Buffl
Buffl
How to Mathe Abi

08) Tools (Geometrie)

RK
by 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.

Gegeben ist der folgende Punkt:


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


Gib den dazugehörigen Ortsvektor an.


Lösung:


OA→=(102)‾‾\underline{\underline{\overrightarrow{OA}=\left(\begin{array}{c} 1 \\ 0 \\ 2\end{array}\right)}}


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B(3;4;−1)B(3;4;-1)


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OB→=(34−1)‾‾\underline{\underline{\overrightarrow{OB}=\left(\begin{array}{c} 3 \\ 4 \\ -1\end{array}\right)}}


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

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


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AB→\overrightarrow{AB}


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


AB→=(235)−(120)\overrightarrow{AB}=\left(\begin{array}{c} 2 \\ 3 \\ 5\end{array}\right)-\left(\begin{array}{c} 1 \\ 2 \\ 0\end{array}\right)


AB→=(115)‾‾\underline{\underline{\overrightarrow{AB}=\left(\begin{array}{c} 1 \\ 1 \\ 5\end{array}\right)}}


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

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D(1;−1;2)D(1;-1;2)


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AD→\overrightarrow{AD}


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AD→=(0−32)‾‾\underline{\underline{\overrightarrow{AD}=\left(\begin{array}{c} 0 \\ -3 \\ 2\end{array}\right)}}


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

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M(8;1;4)M(8;1;4)


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MA→\overrightarrow{MA}


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MA→=(−71−4)‾‾\underline{\underline{\overrightarrow{MA}=\left(\begin{array}{c} -7 \\ 1 \\ -4\end{array}\right)}}


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a→=(21−1);b→=(052)\overrightarrow{a}=\left(\begin{array}{c} 2 \\ 1 \\ -1\end{array}\right); \overrightarrow{b}=\left(\begin{array}{c} 0 \\ 5 \\ 2\end{array}\right)


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c→=2⋅a→−3⋅b→\overrightarrow{c}=2\cdot \overrightarrow{a}-3\cdot \overrightarrow{b}


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c→=(4−13−8)‾‾\underline{\underline{\overrightarrow{c}=\left(\begin{array}{c} 4 \\ -13 \\ -8\end{array}\right)}}


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OP′→\overrightarrow{OP'}



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OP′→=OP→+2⋅PS→‾‾\underline{\underline{\overrightarrow{OP'}=\overrightarrow{OP}+2\cdot \overrightarrow{PS}}}


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a→=(130);b→=(112)\overrightarrow{a}=\left(\begin{array}{c} 1 \\ 3 \\ 0\end{array}\right); \overrightarrow{b}=\left(\begin{array}{c} 1 \\ 1 \\ 2\end{array}\right)


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c→=2⋅a→+b→\overrightarrow{c}=2\cdot \overrightarrow{a}+\overrightarrow{b}


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c→=(372)‾‾\underline{\underline{\overrightarrow{c}=\left(\begin{array}{c} 3 \\ 7 \\ 2\end{array}\right)}}


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OP′→\overrightarrow{OP'}



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OP′→=OP→+2⋅PS→‾‾\underline{\underline{\overrightarrow{OP'}=\overrightarrow{OP}+2\cdot \overrightarrow{PS}}}


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OD→\overrightarrow{OD}



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OD→=OA→+BC→‾‾\underline{\underline{\overrightarrow{OD}=\overrightarrow{OA}+\overrightarrow{BC}}}


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OD→=OC→+BA→‾‾\underline{\underline{\overrightarrow{OD}=\overrightarrow{OC}+\overrightarrow{BA}}}


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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→=(264);b→=(132)\overrightarrow{a}=\left(\begin{array}{c} 2 \\ 6 \\ 4\end{array}\right); \overrightarrow{b}=\left(\begin{array}{c} 1 \\ 3 \\ 2\end{array}\right)


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a→=2⋅b→\overrightarrow{a}=2\cdot\overrightarrow{b}


