vault backup: 2023-04-24 20:21:22
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9406
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9406
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@@ -1,37 +1,5 @@
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# 1
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Um die Koordinaten von Vektor $x$ bezüglich der neuen Basis zu finden, müssen wir zunächst den Vektor $x$ in der Standardbasis berechnen. Dazu verwenden wir die gegebenen Koordinaten $\alpha=1$, $\beta=-1$ und $\gamma=1$ und die Basisvektoren $a$, $b$ und $c$.
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$x = \alpha a + \beta b + \gamma c = 1 \cdot \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} - 1 \cdot \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix} + 1 \cdot \begin{bmatrix} 3 \\ 10 \\ 2 \end{bmatrix} = \begin{bmatrix} 3 \\ 9 \\ 4 \end{bmatrix}$.
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Als nächstes müssen wir den Übergangsmatrix von der alten Basis zur neuen Basis finden. Das bedeutet, wir müssen die Koordinaten der alten Basisvektoren $a$, $b$ und $c$ in Bezug auf die neue Basis $p$, $q$ und $v$ finden. Wir erhalten eine Matrix $M$:
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$M = \begin{bmatrix} P_a & P_b & P_c \end{bmatrix}$,
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wobei $P_a$, $P_b$ und $P_c$ die Koordinaten von $a$, $b$ und $c$ bezüglich der neuen Basis sind. Um diese Koordinaten zu finden, müssen wir das lineare Gleichungssystem lösen:
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$\begin{bmatrix} 0 & 1 & 1 \\ 1 & 1 & 2 \\ 2 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} P_a \\ P_b \\ P_c \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix}$
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Die Lösung dieses Gleichungssystems ergibt:
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$M = \begin{bmatrix} 2 & -2 & 4 \\ 2 & 1 & 1 \\ -1 & 1 & -1 \end{bmatrix}$.
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Um die Koordinaten von Vektor $x$ bezüglich der neuen Basis zu finden, multiplizieren wir die Inverse von Matrix $M$ mit dem Vektor $x$:
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$x' = M^{-1} \cdot x = \begin{bmatrix} 1 & 1 & 3 \\ 0 & 1 & -1 \\ 1 & -1 & 2 \end{bmatrix} \cdot \begin{bmatrix} 3 \\ 9 \\ 4 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$.
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Daher sind die Koordinaten von Vektor $x$ bezüglich der neuen Basis:
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$\alpha' = 2$, $\beta' = -1$, $\gamma' = 1$.
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Da wir nun die Koordinaten von Vektor $x$ bezüglich der neuen Basis gefunden haben, können wir die Koordinaten $\alpha' = 2$, $\beta' = -1$ und $\gamma' = 1$ verwenden, um den Vektor $x$ in der neuen Basis auszudrücken:
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$x = \alpha' p + \beta' q + \gamma' v = 2 \cdot \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} - 1 \cdot \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + 1 \cdot \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ 5 \end{bmatrix}$.
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||||
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Der Vektor $x$ ist also in der neuen Basis gleich:
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$x = \begin{bmatrix} 1 \\ 3 \\ 5 \end{bmatrix}$.
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Zusammenfassend haben wir die Koordinaten von Vektor $x$ in der neuen Basis $p$, $q$ und $v$ gefunden und den Vektor $x$ in dieser Basis ausgedrückt.
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# 2
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@@ -86,3 +54,5 @@ $$\textbf{p} = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix}$$
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Der gesuchte Punkt Vektor $$\textbf{p}$$ der Zentralprojektion des Punktes Vektor $$\textbf{x}$$ in die durch Vektor $$\textbf{a}$$ und Vektor $$\textbf{b}$$ gegebene Projektionsebene durch das Projektionszentrum Vektor $$\textbf{c}$$ ist $$\begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix}^T$$
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@@ -1,18 +0,0 @@
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# Conflicts
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Please resolve them and commit them using the commands `Obsidian Git: Commit all changes` followed by `Obsidian Git: Push`
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(This file will automatically be deleted before commit)
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[[#Additional Instructions]] available below file list
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- Not a file: .obsidian/workspace.json
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- [[KW17]]
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# Additional Instructions
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I strongly recommend to use "Source mode" for viewing the conflicted files. For simple conflicts, in each file listed above replace every occurrence of the following text blocks with the desired text.
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```diff
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<<<<<<< HEAD
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File changes in local repository
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=======
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File changes in remote repository
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>>>>>>> origin/main
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```
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Reference in New Issue
Block a user