mirror of https://github.com/2martens/uni.git
430 lines
9.1 KiB
TeX
430 lines
9.1 KiB
TeX
\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
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\usepackage[T1]{fontenc}
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\usepackage[utf8]{inputenc}
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\usepackage[ngerman]{babel}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{paralist}
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\usepackage{gauss}
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\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
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\usepackage{tikz}
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\usetikzlibrary{matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
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\usepackage{polynom}
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\polyset{style=C, div=:,vars=x}
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\pagenumbering{arabic}
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\def\thesection{\arabic{section})}
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\def\thesubsection{\alph{subsection})}
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\def\thesubsubsection{(\roman{subsubsection})}
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\makeatletter
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\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
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\hskip -\arraycolsep
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\let\@ifnextchar\new@ifnextchar
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\array{#1}}
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\makeatother
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\begin{document}
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\author{Jim Martens}
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\title{Hausaufgaben zum 24./25. Januar}
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\maketitle
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\section{} %1
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\subsection{} %a
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\begin{equation*}
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x =
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\begin{bmatrix}
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2 \\
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-1 \\
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3
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\end{bmatrix} + t \cdot
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\begin{bmatrix}
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5 \\
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-2 \\
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-1
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\end{bmatrix}
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\end{equation*}
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\subsection{} %b
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\subsubsection{} %(i)
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\begin{equation*}
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x =
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\begin{bmatrix}
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5 \\
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1 \\
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2
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\end{bmatrix} + s \cdot
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\begin{bmatrix}
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-8 \\
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0 \\
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2
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\end{bmatrix} + t \cdot
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\begin{bmatrix}
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-3 \\
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-2 \\
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1
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\end{bmatrix}
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\end{equation*}
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\subsubsection{} %(ii)
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\begin{alignat*}{2}
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\underset{0D}{\rightarrow} &=&
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\underset{0A}{\rightarrow} + \underset{AB}{\rightarrow} + \underset{AC}{\rightarrow} \\
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&=&
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\begin{bmatrix}
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-6 \\
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-3 \\
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5
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\end{bmatrix} \\
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x &=&
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\begin{bmatrix}
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-6 \\
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-3 \\
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5
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\end{bmatrix} + s \cdot
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\begin{bmatrix}
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3 \\
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4 \\
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-1
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\end{bmatrix} + t \cdot
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\begin{bmatrix}
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8 \\
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2 \\
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-2
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\end{bmatrix}
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\end{alignat*}
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\subsection{} %c
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\begin{alignat*}{3}
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&& \begin{bmatrix}
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x_{1} \\
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x_{2} \\
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x_{3}
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\end{bmatrix} &=&
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\begin{bmatrix}
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5 \\
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1 \\
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2
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\end{bmatrix} + s \cdot
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\begin{bmatrix}
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-8 \\
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0 \\
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2
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\end{bmatrix} + t \cdot
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\begin{bmatrix}
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-3 \\
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-2 \\
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1
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\end{bmatrix} \\
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\Rightarrow & I & x_{1} &=& 5 - 8s - 3t \\
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& II & x_{2} &=& 1 - 2t \\
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& III & x_{3} &=& 2 + 2s + t \\
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\overset{II}{\Rightarrow} && x_{2} &=& 1 - 2t & \;| -1 \\
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&& x_{2} - 1 &=& -2t & \;| \cdot -\frac{1}{2} \\
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&& -\frac{1}{2}x_{2} + \frac{1}{2} &=& t & \\
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\overset{III}{\Rightarrow} && x_{3} &=& 2 + 2s + t & \\
