3,0) -- (4,1.5);
+\end{tikzpicture}

tikzgrafiken/tikz_3_3.tex

@@ -0,0 +1,20 @@
+%!TEX root=Grafiken.tex
+
+\usetikzlibrary{arrows}
+\usetikzlibrary{snakes}
+
+\begin{tikzpicture}[scale=1.5]
+
+    \draw[help lines] (-5,-5) grid (5,5);
+
+	
+    \draw [thick,-latex] (0,0) -- (3,0);
+	 \draw [thick,-latex] (0,0) -- (1,2);
+	 \draw [thick,-latex] (3,0) -- (4.2,0.5);
+	 \draw [thick,-latex, dashed] (0,0) -- (1.2,0.5);
+	 \draw [thick, dashed] (1.2,0.5) -- (4.2,0.5);
+
+ %       \draw [thick,dashed] (1,1.5) -- (4,1.5);
+ %       \draw [thick, -latex] (0,0) -- (1,1.5);
+%	    \draw [thick,dashed] (3,0) -- (4,1.5);
+\end{tikzpicture}