Ein Vektorraum
R
n
{\displaystyle \mathbb {R} ^{n}}
a
=
(
a
1
,
…
,
a
n
)
=
(
a
1
⋮
a
n
)
;
a
∈
R
n
{\displaystyle \mathbf {a} =(a_{1},\ldots ,a_{n})\ ={\begin{pmatrix}a_{1}\\\vdots \\a_{n}\end{pmatrix}};\quad \mathbf {a} \in \mathbb {R} ^{n}}
mit den Koordinaten des Vektors
a
1
,
…
,
a
n
;
a
i
∈
R
{\displaystyle a_{1},\ldots ,a_{n}\ ;\quad a_{i}\in \mathbb {R} }
Freie Vektoren können beliebig im Raum verschoben werden. Ortsvektoren (gebundene Vektoren) sind ortsgebunden.
a
=
(
a
1
,
…
,
a
n
)
∈
R
n
{\displaystyle \mathbf {a} =(a_{1},\ldots ,a_{n})\in \mathbb {R} ^{n}}
b
=
(
b
1
,
…
,
b
n
)
∈
R
n
{\displaystyle \mathbf {b} =(b_{1},\ldots ,b_{n})\in \mathbb {R} ^{n}}
a
+
b
=
(
a
1
+
b
1
,
…
,
a
n
+
b
n
)
=
(
a
1
+
b
1
⋮
a
n
+
b
n
)
{\displaystyle \mathbf {a} +\mathbf {b} =\left(a_{1}+b_{1},\ldots ,a_{n}+b_{n}\right)={\begin{pmatrix}a_{1}+b_{1}\\\vdots \\a_{n}+b_{n}\end{pmatrix}}}
(
a
+
b
)
+
c
=
a
+
(
b
+
c
)
{\displaystyle (\mathbf {a} +\mathbf {b} )+\mathbf {c} =\mathbf {a} +(\mathbf {b} +\mathbf {c} )}
a
+
b
=
b
+
a
{\displaystyle \mathbf {a} +\mathbf {b} =\mathbf {b} +\mathbf {a} }
Der Nullvektor ist das neutrale Element der Vektoraddition.
o
=
(
0
,
…
,
0
)
∈
R
n
{\displaystyle \mathbf {o} =(0,\ldots ,0)\in \mathbb {R} ^{n}}
o
+
a
=
a
+
o
=
a
{\displaystyle \mathbf {o} +\mathbf {a} =\mathbf {a} +\mathbf {o} =\mathbf {a} }
a
+
c
=
b
{\displaystyle \mathbf {a} +\mathbf {c} =\mathbf {b} }
c
=
b
−
a
{\displaystyle \mathbf {c} =\mathbf {b} -\mathbf {a} }
c
{\displaystyle \mathbf {c} }
Gegeben seien die Vektoren
a
=
(
1
,
−
1
,
5
)
{\displaystyle \mathbf {a} =(1,-1,5)}
b
=
(
2
,
0
,
1
)
{\displaystyle \mathbf {b} =(2,0,1)}
a
+
b
{\displaystyle \mathbf {a} +\mathbf {b} }
a
−
b
{\displaystyle \mathbf {a} -\mathbf {b} }
a
+
b
=
(
1
−
1
5
)
+
(
2
0
1
)
=
(
3
−
1
6
)
{\displaystyle \mathbf {a} +\mathbf {b} ={\begin{pmatrix}1\\-1\\5\end{pmatrix}}+{\begin{pmatrix}2\\0\\1\end{pmatrix}}={\begin{pmatrix}3\\-1\\6\end{pmatrix}}}
a
−
b
=
(
1
−
1
5
)
−
(
2
0
1
)
=
(
−
1
−
1
4
)
{\displaystyle \mathbf {a} -\mathbf {b} ={\begin{pmatrix}1\\-1\\5\end{pmatrix}}-{\begin{pmatrix}2\\0\\1\end{pmatrix}}={\begin{pmatrix}-1\\-1\\4\end{pmatrix}}}
Multiplikation eines Vektors mit einem Skalar
Bearbeiten
Geometrisch entspricht die Multiplikation eines Vektors mit einem Skalar der Streckung eines Vektors.
α
,
β
∈
R
{\displaystyle \alpha ,\;\beta \ \in \mathbb {R} }
α
a
=
(
α
a
1
,
…
,
α
a
n
)
=
(
α
a
1
⋮
α
a
n
)
{\displaystyle \alpha \mathbf {a} =\left(\alpha a_{1},\ldots ,\alpha a_{n}\right)={\begin{pmatrix}\alpha a_{1}\\\vdots \\\alpha a_{n}\end{pmatrix}}}
1
a
=
a
{\displaystyle 1\mathbf {a} =\mathbf {a} }
(
α
β
)
a
=
α
(
β
a
)
{\displaystyle (\alpha \beta )\mathbf {a} =\alpha (\beta \mathbf {a} )}
(
α
+
β
)
a
=
α
a
+
β
a
{\displaystyle (\alpha +\beta )\mathbf {a} =\alpha \mathbf {a} +\beta \mathbf {a} }
α
(
a
+
b
)
=
α
a
+
α
b
{\displaystyle \alpha (\mathbf {a} +\mathbf {b} )=\alpha \mathbf {a} +\alpha \mathbf {b} }
−
a
=
(
−
a
1
,
…
,
−
a
n
)
=
(
−
a
1
⋮
−
a
n
)
{\displaystyle -\mathbf {a} =\left(-a_{1},\ldots ,-a_{n}\right)={\begin{pmatrix}-a_{1}\\\vdots \\-a_{n}\end{pmatrix}}}
a
+
(
−
a
)
=
o
{\displaystyle \mathbf {a} +(-\mathbf {a} )=\mathbf {o} }
Die zwei Vektoren a und b sind dann parallel, wenn gilt
a
=
λ
b
;
λ
∈
R
∧
λ
≠
0
{\displaystyle \mathbf {a} =\lambda \mathbf {b} \ ;\quad \lambda \in \mathbb {R} \wedge \lambda \neq 0}
λ
>
0
{\displaystyle \lambda >0}
λ
<
0
{\displaystyle \lambda <0}
Sind die Vektoren
a
=
(
5
,
1
,
−
1
)
{\displaystyle \mathbf {a} =(5,1,-1)}
b
=
12
(
4
,
1
,
−
1
)
−
(
63
,
15
,
−
15
)
{\displaystyle \mathbf {b} =12(4,1,-1)-(63,15,-15)}
b
=
12
(
4
1
−
1
)
−
(
63
15
−
15
)
=
(
−
15
−
3
3
)
=
−
3
(
5
1
−
1
)
{\displaystyle \mathbf {b} =12{\begin{pmatrix}4\\1\\-1\end{pmatrix}}-{\begin{pmatrix}63\\15\\-15\end{pmatrix}}={\begin{pmatrix}-15\\-3\\3\end{pmatrix}}=-3{\begin{pmatrix}5\\1\\-1\end{pmatrix}}}
⇒
b
=
−
3
a
⇒
a
|
|
b
{\displaystyle \Rightarrow \;\mathbf {b} =-3\mathbf {a} \;\Rightarrow \;\mathbf {a} ||\mathbf {b} }
Gegeben sind zwei Vektoren
a
=
(
a
1
,
…
,
a
n
)
∈
R
n
{\displaystyle \mathbf {a} =(a_{1},\ldots ,a_{n})\in \mathbb {R} ^{n}}
b
=
(
b
1
,
…
,
b
n
)
∈
R
n
{\displaystyle \mathbf {b} =(b_{1},\ldots ,b_{n})\in \mathbb {R} ^{n}}
Das Skalarprodukt ergibt sich zu
a
⋅
b
=
∑
i
=
1
n
a
i
b
i
{\displaystyle \mathbf {a} \cdot \mathbf {b} =\sum _{i=1}^{n}a_{i}b_{i}}
Das Skalarprodukt ordnet einem Paar von Vektoren eine reelle Zahl zu.