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&& x_{3} &=& 2 + 2s + (-\frac{1}{2}x_{2} + \frac{1}{2}) & \;| \text{Kl. aufl. + Zus.} \\
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&& x_{3} &=& 2s - \frac{1}{2}x_{2} + \frac{5}{2} & \;| + \frac{1}{2}x_{2}, - \frac{5}{2} \\
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&& x_{3} + \frac{1}{2}x_{2} - \frac{5}{2} &=& 2s & \;| \cdot \frac{1}{2} \\
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&& \frac{1}{2}x_{3} + \frac{1}{4}x_{2} - \frac{5}{4} &=& s & \\
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\overset{I}{\Rightarrow} && x_{1} &=& 5 - 8s - 3t & \\
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&& x_{1} &=& 5 - 8(\frac{1}{2}x_{3} + \frac{1}{4}x_{2} - \frac{5}{4}) - 3(-\frac{1}{2}x_{2} + \frac{1}{2}) & \;| \text{Kl. aufl.} \\
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&& x_{1} &=& 5 - 4x_{3} - 2x_{2} + 10 + \frac{3}{2}x_{2} - \frac{3}{2} & \;| \text{Zus.} \\
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&& x_{1} &=& - 4x_{3} - \frac{1}{2}x_{2} + \frac{27}{2} & \;| +4x_{3} + \frac{1}{2}x_{2} - \frac{27}{2} \\
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&& x_{1} + \frac{1}{2}x_{2} + 4x_{3} - \frac{27}{2} &=& 0 &
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\end{alignat*}
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\\
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Schnittpunkt mit der $x_{1}$-Achse:\\
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\begin{alignat*}{2}
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x_{1} + \frac{1}{2}x_{2} + 4x_{3} - \frac{27}{2} &=& 0 & \;| x_{2} = 0, x_{3} = 0 \\
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x_{1} - \frac{27}{2} &=& 0 & \;| + \frac{27}{2} \\
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x_{1} &=& \frac{27}{2} &
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\end{alignat*}
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\\
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Schnittpunkt mit der $x_{2}$-Achse:\\
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\begin{alignat*}{2}
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x_{1} + \frac{1}{2}x_{2} + 4x_{3} - \frac{27}{2} &=& 0 & \;| x_{1} = 0, x_{3} = 0 \\
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\frac{1}{2}x_{2} - \frac{27}{2} &=& 0 & \;| + \frac{27}{2} \\
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\frac{1}{2}x_{2} &=& \frac{27}{2} & \;| \cdot 2 \\
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x_{2} &=& 27 &
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\end{alignat*}
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\\
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Schnittpunkt mit der $x_{3}$-Achse:\\
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\begin{alignat*}{2}
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x_{1} + \frac{1}{2}x_{2} + 4x_{3} - \frac{27}{2} &=& 0 & \;| x_{1} = 0, x_{2} = 0 \\
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4x_{3} - \frac{27}{2} &=& 0 & \;| + \frac{27}{2} \\
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4x_{3} &=& \frac{27}{2} & \;| \cdot \frac{1}{4} \\
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x_{3} &=& \frac{27}{8} &
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\end{alignat*}
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\section{} %2
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\begin{alignat*}{3}
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3x_{1} - 2x_{2} + 8x_{3} - 10 &=& 0 & \;| + 2x_{2} \\
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2x_{2} &=& -10 + 3x_{1} + 8x_{3} & \;| \cdot \frac{1}{2} \\
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x_{2} &=& -5 + \frac{3}{2}x_{1} + 4x_{3} &
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\end{alignat*}
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\begin{alignat*}{5}
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\Rightarrow & x_{1} &=&& 0 &\,+\,& 1x_{1} &\,+\,& 0x_{3} \\
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& x_{2} &=&& -5 &\,+\,& \frac{3}{2}x_{1} &\,+\,& 4x_{3}\\
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& x_{3} &=&& 0 &\,+\,& 0x_{1} &\,+\,& 1x_{3}
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\end{alignat*}
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Daraus ergibt sich:\\
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\begin{alignat*}{2}
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\begin{bmatrix}
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x_{1} \\
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x_{2} \\
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x_{3}
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\end{bmatrix} &=&
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\begin{bmatrix}
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0 \\
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-5 \\
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0
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\end{bmatrix} + x_{1}
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\begin{bmatrix}
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1 \\
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\frac{3}{2} \\
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0
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\end{bmatrix} + x_{3}
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\begin{bmatrix}
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0 \\
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4 \\
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1
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\end{bmatrix} \\
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\intertext{$x_{1} = s, x_{3} = t$}
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x &=&
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\begin{bmatrix}
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0 \\
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-5 \\
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0
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\end{bmatrix} + s
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\begin{bmatrix}
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1 \\
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\frac{3}{2} \\
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0
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\end{bmatrix} + t
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\begin{bmatrix}
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0 \\
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4 \\
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1
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\end{bmatrix}
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\end{alignat*}
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\section{} %3
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\subsection{} %a
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\begin{alignat*}{3}
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& a \cdot b &=& (2 \cdot 3) + (-1 \cdot 2) + (5 \cdot (-3)) & \;|\text{Kl. aufl.} \\
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& &=& 6 - 2 - 15 & \;|\text{Zus.} \\
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& &=& -11 & \\
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\intertext{Einsetzen von $z$}
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& a \cdot b &=& (2 \cdot 3) + (-1 \cdot 2) + (5 \cdot z) & \;|\text{Kl. aufl.} \\
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& &=& 6 - 2 + 5z & \;|\text{Zus.} \\
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& &=& 4 + 5z & \\
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\Rightarrow & 0 &=& 4 + 5z & \;| -4 \\
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& 5z &=& -4 & \;| \cdot \frac{1}{5} \\
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& z &=& -\frac{4}{5} &
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\end{alignat*}
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$z$ ist $-\frac{4}{5}$.
