Die Dreiecksberechnung erfolgt in diesem Anwendungsbeispiel hauptsächlich mittels Vektorrechnung.
Näheres zu Dreiecken und zur Vektorrechnung ist folgenden Enzyklopädieartikeln und Büchern zu entnehmen:
Koordinatenwerte ---> Richtungsvektoren Bearbeiten
a
=
p
3
−
p
1
{\displaystyle \mathbf {a} =\mathbf {p} _{3}-\mathbf {p} _{1}}
b
=
p
2
−
p
1
{\displaystyle \mathbf {b} =\mathbf {p} _{2}-\mathbf {p} _{1}}
c
=
b
−
a
{\displaystyle \mathbf {c} =\mathbf {b} -\mathbf {a} }
Seitenlängen und Umfang Bearbeiten
a
n
o
r
m
=
|
a
|
{\displaystyle a_{norm}=|\mathbf {a} |}
b
n
o
r
m
=
|
b
|
{\displaystyle b_{norm}=|\mathbf {b} |}
c
n
o
r
m
=
|
c
|
{\displaystyle c_{norm}=|\mathbf {c} |}
U
=
a
n
o
r
m
+
b
n
o
r
m
+
c
n
o
r
m
{\displaystyle U\ =\ a_{norm}+b_{norm}+c_{norm}}
Bedingung:
a
n
o
r
m
≠
0
{\displaystyle a_{norm}\neq 0}
b
n
o
r
m
≠
0
{\displaystyle b_{norm}\neq 0}
c
n
o
r
m
≠
0
{\displaystyle c_{norm}\neq 0}
∠
a
b
=
arccos
a
⋅
b
a
n
o
r
m
b
n
o
r
m
{\displaystyle \angle ab=\arccos {\frac {\mathbf {a} \cdot \mathbf {b} }{a_{norm}b_{norm}}}}
∠
b
c
=
arccos
b
⋅
c
b
n
o
r
m
c
n
o
r
m
{\displaystyle \angle bc=\arccos {\frac {\mathbf {b} \cdot \mathbf {c} }{b_{norm}c_{norm}}}}
∠
a
c
=
π
−
∠
a
b
−
∠
b
c
{\displaystyle \angle ac=\pi -\angle ab-\angle bc}
Bedingung:
∠
a
b
≠
0
{\displaystyle \angle ab\neq 0}
∠
a
c
≠
0
{\displaystyle \angle ac\neq 0}
∠
b
c
≠
0
{\displaystyle \angle bc\neq 0}
Es gilt
A
P
a
r
a
l
l
e
l
o
g
r
a
m
m
=
|
a
×
b
|
=
a
n
o
r
m
b
n
o
r
m
sin
∠
a
b
{\displaystyle A_{Parallelogramm}=|\mathbf {a} \times \mathbf {b} |=a_{norm}b_{norm}\sin \angle ab}
und somit
A
=
A
p
a
r
a
l
l
e
l
o
g
r
a
m
m
2
=
a
n
o
r
m
b
n
o
r
m
sin
∠
a
b
2
{\displaystyle A={\frac {A_{parallelogramm}}{2}}={\frac {a_{norm}b_{norm}\sin \angle ab}{2}}}
Normalen:
a
⋅
n
=
0
{\displaystyle \mathbf {a} \cdot \mathbf {n} =0}
b
⋅
m
=
0
{\displaystyle \mathbf {b} \cdot \mathbf {m} =0}
|
n
|
=
1
{\displaystyle |\mathbf {n} |=1}
|
m
|
=
1
{\displaystyle |\mathbf {m} |=1}
⇒
{\displaystyle \Rightarrow }
n
1
=
−
a
2
1
a
1
2
+
a
2
2
{\displaystyle n_{1}=-a_{2}{\sqrt {\frac {1}{a_{1}^{2}+a_{2}^{2}}}}}
n
2
=
a
1
1
a
1
2
+
a
2
2
{\displaystyle n_{2}=a_{1}{\sqrt {\frac {1}{a_{1}^{2}+a_{2}^{2}}}}}
m
1
=
−
b
2
1
b
1
2
+
b
2
2
{\displaystyle m_{1}=-b_{2}{\sqrt {\frac {1}{b_{1}^{2}+b_{2}^{2}}}}}
m
2
=
b
1
1
b
1
2
+
b
2
2
{\displaystyle m_{2}=b_{1}{\sqrt {\frac {1}{b_{1}^{2}+b_{2}^{2}}}}}