Das Skalarprodukt unterscheidet sich grundlegend von der Multiplikation reeller Zahlen oder der Multiplikation eines Vektors mit einem Skalar.
f
(
x
)
=
a
1
x
1
+
a
2
x
2
+
…
+
a
n
x
n
=
a
⋅
x
{\displaystyle f(x)=a_{1}x_{1}+a_{2}x_{2}+\ldots +a_{n}x_{n}=\mathbf {a} \cdot \mathbf {x} }
Die Norm (Länge) eines Vektors a ist gegeben durch
‖
a
‖
=
a
⋅
a
=
a
1
2
+
⋯
+
a
n
2
{\displaystyle \|\mathbf {a} \|={\sqrt {\mathbf {a} \cdot \mathbf {a} }}={\sqrt {a_{1}^{2}+\cdots +a_{n}^{2}}}}
‖
α
a
‖
=
|
α
|
‖
a
‖
{\displaystyle \|\alpha \mathbf {a} \|=|\alpha |\|\mathbf {a} \|}
a , so bringt man ihn auf die
Länge Eins
a
0
=
a
‖
a
‖
{\displaystyle \mathbf {a} _{0}={\frac {\mathbf {a} }{\|\mathbf {a} \|}}}
a
⋅
b
=
‖
a
‖
‖
b
‖
cos
(
∠
)
{\displaystyle \mathbf {a} \cdot \mathbf {b} =\|\mathbf {a} \|\|\mathbf {b} \|\cos(\angle )}
wobei
∠
{\displaystyle \angle }
a und b bezeichnet.
a
⋅
b
=
b
⋅
a
{\displaystyle \mathbf {a} \cdot \mathbf {b} =\mathbf {b} \cdot \mathbf {a} }
a
⋅
(
b
⋅
c
)
≠
(
a
⋅
b
)
⋅
c
{\displaystyle \mathbf {a} \cdot (\mathbf {b} \cdot \mathbf {c} )\neq (\mathbf {a} \cdot \mathbf {b} )\cdot \mathbf {c} }
a
⋅
(
b
+
c
)
=
a
⋅
b
+
a
⋅
c
{\displaystyle \mathbf {a} \cdot (\mathbf {b} +\mathbf {c} )=\mathbf {a} \cdot \mathbf {b} +\mathbf {a} \cdot \mathbf {c} }
(
a
⋅
b
)
2
≤
‖
a
‖
2
‖
b
‖
2
{\displaystyle (\mathbf {a} \cdot \mathbf {b} )^{2}\leq \|\mathbf {a} \|^{2}\|\mathbf {b} \|^{2}}
‖
a
+
b
‖
≤
‖
a
‖
+
‖
b
‖
{\displaystyle \|\mathbf {a} +\mathbf {b} \|\leq \|\mathbf {a} \|+\|\mathbf {b} \|}
Es ist folgende Gleichung herzuleiten
‖
a
+
b
‖
2
+
‖
a
−
b
‖
2
=
2
(
‖
a
‖
2
+
‖
b
‖
2
)
{\displaystyle \|\mathbf {a} +\mathbf {b} \|^{2}+\|\mathbf {a} -\mathbf {b} \|^{2}=2\left(\|\mathbf {a} \|^{2}+\|\mathbf {b} \|^{2}\right)}
‖
a
+
b
‖
2
=
(
a
+
b
)
⋅
(
a
+
b
)
{\displaystyle \|\mathbf {a} +\mathbf {b} \|^{2}=(\mathbf {a} +\mathbf {b} )\cdot (\mathbf {a} +\mathbf {b} )}
‖
a
−
b
‖
2
=
(
a
−
b
)
⋅
(
a
−
b
)
{\displaystyle \|\mathbf {a} -\mathbf {b} \|^{2}=(\mathbf {a} -\mathbf {b} )\cdot (\mathbf {a} -\mathbf {b} )}
‖
a
+
b
‖
2
+
‖
a
−
b
‖
2
=
a
a
+
a
b
+
a
b
+
b
b
+
a
a
−
a
b
−
a
b
+
b
b
=
2
a
a
+
2
b
b
{\displaystyle \|\mathbf {a} +\mathbf {b} \|^{2}+\|\mathbf {a} -\mathbf {b} \|^{2}=\mathbf {a} \mathbf {a} +\mathbf {a} \mathbf {b} +\mathbf {a} \mathbf {b} +\mathbf {b} \mathbf {b} +\mathbf {a} \mathbf {a} -\mathbf {a} \mathbf {b} -\mathbf {a} \mathbf {b} +\mathbf {b} \mathbf {b} =2\mathbf {a} \mathbf {a} +2\mathbf {b} \mathbf {b} }
‖
a
+
b
‖
2
+
‖
a
−
b
‖
2
=
2
(
‖
a
‖
2
+
‖
b
‖
2
)
{\displaystyle \|\mathbf {a} +\mathbf {b} \|^{2}+\|\mathbf {a} -\mathbf {b} \|^{2}=2\left(\|\mathbf {a} \|^{2}+\|\mathbf {b} \|^{2}\right)}
Zwei Vektoren a , b heißen orthogonal (= senkrecht), wenn
a
⋅
b
=
0
{\displaystyle \mathbf {a} \cdot \mathbf {b} =0}
Man schreibt dies auch als
a
⊥
b
{\displaystyle \mathbf {a} \bot \mathbf {b} }
Es gilt dann der Satz von Pythagoras
‖
a
+
b
‖
2
=
‖
a
‖
2
+
‖
b
‖
2
{\displaystyle \|\mathbf {a} +\mathbf {b} \|^{2}=\|\mathbf {a} \|^{2}+\|\mathbf {b} \|^{2}}
Die Vektoren
a
i
∈
R
n
{\displaystyle \mathbf {a} _{i}\in \mathbb {R} ^{n}}
∀
i
a
i
≠
o
{\displaystyle \forall i\quad \mathbf {a} _{i}\neq \mathbf {o} }
∀
(
i
≠
j
)
a
i
⋅
a
j
=
0
{\displaystyle \forall (i\neq j)\quad \mathbf {a} _{i}\cdot \mathbf {a} _{j}=0}
Gilt zusätzlich noch
∀
i
:
‖
a
i
‖
=
1
{\displaystyle \forall i:\;\|\mathbf {a} _{i}\|=1}
dann bilden die Vektoren ein Orthonormalsystem.