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\subsection{} %b
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\begin{alignat*}{2}
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|a| &=& \sqrt{2^{2} + (-1)^{2} + 5^{2}} \\
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&=& \sqrt{4 + 1 + 25} \\
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&=& \sqrt{30} \\
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&\approx & 5,48
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\end{alignat*}
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\\
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Errechnung von d:\\
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\begin{alignat*}{2}
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& \begin{bmatrix}
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d_{1} \\
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d_{2} \\
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d_{3}
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\end{bmatrix} &=& \frac{1}{\sqrt{30}} \cdot
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\begin{bmatrix}
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2 \\
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-1 \\
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5
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\end{bmatrix} \\
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\Rightarrow I: & d_{1} &=& \frac{2}{\sqrt{30}} \\
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II: & d_{2} &=& -\frac{1}{\sqrt{30}} \\
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III: & d_{3} &=& \frac{5}{\sqrt{30}}
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\end{alignat*}
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\subsection{} %c
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\begin{alignat*}{2}
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|P_{1}P_{2}| &=& \sqrt{(0-4)^{2} + (3-2)^{2} + (1-1)^{2}} \\
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&=& \sqrt{-4^{2} + {1}^{2}} \\
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&=& \sqrt{16 + 1} \\
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&=& \sqrt{17} \\
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&\approx & 4,12
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\end{alignat*}
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\subsection{} %d
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\begin{alignat*}{2}
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cos \phi &=& \frac{u \cdot v}{|u| \cdot |v|} \\
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&=& \frac{2}{4 \cdot \sqrt{5}} \\
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&=& \frac{1}{2 \cdot \sqrt{5}} \\
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&\approx & 0,22 \\
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arccos(0,22) &\approx & 77,29^{\circ}
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\end{alignat*}
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\section{} %4
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\subsection{} %a
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Es gilt: \\
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\begin{alignat*}{2}
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x \cdot u &=& 0 \\
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x \cdot v &=& 0 \\
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\begin{bmatrix}
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x_{1} \\
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x_{2} \\
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x_{3}
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\end{bmatrix} \cdot
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\begin{bmatrix}
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1 \\
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2 \\
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3
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\end{bmatrix} &=& 0 \\
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\begin{bmatrix}
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x_{1} \\
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x_{2} \\
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x_{3}
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\end{bmatrix} \cdot
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\begin{bmatrix}
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-4 \\
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-7 \\
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5
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\end{bmatrix} &=& 0 \\
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\intertext{In LGS umformen}
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I: 1x_{1} + 2x_{2} + 3x_{3} &=& 0 \\
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II: -4x_{1} - 7x_{2} + 5x_{3} &=& 0 \\
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\intertext{II = II + 4I}
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1x_{1} + 2x_{2} + 3x_{3} &=& 0 \\
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0x_{1} + 1x_{2} + 17x_{3} &=& 0 \\
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\intertext{II = II - $17x_{3}$}
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x_{2} &=& -17x_{3} \\
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\intertext{In I einsetzen}
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1x_{1} + 2 \cdot (-17x_{3}) + 3x_{3} &=& 0 \\
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x_{1} - 31x_{3} &=& 0 \\
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\intertext{I = I + $31x_{3}$}
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x_{1} &=& 31x_{3} \\
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\intertext{Sei $x_{3} = t, t \in \mathbb{R}$}
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x_{1} &=& 31t \\
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\intertext{Daraus ergibt sich für $x_{2}$}
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x_{2} &=& -17t \\
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\intertext{Daraus folgt für $x$}
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x &=& \begin{bmatrix}
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31t \\
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-17t \\
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t
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\end{bmatrix}, \; t \in \mathbb{R} \\
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&=& t \cdot \begin{bmatrix}
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31 \\
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-17 \\
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1
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\end{bmatrix}, \; t \in \mathbb{R}
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\end{alignat*}
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\subsection{} %b
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Es gilt:\\
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\begin{alignat*}{2}
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x \cdot u &=& 0 \\
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\begin{bmatrix}
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x_{1} \\
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x_{2} \\
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x_{3}
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\end{bmatrix} \cdot
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\begin{bmatrix}
