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How to Mathe Abi

13) Zusammenfassung (Geometrie)

RK
von Rick K.

Definiere den folgenden Begriff:


VEKTOR

Vektor:


Eine Klasse von Pfeilen gleicher Länge und gleicher Richtung heißt Vektor.


Vektoren sind also durch ihre Länge und ihre Richtung definiert, die Lage im Raum hingegen ist nicht fixiert.

Was charakterisiert den Ortsvektor zu einem beliebigen Punkt P?

Der Ortsvektor zum Punkt P startet im Koordinatenursprung O und zeigt auf den Punkt P.

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AB→=OB→−OA→‾‾\underline{\underline{\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}}}


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a→;b→\overrightarrow{a}; \overrightarrow{b}


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∃!m∈R→a→=m⋅b→\exists! m \in \mathbb{R} \rightarrow \overrightarrow{a}=m\cdot \overrightarrow{b}


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a→=(xyz)\overrightarrow{a}=\left(\begin{array}{c} x \\ y \\ z\end{array}\right)


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∣a→∣=x2+y2+z2‾‾\underline{\underline{|\overrightarrow{a}|=\sqrt{x^2+y^2+z^2}}}


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a→⊥b→⟺a→∘b→=0\overrightarrow{a} \perp \overrightarrow{b} \Longleftrightarrow \overrightarrow{a} \circ \overrightarrow{b}=0


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α=arccos⁡a→∘b→∣a→∣⋅∣b→∣‾‾\underline{\underline{\alpha=\arccos\frac{\overrightarrow{a} \circ \overrightarrow{b}}{|\overrightarrow{a}|\cdot |\overrightarrow{b}| }}}


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g:x→=a→+r⋅b→g: \overrightarrow{x}=\overrightarrow{a}+r\cdot \overrightarrow{b}

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r∈Rr \in \mathbb{R}


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n→=(001)‾‾\underline{\underline{\overrightarrow{n}=\left(\begin{array}{c} 0 \\ 0 \\ 1\end{array}\right)}}


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g:x→=a→+r⋅b→g: \overrightarrow{x}=\overrightarrow{a}+r\cdot \overrightarrow{b}


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α=cos⁡−1∣b→∘c→∣∣b→∣⋅∣c→∣‾‾\underline{\underline{\alpha=\cos^{-1}\frac{|\overrightarrow{b}\circ \overrightarrow{c}|}{|\overrightarrow{b}|\cdot |\overrightarrow{c}|}}}


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α=cos⁡−1∣n1→∘n2→∣∣n1→∣⋅∣n2→∣‾‾\underline{\underline{\alpha=\cos^{-1}\frac{|\overrightarrow{n_1}\circ \overrightarrow{n_2}|}{|\overrightarrow{n_1}|\cdot |\overrightarrow{n_2}|}}}


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α=sin⁡−1∣a→∘n→∣∣a→∣⋅∣n→∣‾‾\underline{\underline{\alpha=\sin^{-1}\frac{|\overrightarrow{a}\circ \overrightarrow{n}|}{|\overrightarrow{a}|\cdot |\overrightarrow{n}|}}}


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E:n1⋅x+n2⋅y+n3⋅z=dE:n_1\cdot x + n_2\cdot y + n_3\cdot z = d


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dist(P;E)=∣n1⋅xP+n2⋅yP+n3⋅zP−d∣n12+n22+n32dist(P;E)=\frac{|n_1\cdot x_P+n_2\cdot y_P+n_3\cdot z_P-d| }{\sqrt{n_1^2+n_2^2+n_3^2}}


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