Gram-Schmidtsches Orthogonalisierungsverfahren
Bearbeiten
Wikipedia: Gram-Schmidtsches Orthogonalisierungsverfahren
Es ist der Satz von Thales "Jeder Peripheriewinkel über dem Durchmesser eines Kreises beträgt 90° " zu überprüfen.
a
=
r
+
c
{\displaystyle \mathbf {a} =\mathbf {r} +\mathbf {c} }
b
=
r
−
c
{\displaystyle \mathbf {b} =\mathbf {r} -\mathbf {c} }
‖
r
‖
=
‖
c
‖
{\displaystyle \|\mathbf {r} \|=\|\mathbf {c} \|}
a
⋅
b
=
(
r
+
c
)
⋅
(
r
−
c
)
=
r
r
+
r
c
−
r
c
−
c
c
=
‖
r
‖
2
−
‖
c
‖
2
=
0
{\displaystyle \mathbf {a} \cdot \mathbf {b} =(\mathbf {r} +\mathbf {c} )\cdot (\mathbf {r} -\mathbf {c} )=\mathbf {r} \mathbf {r} +\mathbf {r} \mathbf {c} -\mathbf {r} \mathbf {c} -\mathbf {c} \mathbf {c} =\|\mathbf {r} \|^{2}-\|\mathbf {c} \|^{2}=0}
⇒
a
⊥
b
{\displaystyle \Rightarrow \;\mathbf {a} \bot \mathbf {b} }
Den Vektor
b
=
α
1
a
1
+
…
+
α
n
a
n
{\displaystyle \mathbf {b} =\alpha _{1}\mathbf {a} _{1}+\ldots +\alpha _{n}\mathbf {a} _{n}}
nennt man Linearkombination aus den Vektoren
a
1
,
…
,
a
n
∈
R
n
{\displaystyle \mathbf {a} _{1},\ldots ,\mathbf {a} _{n}\;\in \mathbb {R} ^{n}}
b
=
0
{\displaystyle \mathbf {b} =0}
a
1
,
…
,
a
n
{\displaystyle \mathbf {a} _{1},\ldots ,\mathbf {a} _{n}}
linear abhängig, wenn
∃
i
:
α
i
≠
0
{\displaystyle \exists i:\;\alpha _{i}\neq 0}
linear unabhängig, wenn
∀
i
:
α
i
=
0
{\displaystyle \forall i:\;\alpha _{i}=0}
Die Einheitsvektoren (Basisvektoren)
e
1
=
(
1
0
⋮
0
)
,
⋯
,
e
n
=
(
0
0
⋮
1
)
{\displaystyle \mathbf {e} _{1}={\begin{pmatrix}1\\0\\\vdots \\0\end{pmatrix}},\;\cdots \;,\mathbf {e} _{n}={\begin{pmatrix}0\\0\\\vdots \\1\end{pmatrix}}}
nennt man natürliche Basis des
R
n
{\displaystyle \mathbb {R} ^{n}}
Für diese Basisvektoren gilt das sogenannte Kronecker-Delta
δ
i
j
=
e
i
⋅
e
j
=
{
1
,
wenn
i
=
k
0
,
wenn
i
≠
k
{\displaystyle \delta _{ij}=\mathbf {e} _{i}\cdot \mathbf {e} _{j}={\begin{cases}1,&{\mbox{wenn}}\ i=k\\0,&{\mbox{wenn}}\ i\neq k\end{cases}}}
x
=
(
x
1
,
…
,
x
n
)
∈
R
n
{\displaystyle \mathbf {x} =\left(x_{1},\ldots ,x_{n}\right)\in \mathbb {R} ^{n}}
x
=
x
1
e
1
+
…
+
x
n
e
n
{\displaystyle \mathbf {x} =x_{1}\mathbf {e} _{1}+\ldots +x_{n}\mathbf {e} _{n}}
x
1
e
1
,
…
,
x
n
e
n
{\displaystyle x_{1}\mathbf {e} _{1},\ldots ,x_{n}\mathbf {e} _{n}}
R
n
{\displaystyle \mathbb {R} ^{n}}
x
⋅
e
i
=
‖
x
‖
cos
α
i
{\displaystyle \mathbf {x} \cdot \mathbf {e} _{i}=\|\mathbf {x} \|\cos \alpha _{i}}
daraus folgt
cos
α
i
=
x
⋅
e
i
‖
x
‖
=
x
i
‖
x
‖
{\displaystyle \cos \alpha _{i}={\frac {\mathbf {x} \cdot \mathbf {e} _{i}}{\|\mathbf {x} \|}}={\frac {x_{i}}{\|\mathbf {x} \|}}}
Gesucht ist der Winkel, den der Vektor
a
=
(
3
,
−
1
,
−
5
)
{\displaystyle \mathbf {a} =(3,-1,-5)}
e
1
{\displaystyle \mathbf {e} _{1}}
a
⋅
e
1
=
‖
a
‖
cos
∠
{\displaystyle \mathbf {a} \cdot \mathbf {e} _{1}=\|\mathbf {a} \|\cos \angle }
cos
∠
=
a
⋅
e
1
‖
a
‖
=
3
35
{\displaystyle \cos \angle ={\frac {\mathbf {a} \cdot \mathbf {e} _{1}}{\|\mathbf {a} \|}}={\frac {3}{\sqrt {35}}}}
∠
=
1
,
04
r
a
d
{\displaystyle \angle =1,04\ {\rm {rad}}}
a
⋅
b
=
‖
b
‖
‖
a
‖
cos
∠
⏟
‖
a
b
‖