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2 \\
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4 \\
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1
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\end{bmatrix} &=& 0 \\
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\intertext{In LGS umwandeln}
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2x_{1} + 4x_{2} + x_{3} &=& 0 \\
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\intertext{Umstellen nach $x_{3}$}
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x_{3} &=& -2x_{1} - 4x_{2} \\
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\intertext{Sei $x_{1} = s, s \in \mathbb{R}$ und $x_{2} = t, t \in \mathbb{R}$}
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x_{3} &=& -2s - 4t \\
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\intertext{Daraus ergibt sich für $x$}
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x &=& \begin{bmatrix}
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s \\
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t \\
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-2s - 4t
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\end{bmatrix}, \; s,t \in \mathbb{R} \\
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&=& s \cdot \begin{bmatrix}
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1 \\
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0 \\
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-2
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\end{bmatrix} + t \cdot
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\begin{bmatrix}
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0 \\
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1 \\
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-4
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\end{bmatrix}, \; s,t \in \mathbb{R}
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\end{alignat*}
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\subsection{} %c
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Es gilt: \\
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\begin{alignat*}{2}
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x \cdot u &=& 0 \\
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x \cdot v &=& 0 \\
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\begin{bmatrix}
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x_{1} \\
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x_{2} \\
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x_{3} \\
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x_{4}
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\end{bmatrix} \cdot
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\begin{bmatrix}
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1 \\
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2 \\
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-1 \\
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2
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\end{bmatrix} &=& 0 \\
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\begin{bmatrix}
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x_{1} \\
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x_{2} \\
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x_{3} \\
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x_{4}
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\end{bmatrix} \cdot
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\begin{bmatrix}
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1 \\
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1 \\
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4 \\
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2
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\end{bmatrix} &=& 0 \\
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\intertext{In LGS umformen}
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I: 1x_{1} + 2x_{2} - 1x_{3} + 2x_{4} &=& 0 \\
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II: 1x_{1} + 1x_{2} + 4x_{3} + 2x_{4} &=& 0 \\
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\intertext{II = II - I}
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1x_{1} + 2x_{2} - 1x_{3} + 2x_{4} &=& 0 \\
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0x_{1} - 1x_{2} + 5x_{3} + 0x_{4} &=& 0 \\
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\intertext{II = II + $x_{2}$}
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x_{2} &=& 5x_{3} \\
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\intertext{In I einsetzen}
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x_{1} + 2 \cdot (5x_{3}) - x_{3} + 2x_{4} &=& 0 \\
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x_{1} + 9x_{3} + 2x_{4} &=& 0 \\
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\intertext{Nach $x_{1}$ umstellen}
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x_{1} &=& -9x_{3} - 2x_{4} \\
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\intertext{Sei $x_{3} = s, s \in \mathbb{R}$ und $x_{4} = t, t \in \mathbb{R}$}
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x_{1} &=& -9s - 2t \\
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\intertext{Daraus ergibt sich für $x_{2}$}
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x_{2} &=& 5s \\
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\intertext{Daraus folgt für $x$}
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x &=& \begin{bmatrix}
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-9s - 2t \\
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5s \\
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s \\
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t
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\end{bmatrix}, \; s,t \in \mathbb{R} \\
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&=& s \cdot \begin{bmatrix}
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-9 \\
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5 \\
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1 \\
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0
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\end{bmatrix} + t \cdot
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\begin{bmatrix}
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-2 \\
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0 \\
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0 \\
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1
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\end{bmatrix}, \; s,t \in \mathbb{R}
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\end{alignat*}
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\subsection{} %d
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\begin{alignat*}{2}
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|u| &=& \sqrt{1^{2} + 2^{2} + (-1)^{2} + 2^{2}} \\
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&=& \sqrt{1 + 4 + 1 + 4} \\
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&=& \sqrt{10} \\
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&\approx & 3,16 \\
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|v| &=& \sqrt{1^{2} + 1^{2} + 4^{2} + 2^{2}} \\
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&=& \sqrt{1 + 1 + 16 + 4} \\
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&=& \sqrt{22} \\
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&\approx & 4,69
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\end{alignat*}
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\end{document}
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