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How to Mathe Abi

10) Geraden und Ebenen (Geometrie)

RK
by Rick K.

Gegeben ist die folgende Geradengleichung:


g:x→=(130)+r⋅(012)g: \overrightarrow{x}=\left(\begin{array}{c} 1 \\ 3 \\ 0\end{array}\right)+r\cdot \left(\begin{array}{c} 0 \\ 1 \\ 2\end{array}\right)


wobei:


r∈Rr \in \mathbb{R}



Benenne den Stützvektor.


Stützvektor:


a→=(130)‾‾\underline{\underline{\overrightarrow{a}=\left(\begin{array}{c} 1 \\ 3 \\ 0\end{array}\right)}}


Gegeben sind die folgenden beiden Punkte:


A(1;2;0)A(1;2;0)

und

B(2;3;5)B(2;3;5)


Existiert eine eindeutige Gerade, die durch beide Punkte verläuft?

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g:x→=(221)+r⋅(012)g: \overrightarrow{x}=\left(\begin{array}{c} 2 \\ 2 \\ 1\end{array}\right)+r\cdot \left(\begin{array}{c} 0 \\ 1 \\ 2\end{array}\right)

ozdyem


r∈Rr \in \mathbb{R}



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b→=(012)‾‾\underline{\underline{\overrightarrow{b}=\left(\begin{array}{c} 0 \\ 1 \\ 2\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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r∈Rr \in \mathbb{R}


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A(1;2;0)A(1;2;0)

sus

C(1;0;0)C(1;0;0)


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g:x→=(221)+r⋅(001)g: \overrightarrow{x}=\left(\begin{array}{c} 2 \\ 2 \\ 1\end{array}\right)+r\cdot \left(\begin{array}{c} 0 \\ 0 \\ 1\end{array}\right)


rotobb


r∈Rr \in \mathbb{R}


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g:x→=r⋅(100)g: \overrightarrow{x}=r\cdot \left(\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right)


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r∈Rr \in \mathbb{R}


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h:x→=r⋅(010)h: \overrightarrow{x}=r\cdot \left(\begin{array}{c} 0 \\ 1 \\ 0\end{array}\right)


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r∈Rr \in \mathbb{R}


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g:x→=(320)+r⋅(150)g: \overrightarrow{x}=\left(\begin{array}{c} 3 \\ 2 \\ 0\end{array}\right)+r\cdot \left(\begin{array}{c} 1 \\ 5 \\ 0\end{array}\right)


ztvorm


r∈Rr \in \mathbb{R}


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g:x→=(123)+r⋅(001)g: \overrightarrow{x}=\left(\begin{array}{c} 1 \\ 2 \\ 3\end{array}\right)+r\cdot \left(\begin{array}{c} 0 \\ 0 \\ 1\end{array}\right)


oipado


r∈Rr \in \mathbb{R}


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E:x→=(231)+r⋅(113)+s⋅(280)E: \overrightarrow{x}=\left(\begin{array}{c} 2 \\ 3 \\ 1\end{array}\right)+r\cdot \left(\begin{array}{c} 1 \\ 1 \\ 3\end{array}\right)+s\cdot \left(\begin{array}{c} 2 \\ 8 \\ 0\end{array}\right)



biejgf


r,s∈Rr,s \in \mathbb{R}


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Boyaqls


a→=(231)‾‾\underline{\underline{\overrightarrow{a}=\left(\begin{array}{c} 2 \\ 3 \\ 1\end{array}\right)}}


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E:x→=(170)+r⋅(253)+s⋅(192)E: \overrightarrow{x}=\left(\begin{array}{c} 1 \\ 7 \\ 0\end{array}\right)+r\cdot \left(\begin{array}{c} 2 \\ 5 \\ 3\end{array}\right)+s\cdot \left(\begin{array}{c} 1 \\ 9 \\ 2\end{array}\right)



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r,s∈Rr,s \in \mathbb{R}


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b→=(253);c→=(192)‾‾\underline{\underline{\overrightarrow{b}=\left(\begin{array}{c} 2 \\ 5 \\ 3\end{array}\right); \overrightarrow{c}=\left(\begin{array}{c} 1 \\ 9 \\ 2\end{array}\right)}}


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E:2x+3y+4z=0E:2x+3y+4z=0


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n→=(234)‾‾\underline{\underline{\overrightarrow{n}=\left(\begin{array}{c} 2 \\ 3 \\ 4\end{array}\right)}}


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F:1x−3y+5z=0F:1x-3y+5z=0


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n→=(1−35)‾‾\underline{\underline{\overrightarrow{n}=\left(\begin{array}{c} 1 \\ -3 \\ 5\end{array}\right)}}


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G:2y+3z=0G:2y+3z=0


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n→=(023)‾‾\underline{\underline{\overrightarrow{n}=\left(\begin{array}{c} 0 \\ 2 \\ 3\end{array}\right)}}


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H:2x+7z=0H:2x+7z=0


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n→=(207)‾‾\underline{\underline{\overrightarrow{n}=\left(\begin{array}{c} 2 \\ 0 \\ 7\end{array}\right)}}


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I:z=0I:z=0


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


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n→=(010)‾‾\underline{\underline{\overrightarrow{n}=\left(\begin{array}{c} 0 \\ 1 \\ 0\end{array}\right)}}


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


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n→=(100)‾‾\underline{\underline{\overrightarrow{n}=\left(\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right)}}


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


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


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n→=(100)\overrightarrow{n}=\left(\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right)


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n→=(040)\overrightarrow{n}=\left(\begin{array}{c} 0 \\ 4 \\ 0\end{array}\right)


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n→=(011)\overrightarrow{n}=\left(\begin{array}{c} 0 \\ 1 \\ 1\end{array}\right)


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Author

Rick K.

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