{\displaystyle \mathbf {a} \cdot \mathbf {b} =\|\mathbf {b} \|\underbrace {\|\mathbf {a} \|\cos \angle } _{\|\mathbf {a} _{\mathbf {b} }\|}}
a auf b
‖
a
b
‖
=
a
⋅
b
‖
b
‖
{\displaystyle \|\mathbf {a} _{\mathbf {b} }\|={\frac {\mathbf {a} \cdot \mathbf {b} }{\|\mathbf {b} \|}}}
a
b
=
‖
a
b
‖
b
0
{\displaystyle \mathbf {a} _{\mathbf {b} }=\|\mathbf {a} _{\mathbf {b} }\|\ \mathbf {b} _{0}}
b
0
=
b
‖
b
‖
{\displaystyle \mathbf {b} _{0}={\frac {\mathbf {b} }{\|\mathbf {b} \|}}}
a
b
=
a
⋅
b
‖
b
‖
2
b
{\displaystyle \mathbf {a} _{\mathbf {b} }={\frac {\mathbf {a} \cdot \mathbf {b} }{\|\mathbf {b} \|^{2}}}\mathbf {b} }
Gegeben ist ein Vektor
a
=
(
3
,
−
1
,
−
5
)
{\displaystyle \mathbf {a} =(3,-1,-5)}
a auf die
e
1
e
2
{\displaystyle \mathbf {e} _{1}\mathbf {e} _{2}}
e
1
{\displaystyle \mathbf {e} _{1}}
a
1
=
a
⋅
e
1
‖
e
1
‖
e
1
=
a
1
e
1
=
(
3
,
0
,
0
)
{\displaystyle \mathbf {a} _{1}={\frac {\mathbf {a} \cdot \mathbf {e} _{1}}{\|\mathbf {e} _{1}\|}}\mathbf {e} _{1}=a_{1}\mathbf {e} _{1}=(3,0,0)}
a
2
=
a
⋅
e
2
‖
e
2
‖
e
2
=
a
2
e
2
=
(
0
,
−
1
,
0
)
{\displaystyle \mathbf {a} _{2}={\frac {\mathbf {a} \cdot \mathbf {e} _{2}}{\|\mathbf {e} _{2}\|}}\mathbf {e} _{2}=a_{2}\mathbf {e} _{2}=(0,-1,0)}
a
12
=
a
1
+
a
2
=
(
3
,
−
1
,
0
)
{\displaystyle \mathbf {a} _{12}=\mathbf {a} _{1}+\mathbf {a} _{2}=(3,-1,0)}
Dieses Zwischenergebnis hätte man natürlich auch direkt aus der Skizze ermitteln können, dann wäre aber der Witz dieses Beispiels verloren gegangen.
cos
∠
=
a
12
e
1
‖
a
12
‖
=
3
10
{\displaystyle \cos \angle ={\frac {\mathbf {a} _{12}\mathbf {e} _{1}}{\|\mathbf {a} _{12}\|}}={\frac {3}{\sqrt {10}}}}
∠
=
0
,
32
r
a
d
{\displaystyle \angle =0,32\ {\rm {rad}}}
Das Vektorprodukt zweier linear unabhängiger Vektoren ist nur für den euklidischen Raum
R
3
{\displaystyle \mathbb {R} ^{3}}
a
×
b
=
(
a
2
b
3
−
a
3
b
2
,
a
3
b
1
−
a
1
b
3
,
a
1
b
2
−
a
2
b
1
)
{\displaystyle \mathbf {a} \times \mathbf {b} =\left(a_{2}b_{3}-a_{3}b_{2},\ a_{3}b_{1}-a_{1}b_{3},\ a_{1}b_{2}-a_{2}b_{1}\right)}
Die Vektoren
a
,
b
,
a
×
b
{\displaystyle \mathbf {a} ,\mathbf {b} ,\mathbf {a} \times \mathbf {b} }
Das Vektorprodukt ist nicht kommutativ:
a
×
b
=
−
(
b
×
a
)
{\displaystyle \mathbf {a} \times \mathbf {b} =-(\mathbf {b} \times \mathbf {a} )}
Orthogonalität:
(
a
⊥
a
×
b
)
∧
(
b
⊥
a
×
b
)
{\displaystyle (\mathbf {a} \;\bot \;\mathbf {a} \times \mathbf {b} )\wedge (\mathbf {b} \;\bot \;\mathbf {a} \times \mathbf {b} )}
Parallelität:
a
×
b
=
o
⇔
{
a
=
o
∨
b
=
o
a
‖
b
{\displaystyle \mathbf {a} \times \mathbf {b} =\mathbf {o} \Leftrightarrow {\begin{cases}\mathbf {a} =\mathbf {o} \;\vee \;\mathbf {b} =\mathbf {o} \\\mathbf {a} \|\mathbf {b} \end{cases}}}
Lagrangesche Identität:
‖
a
×
b
‖
2
=
‖
a
‖
2
‖
b
‖
2
−
(
a
⋅
b
)
2
{\displaystyle \|\mathbf {a} \times \mathbf {b} \|^{2}=\|\mathbf {a} \|^{2}\|\mathbf {b} \|^{2}-(\mathbf {a} \cdot \mathbf {b} )^{2}}
Parallelogramm:
F
◊
=
‖
a
‖
‖
b
‖
sin
∠
=
‖
a
×
b
‖
{\displaystyle F_{\Diamond }=\|\mathbf {a} \|\|\mathbf {b} \|\sin \angle =\|\mathbf {a} \times \mathbf {b} \|}
a
,
b
{\displaystyle \mathbf {a} ,\mathbf {b} }
∠
{\displaystyle \angle }
a und b eingeschlossene Winkel.
