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How to Mathe Abi

11) Lagebeziehungen (Geometrie)

RK
by Rick K.

Welche Lagebeziehungen sind zwischen Punkt und Punkt möglich?

Lagebeziehung Punkt-Punkt:


1) Die Punkte sind identisch.


2) Die Punkte sind nicht identisch.

Was bedeutet die folgende Aussage:


P∈gP \in g


Lösung:


Der Punkt P liegt auf der Gerade g bzw. der Punkt P ist Element der Gerade g.

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P∉gP \notin g


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P(xP;yP;zP)P(x_P;y_P;z_P)


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


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P(2;4;6)P(2;4;6)


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g:x→=(110)+r⋅(136)g: \overrightarrow{x}=\left(\begin{array}{c} 1 \\ 1 \\ 0\end{array}\right) + r\cdot \left(\begin{array}{c} 1 \\ 3 \\ 6\end{array}\right)


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Q(3;6;1)Q(3;6;1)


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h:x→=(221)+r⋅(151)h: \overrightarrow{x}=\left(\begin{array}{c} 2 \\ 2 \\ 1\end{array}\right) + r\cdot \left(\begin{array}{c} 1 \\ 5 \\ 1\end{array}\right)


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R(7;8;4)R(7;8;4)


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i:x→=(340)+r⋅(111)i: \overrightarrow{x}=\left(\begin{array}{c} 3 \\ 4 \\ 0\end{array}\right) + r\cdot \left(\begin{array}{c} 1 \\ 1 \\ 1\end{array}\right)


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P(1;2;1)P(1;2;1)


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E:2x+y−z=3E: 2x+y-z=3


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P(1;2;3)P(1;2;3)


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E:2x−4y+z=dE:2x-4y+z=d


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P∈EP \in E


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d=−3‾‾\underline{\underline{d=-3}}


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2⋅1−4⋅2+3=−32\cdot 1-4\cdot 2+3=-3



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P(0;2;5)P(0;2;5)


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E:4x−4y−2z=dE:4x−4y-2z=d


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P∈EP∈E


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d=−18‾‾\underline{\underline{d=-18}}


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4⋅0−4⋅2−2⋅5=−184\cdot 0-4\cdot 2-2\cdot 5=-18



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P(5;1;3)P(5;1;3)


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E:3x+y−3z=dE:3x+y-3z=d


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P∈EP∈E


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d=7‾‾\underline{\underline{d=7}}


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3⋅5+1−3⋅3=73\cdot 5+1-3\cdot 3=7



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g∥hg \parallel h


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g⊥hg \perp h


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g∩h=∅g \cap h = \emptyset


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g∩h={O}g \cap h = \{O\}


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g≡hg \equiv h


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g:x→=(120)+r⋅(110)g: \overrightarrow{x}=\left(\begin{array}{c} 1 \\ 2 \\ 0\end{array}\right)+r\cdot \left(\begin{array}{c} 1 \\ 1 \\ 0\end{array}\right)


h:x→=(320)+s⋅(140)h: \overrightarrow{x}=\left(\begin{array}{c} 3 \\ 2 \\ 0\end{array}\right) + s\cdot \left(\begin{array}{c} 1 \\ 4 \\ 0\end{array}\right)


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r,s∈Rr,s \in \mathbb{R}


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g:x→=(207)+r⋅(114)g: \overrightarrow{x}=\left(\begin{array}{c} 2 \\ 0 \\ 7\end{array}\right)+r\cdot \left(\begin{array}{c} 1 \\ 1 \\ 4\end{array}\right)


h:x→=(−1312)+s⋅(114)h: \overrightarrow{x}=\left(\begin{array}{c} -1 \\ 3 \\ 12\end{array}\right) + s\cdot \left(\begin{array}{c} 1 \\ 1 \\ 4\end{array}\right)


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r,s∈Rr,s \in \mathbb{R}


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g:x→=(211)+r⋅(324)g: \overrightarrow{x}=\left(\begin{array}{c} 2 \\ 1 \\ 1\end{array}\right)+r\cdot \left(\begin{array}{c} 3 \\ 2 \\ 4\end{array}\right)


h:x→=(211)+s⋅(120)h: \overrightarrow{x}=\left(\begin{array}{c} 2 \\ 1 \\ 1\end{array}\right) + s\cdot \left(\begin{array}{c} 1 \\ 2 \\ 0\end{array}\right)


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r,s∈Rr,s \in \mathbb{R}


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g:x→=(324)+r⋅(137)g: \overrightarrow{x}=\left(\begin{array}{c} 3 \\ 2 \\ 4\end{array}\right)+r\cdot \left(\begin{array}{c} 1 \\ 3 \\ 7\end{array}\right)


h:x→=(324)+s⋅(801)h: \overrightarrow{x}=\left(\begin{array}{c} 3 \\ 2 \\ 4\end{array}\right) + s\cdot \left(\begin{array}{c} 8 \\ 0 \\ 1\end{array}\right)


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r,s∈Rr,s \in \mathbb{R}


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g:x→=(211)+r⋅(123)g: \overrightarrow{x}=\left(\begin{array}{c} 2 \\ 1 \\ 1\end{array}\right)+r\cdot \left(\begin{array}{c} 1 \\ 2 \\ 3\end{array}\right)


h:x→=(211)+s⋅(246)h: \overrightarrow{x}=\left(\begin{array}{c} 2 \\ 1 \\ 1\end{array}\right) + s\cdot \left(\begin{array}{c} 2 \\ 4 \\ 6\end{array}\right)


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r,s∈Rr,s \in \mathbb{R}


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g:x→=(211)+r⋅(10−1)g: \overrightarrow{x}=\left(\begin{array}{c} 2 \\ 1 \\ 1\end{array}\right)+r\cdot \left(\begin{array}{c} 1 \\ 0 \\ -1\end{array}\right)


h:x→=(211)+s⋅(242)h: \overrightarrow{x}=\left(\begin{array}{c} 2 \\ 1 \\ 1\end{array}\right) + s\cdot \left(\begin{array}{c} 2 \\ 4 \\ 2\end{array}\right)


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r,s∈Rr,s \in \mathbb{R}


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g:x→=(120)+r⋅(001)g: \overrightarrow{x}=\left(\begin{array}{c} 1 \\ 2 \\ 0\end{array}\right)+r\cdot \left(\begin{array}{c} 0 \\ 0 \\ 1\end{array}\right)


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h:x→=(134)+r⋅(040)h: \overrightarrow{x}=\left(\begin{array}{c} 1 \\ 3 \\ 4\end{array}\right)+r\cdot \left(\begin{array}{c} 0 \\ 4 \\ 0\end{array}\right)


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


h:x→=a→+s⋅d→h: \overrightarrow{x}=\overrightarrow{a}+s\cdot \overrightarrow{d}


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r,s∈Rr,s \in \mathbb{R}


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


Das stsnba Poklmohdhiuhsbir


n1→;n2→\overrightarrow{n_1}; \overrightarrow{n_2}


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


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n→\overrightarrow{n}


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





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Author

Rick K.

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