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a→=(12−2);b→=(24−4)\overrightarrow{a}=\left(\begin{array}{c} 1 \\ 2 \\ -2\end{array}\right); \overrightarrow{b}=\left(\begin{array}{c} 2 \\ 4 \\ -4\end{array}\right)


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a→=12⋅b→\overrightarrow{a}=\frac{1}{2}\cdot \overrightarrow{b}


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a→=(324);b→=(644)\overrightarrow{a}=\left(\begin{array}{c} 3 \\ 2 \\ 4\end{array}\right); \overrightarrow{b}=\left(\begin{array}{c} 6 \\ 4 \\ 4\end{array}\right)


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a→=m⋅b→\overrightarrow{a}=m\cdot \overrightarrow{b}


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b→=(5−24)\overrightarrow{b}=\left(\begin{array}{c} 5 \\ -2 \\ 4\end{array}\right)


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a→=(10−48)‾‾ \underline{\underline{\overrightarrow{a}=\left(\begin{array}{c} 10 \\ -4 \\ 8\end{array}\right)}}


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m=2m=2


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a→=−1⋅(−a→)\overrightarrow{a}=-1\cdot (-\overrightarrow{a})


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∃!m∈R→a→=m⋅(−a→)\exists! m \in \mathbb{R} \rightarrow \overrightarrow{a}=m\cdot (\overrightarrow{-a})


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m=−1m=-1





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a→=(110)\overrightarrow{a}=\left(\begin{array}{c} 1 \\ 1 \\ 0\end{array}\right)


Chnytar


∣a→∣=2‾‾\underline{\underline{|\overrightarrow{a}|=\sqrt{2}}}


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b→=(304)\overrightarrow{b}=\left(\begin{array}{c} 3 \\ 0 \\ 4\end{array}\right)


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∣b→∣=5‾‾\underline{\underline{|\overrightarrow{b}|=5}}


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c→=(612)\overrightarrow{c}=\left(\begin{array}{c} 6 \\ 1 \\ 2\end{array}\right)


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∣c→∣=41‾‾\underline{\underline{|\overrightarrow{c}|=\sqrt{41}}}


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d→=(−41−2)\overrightarrow{d}=\left(\begin{array}{c} -4 \\ 1 \\ -2\end{array}\right)


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∣d→∣=21‾‾\underline{\underline{|\overrightarrow{d}|=\sqrt{21}}}


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e→=(1153)\overrightarrow{e}=\left(\begin{array}{c} 11 \\ 5 \\ 3\end{array}\right)


Apppczu


∣e→∣=155‾‾\underline{\underline{|\overrightarrow{e}|=\sqrt{155}}}


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


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AKreis=π⋅r2A_{Kreis}=\pi\cdot r^2

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r=∣r→∣r=|\overrightarrow{r}|


A=π⋅1=π(FE)‾‾\underline{\underline{A=\pi\cdot 1=\pi (FE)}}


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a→=(130);b→=(112)\overrightarrow{a}=\left(\begin{array}{c} 1 \\ 3 \\ 0\end{array}\right); \overrightarrow{b}=\left(\begin{array}{c} 1 \\ 1 \\ 2\end{array}\right)


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a→∘b→=4‾‾\underline{\underline{\overrightarrow{a} \circ \overrightarrow{b}= 4}}


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c→=(2−31);d→=(452)\overrightarrow{c}=\left(\begin{array}{c} 2 \\ -3 \\ 1\end{array}\right); \overrightarrow{d}=\left(\begin{array}{c} 4 \\ 5 \\ 2\end{array}\right)


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c→∘d→=−5‾‾\underline{\underline{\overrightarrow{c} \circ \overrightarrow{d}=-5}}


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e→=(5−13);f→=(261)\overrightarrow{e}=\left(\begin{array}{c} 5 \\ -1 \\ 3\end{array}\right); \overrightarrow{f}=\left(\begin{array}{c} 2 \\ 6 \\ 1\end{array}\right)


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e→∘f→=7‾‾\underline{\underline{\overrightarrow{e} \circ \overrightarrow{f}=7}}