Distributivgesetz:
(
a
+
b
)
×
c
=
(
a
×
b
)
+
(
b
×
c
)
{\displaystyle (\mathbf {a} +\mathbf {b} )\times \mathbf {c} =(\mathbf {a} \times \mathbf {b} )+(\mathbf {b} \times \mathbf {c} )}
a
×
(
b
×
c
)
=
(
a
⋅
c
)
b
+
(
a
⋅
b
)
c
{\displaystyle \mathbf {a} \times (\mathbf {b} \times \mathbf {c} )=(\mathbf {a} \cdot \mathbf {c} )\mathbf {b} +(\mathbf {a} \cdot \mathbf {b} )\mathbf {c} }
A
×
B
=
(
a
1
a
2
a
3
)
×
(
b
1
b
2
b
3
)
=
(
a
2
b
3
−
a
3
b
2
a
3
b
1
−
a
1
b
3
a
1
b
2
−
a
2
b
1
)
{\displaystyle \mathbf {A} \times \mathbf {B} ={\begin{pmatrix}a_{1}\\a_{2}\\a_{3}\end{pmatrix}}\times {\begin{pmatrix}b_{1}\\b_{2}\\b_{3}\end{pmatrix}}={\begin{pmatrix}a_{2}b_{3}-a_{3}b_{2}\\a_{3}b_{1}-a_{1}b_{3}\\a_{1}b_{2}-a_{2}b_{1}\end{pmatrix}}}
A
×
A
=
(
a
2
a
3
−
a
3
a
2
a
3
a
1
−
a
1
a
3
a
1
a
2
−
a
2
a
1
)
=
(
0
0
0
)
{\displaystyle \mathbf {A} \times \mathbf {A} ={\begin{pmatrix}a_{2}a_{3}-a_{3}a_{2}\\a_{3}a_{1}-a_{1}a_{3}\\a_{1}a_{2}-a_{2}a_{1}\end{pmatrix}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}}
B
×
A
=
(
b
2
a
3
−
b
3
a
2
b
3
a
1
−
b
1
a
3
b
1
a
2
−
b
2
a
1
)
=
−
(
a
2
b
3
−
a
3
b
2
a
3
b
1
−
a
1
b
3
a
1
b
2
−
a
2
b
1
)
=
−
A
×
B
{\displaystyle \mathbf {B} \times \mathbf {A} ={\begin{pmatrix}b_{2}a_{3}-b_{3}a_{2}\\b_{3}a_{1}-b_{1}a_{3}\\b_{1}a_{2}-b_{2}a_{1}\end{pmatrix}}=-{\begin{pmatrix}a_{2}b_{3}-a_{3}b_{2}\\a_{3}b_{1}-a_{1}b_{3}\\a_{1}b_{2}-a_{2}b_{1}\end{pmatrix}}=-\mathbf {A} \times \mathbf {B} }
B
(
A
⋅
C
)
−
C
(
A
⋅
B
)
=
(
a
1
b
1
c
1
+
a
2
b
1
c
2
+
a
3
b
1
c
3
a
1
b
2
c
1
+
a
2
b
2
c
2
+
a
3
b
2
c
3
a
1
b
3
c
1
+
a
2
b
3
c
2
+
a
3
b
3
c
3
)
−
(
a
1
b
1
c
1
+
a
2
b
2
c
1
+
a
3
b
3
c
1
a
1
b
1
c
2
+
a
2
b
2
c
2
+
a
3
b
3
c
2
a
1
b
1
c
3
+
a
2
b
2
c
3
+
a
3
b
3
c
3
)
=
(
a
2
b
1
c
2
+
a
3
b
1
c
3
−
a
2
b
2
c
1
−
a
3
b
3
c
1
a
1
b
2
c
1
+
a
3
b
2
c
3
−
a
1
b
1
c
2
−
a
3
b
3
c
2
a
1
b
3
c
1
+
a
2
b
3
c
2
−
a
1
b
1
c
3
−
a
2
b
2
c
3
)
{\displaystyle {\begin{matrix}\mathbf {B} (\mathbf {A} \cdot \mathbf {C} )-\mathbf {C} (\mathbf {A} \cdot \mathbf {B} )&=&{\begin{pmatrix}a_{1}b_{1}c_{1}+a_{2}b_{1}c_{2}+a_{3}b_{1}c_{3}\\a_{1}b_{2}c_{1}+a_{2}b_{2}c_{2}+a_{3}b_{2}c_{3}\\a_{1}b_{3}c_{1}+a_{2}b_{3}c_{2}+a_{3}b_{3}c_{3}\\\end{pmatrix}}-{\begin{pmatrix}a_{1}b_{1}c_{1}+a_{2}b_{2}c_{1}+a_{3}b_{3}c_{1}\\a_{1}b_{1}c_{2}+a_{2}b_{2}c_{2}+a_{3}b_{3}c_{2}\\a_{1}b_{1}c_{3}+a_{2}b_{2}c_{3}+a_{3}b_{3}c_{3}\end{pmatrix}}\\&=&{\begin{pmatrix}a_{2}b_{1}c_{2}+a_{3}b_{1}c_{3}-a_{2}b_{2}c_{1}-a_{3}b_{3}c_{1}\\a_{1}b_{2}c_{1}+a_{3}b_{2}c_{3}-a_{1}b_{1}c_{2}-a_{3}b_{3}c_{2}\\a_{1}b_{3}c_{1}+a_{2}b_{3}c_{2}-a_{1}b_{1}c_{3}-a_{2}b_{2}c_{3}\end{pmatrix}}\end{matrix}}}
A
×
(
B
×
C
)
=
(
a
1
a
2
a
3
)
×
(
b
2
c
3
−
b
3
c
2
b
3
c
1
−
b
1
c
3
b
1
c
2
−
b
2
c
1
)
=
(
a
2
b
1
c
2
−
a
2
b
2
c
1
−
a
3
b
3
c
1
+
a
3
b
1
c
3
a
3
b
2
c
3
−
a
3
b
3
c
2
−
a
1
b
1
c
2
+
a
1
b
2
c
1
a
1
b
3
c
1
−
a
1
b
1
c
3
−
a
2
b
2
c
3
+
a
2
b
3
c
2
)
=
B
(
A
⋅
C
)
−
C
(
A
⋅
B
)
{\displaystyle {\begin{matrix}\mathbf {A} \times (\mathbf {B} \times \mathbf {C} )&=&{\begin{pmatrix}a_{1}\\a_{2}\\a_{3}\end{pmatrix}}\times {\begin{pmatrix}b_{2}c_{3}-b_{3}c_{2}\\b_{3}c_{1}-b_{1}c_{3}\\b_{1}c_{2}-b_{2}c_{1}\end{pmatrix}}&=&{\begin{pmatrix}a_{2}b_{1}c_{2}-a_{2}b_{2}c_{1}-a_{3}b_{3}c_{1}+a_{3}b_{1}c_{3}\\a_{3}b_{2}c_{3}-a_{3}b_{3}c_{2}-a_{1}b_{1}c_{2}+a_{1}b_{2}c_{1}\\a_{1}b_{3}c_{1}-a_{1}b_{1}c_{3}-a_{2}b_{2}c_{3}+a_{2}b_{3}c_{2}\end{pmatrix}}\\&=&\mathbf {B} (\mathbf {A} \cdot \mathbf {C} )-\mathbf {C} (\mathbf {A} \cdot \mathbf {B} )&&\end{matrix}}}
ϵ
i
j
k
{\displaystyle \epsilon _{ijk}}
(
A
×
B
)
i
=
ϵ
i
j
k
A
j
B
k
{\displaystyle \left(\mathbf {A} \times \mathbf {B} \right)_{i}=\epsilon _{ijk}\mathbf {A} _{j}\mathbf {B} _{k}}
Gesucht ist ein normierter Normalenvektor auf die beiden Vektoren
a
=
(
1
,
1
,
0
)
,
b
=
(
5
,
4
,
1
)
{\displaystyle \mathbf {a} =(1,1,0),\ \mathbf {b} =(5,4,1)}
n
=
a
×
b
=
(
a
2
b
3
−
a
3
b
2
,
a
3
b
1
−
a
1
b
3
,
a
1
b
2
−
a
2
b
1
)
=
(
1
,
−
1
,
−
1
)
{\displaystyle \mathbf {n} =\mathbf {a} \times \mathbf {b} =\left(a_{2}b_{3}-a_{3}b_{2},\ a_{3}b_{1}-a_{1}b_{3},\ a_{1}b_{2}-a_{2}b_{1}\right)=(1,-1,-1)}
‖
n
‖
=
3
{\displaystyle \|\mathbf {n} \|={\sqrt {3}}}
n
0
=
1
3
(
1
,
−
1
,
−
1
)
{\displaystyle \mathbf {n} _{0}={\frac {1}{\sqrt {3}}}(1,-1,-1)}
Das Spatprodukt der Vektoren
a
,
b
,
c
∈
R
3
{\displaystyle \mathbf {a} ,\mathbf {b} ,\mathbf {c} \in \mathbb {R} ^{3}}
h
a
,
b
,
c
i
=
(
a
×
b
)
⋅
c
{\displaystyle {\mathcal {h}}\mathbf {a} ,\mathbf {b} ,\mathbf {c} {\mathcal {i}}=(\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} }
Unabhängigkeit: Die Vektoren a , b , c sind linear unabhängig, wenn
h
a
,
b
,
c
i
≠
0
{\displaystyle {\mathcal {h}}\mathbf {a} ,\mathbf {b} ,\mathbf {c} {\mathcal {i}}\neq 0}
Rechtssystem:
h
a
,
b
,
c
i
>
0
{\displaystyle {\mathcal {h}}\mathbf {a} ,\mathbf {b} ,\mathbf {c} {\mathcal {i}}>0}
Linkssystem:
h
a
,
b
,
c
i
<
0
{\displaystyle {\mathcal {h}}\mathbf {a} ,\mathbf {b} ,\mathbf {c} {\mathcal {i}}<0}
Spatvolumen:
V
s
p
a
t
=
|
h
a
,
b
,
c
i
|
{\displaystyle V_{spat}=|{\mathcal {h}}\mathbf {a} ,\mathbf {b} ,\mathbf {c} {\mathcal {i}}|}
Vertauschungssatz:
h
a
,
b
,
c
i
=
h
b
,
c
,
a
i
=
h
c
,
a
,
b
i
=
−
h
b
,
a
,
c
i
=
−
h
a
,
c
,
b
i
=
−
h
c
,
b
,
a
i
{\displaystyle {\mathcal {h}}\mathbf {a} ,\mathbf {b} ,\mathbf {c} {\mathcal {i}}={\mathcal {h}}\mathbf {b} ,\mathbf {c} ,\mathbf {a} {\mathcal {i}}={\mathcal {h}}\mathbf {c} ,\mathbf {a} ,\mathbf {b} {\mathcal {i}}=-{\mathcal {h}}\mathbf {b} ,\mathbf {a} ,\mathbf {c} {\mathcal {i}}=-{\mathcal {h}}\mathbf {a} ,\mathbf {c} ,\mathbf {b} {\mathcal {i}}=-{\mathcal {h}}\mathbf {c} ,\mathbf {b} ,\mathbf {a} {\mathcal {i}}}
A (Ortsvektor a ) und ein Richtungsvektor c . Der Punkt X liegt auf der Geraden g (
X
∈
g
{\displaystyle X\in g}
g
:
x
=
a
+
λ
c
,
λ
∈
R
∧
a
,
c
∈
R
n
{\displaystyle g:\;\mathbf {x} =\mathbf {a} +\lambda \mathbf {c} ,\quad \lambda \in \mathbb {R} \;\wedge \;\mathbf {a} ,\mathbf {c} \in \mathbb {R} ^{n}}
Obige Gleichung nennt man auch eine Parameterdarstellung (Parameter
λ
{\displaystyle \lambda }
A und B gegeben (Ortsvektoren a ,b ). Der Punkt X liegt auf der Geraden g (
X
∈
g
{\displaystyle X\in g}
g
:
x
=
a
+
λ
(
b
−
a
)
,
λ
∈
R
∧
a
,
b
∈
R
n
{\displaystyle g:\;\mathbf {x} =\mathbf {a} +\lambda (\mathbf {b} -\mathbf {a} ),\quad \lambda \in \mathbb {R} \;\wedge \;\mathbf {a} ,\mathbf {b} \in \mathbb {R} ^{n}}
Die Punkte A , B und X heißen kollinear, d.h. diese drei Punkte liegen auf der gleichen Geraden.