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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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a→=(1−13);b→=(211)\overrightarrow{a}=\left(\begin{array}{c} 1 \\ -1 \\ 3\end{array}\right); \overrightarrow{b}=\left(\begin{array}{c} 2 \\ 1 \\ 1\end{array}\right)


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AC→⊥AB→⟺AC→∘AB→=0\overrightarrow{AC} \perp \overrightarrow{AB} \Longleftrightarrow \overrightarrow{AC} \circ \overrightarrow{AB}=0


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AC→∘AB→‾‾ \underline{\underline{\overrightarrow{AC}\circ \overrightarrow{AB}}}


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a→=(0−13);b→=(2−11)\overrightarrow{a}=\left(\begin{array}{c} 0 \\ -1 \\ 3\end{array}\right); \overrightarrow{b}=\left(\begin{array}{c} 2 \\ -1 \\ 1\end{array}\right)


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BM→∘a→\overrightarrow{BM}\circ \overrightarrow{a}


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a→=(3−1−1);b→=(2−17)\overrightarrow{a}=\left(\begin{array}{c} 3 \\ -1 \\ -1\end{array}\right); \overrightarrow{b}=\left(\begin{array}{c} 2 \\ -1 \\ 7\end{array}\right)


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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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α\alpha



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



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A=∣a→×b→∣A=|\overrightarrow{a} \times \overrightarrow{b}|


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n→=a→×b→‾‾\underline{\underline{\overrightarrow{n}=\overrightarrow{a}\times \overrightarrow{b}}}


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A=∣a→×b→∣2‾‾\underline{\underline{A=\frac{|\overrightarrow{a}\times \overrightarrow{b}|}{2}}}


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(a→×b→)∘c→(\overrightarrow{a} \times \overrightarrow{b}) \circ \overrightarrow{c}



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a→×(b→×c→)\overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c})



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(a→∘b→)∘c→(\overrightarrow{a} \circ \overrightarrow{b}) \circ \overrightarrow{c}



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(a→×b→)∘(c→×d→)(\overrightarrow{a} \times \overrightarrow{b}) \circ (\overrightarrow{c} \times \overrightarrow{d})



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(λ×b→)∘(c→×d→)(\lambda \times \overrightarrow{b}) \circ (\overrightarrow{c} \times \overrightarrow{d})


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λ∈R\lambda \in \mathbb{R}



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A=∣a→×b→∣2‾‾\underline{\underline{A=\frac{|\overrightarrow{a}\times \overrightarrow{b}|}{2}}}


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2⋅1+2⋅0+4⋅32\cdot 1 + 2\cdot 0 + 4\cdot 3


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12+22+42\sqrt{1^2+2^2+4^2}


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λ⋅a→+μ⋅b→\lambda \cdot \overrightarrow{a} + \mu \cdot \overrightarrow{b}


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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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a→∘b→=(211)∘(431)\overrightarrow{a}\circ\overrightarrow{b}=\left(\begin{array}{c} 2 \\ 1 \\ 1\end{array}\right)\circ\left(\begin{array}{c} 4 \\ 3 \\ 1\end{array}\right)



Tylkybl


a→∘b→=(211)∘(431)=2⋅4+1⋅3+1⋅1\overrightarrow{a}\circ\overrightarrow{b}=\left(\begin{array}{c} 2 \\ 1 \\ 1\end{array}\right)\circ\left(\begin{array}{c} 4 \\ 3 \\ 1\end{array}\right) = 2\cdot 4+1\cdot 3+1\cdot 1


a→∘b→=12‾‾\underline{\underline{\overrightarrow{a}\circ\overrightarrow{b}=12}}


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a→=(120);b→(4−18)\overrightarrow{a}=\left(\begin{array}{c} 1 \\ 2 \\ 0\end{array}\right); \overrightarrow{b}\left(\begin{array}{c} 4 \\ -1 \\ 8\end{array}\right)


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a→=(21−1);b→(−42−1)\overrightarrow{a}=\left(\begin{array}{c} 2 \\ 1 \\ -1\end{array}\right); \overrightarrow{b}\left(\begin{array}{c} -4 \\ 2 \\ -1\end{array}\right)


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