Ebene durch einen Punkt und zwei Richtungen
Bearbeiten
A (Ortsvektor a ) und zwei Richtungsvektoren c ,d . Der Punkt X liegt in der Ebene E (
X
∈
E
{\displaystyle X\in E}
E
:
x
=
a
+
λ
c
+
μ
d
,
λ
,
μ
∈
R
∧
a
,
c
,
d
∈
R
n
{\displaystyle E:\;\mathbf {x} =\mathbf {a} +\lambda \mathbf {c} +\mu \mathbf {d} ,\quad \lambda ,\mu \in \mathbb {R} \;\wedge \;\mathbf {a} ,\mathbf {c} ,\mathbf {d} \in \mathbb {R} ^{n}}
Die Ebene E wird also durch die zwei Geraden g und h aufgespannt.
A , B und C (Ortsvektoren a ,b ,c ). Der Punkt X liegt in der der Ebene E (
X
∈
E
{\displaystyle X\in E}
E
:
x
=
a
+
λ
(
b
−
a
)
+
μ
(
c
−
a
)
,
λ
,
μ
∈
R
∧
a
,
b
,
c
∈
R
n
{\displaystyle E:\;\mathbf {x} =\mathbf {a} +\lambda (\mathbf {b} -\mathbf {a} )+\mu (\mathbf {c} -\mathbf {a} ),\quad \lambda ,\mu \in \mathbb {R} \;\wedge \;\mathbf {a} ,\mathbf {b} ,\mathbf {c} \in \mathbb {R} ^{n}}
V
S
p
a
t
=
|
h
x
−
a
,
b
−
a
,
c
−
a
i
|
{\displaystyle V_{Spat}=|{\mathcal {h}}\mathbf {x} -\mathbf {a} ,\;\mathbf {b} -\mathbf {a} ,\;\mathbf {c} -\mathbf {a} {\mathcal {i}}|}
B , C und X in einer Ebene liegen, ist
V
S
p
a
t
=
0
{\displaystyle V_{Spat}=0}
E
:
h
x
−
a
,
b
−
a
,
c
−
a
i
=
0
{\displaystyle E:\;{\mathcal {h}}\mathbf {x} -\mathbf {a} ,\;\mathbf {b} -\mathbf {a} ,\;\mathbf {c} -\mathbf {a} {\mathcal {i}}=0}
n sei ein Normalenvektor zur Ebene E . Es gilt
n
⊥
(
x
−
a
)
{\displaystyle \mathbf {n} \;\bot \;(\mathbf {x} -\mathbf {a} )}
n
⋅
(
x
−
a
)
=
0
{\displaystyle \mathbf {n} \cdot (\mathbf {x} -\mathbf {a} )=0}
Dividiert man diese Gleichung durch die Norm des Normalenvektors
‖
n
‖
{\displaystyle \|\mathbf {n} \|}
E
:
n
0
⋅
(
x
−
a
)
=
0
{\displaystyle E:\mathbf {n} _{0}\cdot (\mathbf {x} -\mathbf {a} )=0}
Aus der Parameterdarstellung
E
:
x
=
a
+
λ
b
+
μ
c
;
a
,
b
,
c
,
x
∈
R
3
{\displaystyle E:\mathbf {x} =\mathbf {a} +\lambda \mathbf {b} +\mu \mathbf {c} \ ;\quad \mathbf {a} ,\mathbf {b} ,\mathbf {c} ,\mathbf {x} \in \mathbb {R} ^{3}}
erhält man in Koordinaten angeschrieben
x
1
=
a
1
+
λ
b
1
+
μ
c
1
x
2
=
a
2
+
λ
b
2
+
μ
c
2
x
3
=
a
3
+
λ
b
3
+
μ
c
3
{\displaystyle {\begin{matrix}x_{1}=a_{1}+\lambda b_{1}+\mu c_{1}\\x_{2}=a_{2}+\lambda b_{2}+\mu c_{2}\\x_{3}=a_{3}+\lambda b_{3}+\mu c_{3}\\\end{matrix}}}
Eliminiert man
λ
,
μ
{\displaystyle \lambda ,\ \mu }
E
:
a
x
+
b
y
+
c
z
=
d
{\displaystyle E:\;ax+by+cz=d}
Bearbeiten
R
3
{\displaystyle \mathbb {R} ^{3}}
g
:
x
1
=
a
+
λ
b
{\displaystyle g:\mathbf {x} _{1}=\mathbf {a} +\lambda \mathbf {b} }
h
:
x
2
=
a
+
μ
c
{\displaystyle h:\mathbf {x} _{2}=\mathbf {a} +\mu \mathbf {c} }
g und h kann man über das Skalarprodukt gewinnen
cos
∠
=
x
1
⋅
x
2
‖
x
1
‖
‖
x
2
‖
{\displaystyle \cos \angle ={\frac {\mathbf {x} _{1}\cdot \mathbf {x} _{2}}{\|\mathbf {x} _{1}\|\|\mathbf {x} _{2}\|}}}
P und eine Ebene E. D sei der Durchstoßpunkt von n durch die Ebene E .
P zu Ebene E ist
δ
=
P
D
¯
=
‖
p
−
d
‖
{\displaystyle \delta ={\overline {PD}}=\|\mathbf {p} -\mathbf {d} \|}
(
d
−
a
)
⋅
n
=
0
{\displaystyle (\mathbf {d} -\mathbf {a} )\cdot \mathbf {n} =0}
Weiters ist
d
=
p
+
λ
n
{\displaystyle \mathbf {d} =\mathbf {p} +\lambda \mathbf {n} }
λ
{\displaystyle \lambda }
(
p
+
λ
n
−
a
)
n
=
0
{\displaystyle (\mathbf {p} +\lambda \mathbf {n} -\mathbf {a} )\mathbf {n} =0}
(
p
−
a
)
n
+
λ
n
n
=
0
{\displaystyle (\mathbf {p} -\mathbf {a} )\mathbf {n} +\lambda \mathbf {n} \mathbf {n} =0}
λ
=
−
(
p
−
a
)
n
n
⋅
n
=
−
(
p
−
a
)
n
‖
n
‖
2
{\displaystyle \lambda =-{\frac {(\mathbf {p} -\mathbf {a} )\mathbf {n} }{\mathbf {n} \cdot \mathbf {n} }}=-{\frac {(\mathbf {p} -\mathbf {a} )\mathbf {n} }{\|\mathbf {n} \|^{2}}}}
δ
=
‖
p
−
p
+
(
p
−
a
)
n
‖
n
‖
2
n
‖
=
‖
(
p
−
a
)
n
0
⏟
k
n
0
‖
=
|
k
|
‖
n
0
‖
=
|
k
|
=
|
(
p
−
a
)
n
0
|
{\displaystyle \delta =\|\mathbf {p} -\mathbf {p} +{\frac {(\mathbf {p} -\mathbf {a} )\mathbf {n} }{\|\mathbf {n} \|^{2}}}\mathbf {n} \|=\|\underbrace {(\mathbf {p} -\mathbf {a} )\mathbf {n} _{0}} _{k}\mathbf {n_{0}} \|=|k|\|\mathbf {n} _{0}\|=|k|=|(\mathbf {p} -\mathbf {a} )\mathbf {n} _{0}|}
R
3
{\displaystyle \mathbb {R} ^{3}}
g
=
{
E
1
:
a
1
x
1
+
a
2
x
2
+
a
3
x
3
=
c
E
2
:
b
1
x
1
+
b
2
x
2
+
b
3
x
3
=
d
{\displaystyle g={\begin{cases}E_{1}:\ a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}=c\\E_{2}:\ b_{1}x_{1}+b_{2}x_{2}+b_{3}x_{3}=d\end{cases}}}
darstellen. Natürlich dürfen die Ebenen hierbei nicht parallel liegen.
x
=
a
+
λ
b
{\displaystyle \mathbf {x} =\mathbf {a} +\lambda \mathbf {b} }
a
=
(
1
,
1
,
0
)
,
b
=
(
3
,
4
,
−
2
)
,
λ
∈
R
{\displaystyle \mathbf {a} =(1,1,0),\ \mathbf {b} =(3,4,-2),\ \lambda \in \mathbb {R} }
g als Schnittgerade besitzen.
x
1
=
1
+
3
λ
{\displaystyle x_{1}=\ 1+3\lambda }
x
2
=
1
+
4
λ
{\displaystyle x_{2}=\ 1+4\lambda }
x
3
=
−
2
λ
{\displaystyle x_{3}=\ -2\lambda }
λ
=
−
x
3
2
{\displaystyle \lambda =-{\frac {x_{3}}{2}}}
E
1
:
x
1
+
3
2
x
3
=
1
{\displaystyle E_{1}:\ x_{1}+{\frac {3}{2}}x_{3}=1}
E
2
:
x
2
+
2
x
3
=
1
{\displaystyle E_{2}:\ x_{2}+2x_{3}=1}
Beispiel: Gegeben seien zwei Ebenen
E
1
:
x
1
+
x
2
+
x
3
=
1
;
E
2
:
x
1
+
x
3
=
−
3
{\displaystyle E_{1}:\;x_{1}+x_{2}+x_{3}=1;\;E_{2}:\;x_{1}+x_{3}=-3}
x
1
=
λ
{\displaystyle x_{1}=\ \lambda }
und berechnen sukzessive
x
3
=
−
3
−
λ
{\displaystyle x_{3}=\ -3-\lambda }
x
2
=
4
{\displaystyle x_{2}=\ 4}
g
:
x
=
(
x
1
x
2
x
3
)
=
(
0
4
−
3
)
+
λ
(
1
0
−
1
)
{\displaystyle g:\;\mathbf {x} ={\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}={\begin{pmatrix}0\\4\\-3\end{pmatrix}}+\lambda {\begin{pmatrix}1\\0\\-1\end{pmatrix}}}
Bearbeiten
Es gibt drei Möglichkeiten:
Die Gerade g schneidet die Ebene E in einem Punkt
Die Gerade liegt parallel zur Ebene, aber nicht in der Ebene
Die Gerade liegt in der Ebene
g
:
b
=
(
1
0
0
)
+
λ
(
−
1
5
0
)
;
α
=
R
{\displaystyle g:\;\mathbf {b} ={\begin{pmatrix}1\\0\\0\end{pmatrix}}+\lambda {\begin{pmatrix}-1\\5\\0\end{pmatrix}}\ ;\;\alpha =\mathbb {R} }
E
:
x
1
+
x
2
+
x
3
=
0
{\displaystyle E:\;x_{1}+x_{2}+x_{3}=0}
g :
x
2
=
5
(
1
−
x
1
)
{\displaystyle x_{2}=\ 5\left(1-x_{1}\right)}
x
3
=
0
{\displaystyle x_{3}=\ 0}
E :
x
1
+
x
2
+
x
3
=
0
{\displaystyle x_{1}+x_{2}+x_{3}=\ 0}
Durchstoßpunkt
D
=
(
5
4
−
5
4
0
)
{\displaystyle D={\begin{pmatrix}{\frac {5}{4}}\\-{\frac {5}{4}}\\0\end{pmatrix}}}
Winkel zwischen einer Geraden und einer Ebene
Bearbeiten
a
⋅
n
=
‖
a
‖
‖
n
‖
cos
(
π
2
−
α
)
{\displaystyle \mathbf {a} \cdot \mathbf {n} =\|\mathbf {a} \|\|\mathbf {n} \|\cos \left({\frac {\pi }{2}}-\alpha \right)}
Daraus kann man leicht den Winkel
α
{\displaystyle \alpha }
Ähnlich funktioniert die Bestimmung eines Winkels zwischen zwei Ebenen. Dort verwendet man eben die Normalenvektoren der beiden Ebenen zur Winkelberechnung.