g
1
:
x
1
=
a
2
+
t
1
n
{\displaystyle g1:\ \mathbf {x} _{1}={\frac {\mathbf {a} }{2}}+t_{1}\mathbf {n} }
g
2
:
x
2
=
b
2
+
t
2
m
{\displaystyle g2:\ \mathbf {x} _{2}={\frac {\mathbf {b} }{2}}+t_{2}\mathbf {m} }
x
=
x
1
=
x
2
{\displaystyle \mathbf {x} =\mathbf {x} _{1}=\mathbf {x} _{2}}
⇒
{\displaystyle \Rightarrow }
t
2
=
n
1
(
a
2
−
b
2
)
+
n
2
(
b
1
−
a
1
)
2
(
m
2
n
1
−
m
1
n
2
)
{\displaystyle t_{2}={\frac {n_{1}\left(a_{2}-b_{2}\right)+n_{2}\left(b_{1}-a_{1}\right)}{2\left(m_{2}n_{1}-m_{1}n_{2}\right)}}}
Bedingung:
m
2
n
1
−
m
1
n
2
≠
0
{\displaystyle m_{2}n_{1}-m_{1}n_{2}\neq 0}
x
o
=
p
1
+
b
2
+
t
2
m
{\displaystyle \mathbf {x} _{o}=\mathbf {p} _{1}+{\frac {\mathbf {b} }{2}}+t_{2}\mathbf {m} }
r
o
=
|
b
2
+
t
2
m
|
{\displaystyle r_{o}=\left|{\frac {\mathbf {b} }{2}}+t_{2}\mathbf {m} \right|}
v
=
1
2
(
a
a
n
o
r
m
+
b
b
n
o
r
m
)
{\displaystyle \mathbf {v} ={\frac {1}{2}}\left({\frac {\mathbf {a} }{a_{norm}}}+{\frac {\mathbf {b} }{b_{norm}}}\right)}
w
=
1
2
(
b
b
n
o
r
m
+
c
c
n
o
r
m
)
{\displaystyle \mathbf {w} ={\frac {1}{2}}\left({\frac {\mathbf {b} }{b_{norm}}}+{\frac {\mathbf {c} }{c_{norm}}}\right)}
g
1
:
x
1
=
t
1
v
{\displaystyle g1:\ \mathbf {x} _{1}=t_{1}\mathbf {v} }
g
2
:
x
2
=
b
+
t
2
v
{\displaystyle g2:\ \mathbf {x} _{2}=\mathbf {b} +t_{2}\mathbf {v} }
Der Inkreismittelpunkt ergibt sich als Schnittpunkt dieser beiden Geraden:
x
=
x
1
=
x
2
{\displaystyle \mathbf {x} =\mathbf {x} _{1}=\mathbf {x} _{2}}
⇒
{\displaystyle \Rightarrow }
t
2
=
b
1
v
2
−
b
2
v
1
w
2
v
1
−
w
1
v
2
{\displaystyle t_{2}={\frac {b_{1}v_{2}-b_{2}v_{1}}{w_{2}v_{1}-w_{1}v_{2}}}}
Bedingung:
w
2
v
1
−
w
1
v
2
≠
0
{\displaystyle w_{2}v_{1}-w_{1}v_{2}\neq 0}
x
i
=
p
2
+
t
2
w
{\displaystyle \mathbf {x} _{i}=\mathbf {p} _{2}+t_{2}\mathbf {w} }
r
i
=
2
A
a
n
o
r
m
+
b
n
o
r
m
+
c
n
o
r
m
{\displaystyle r_{i}={\frac {2A}{a_{norm}+b_{norm}+c_{norm}}}}
v
=
a
+
c
2
{\displaystyle \mathbf {v} =\mathbf {a} +{\frac {\mathbf {c} }{2}}}
w
=
b
2
−
a
{\displaystyle \mathbf {w} ={\frac {\mathbf {b} }{2}}-\mathbf {a} }
g
1
:
x
1
=
t
1
v
{\displaystyle g1:\ \mathbf {x} _{1}=t_{1}\mathbf {v} }
g
2
:
x
2
=
a
+
t
2
w
{\displaystyle g2:\ \mathbf {x} _{2}=\mathbf {a} +t_{2}\mathbf {w} }
Der Dreiecksschwerpunkt ergibt sich als Schnittpunkt dieser beiden Geraden:
x
=
x
1
=
x
2
{\displaystyle \mathbf {x} =\mathbf {x} _{1}=\mathbf {x} _{2}}
t
2
=
a
1
v
2
−
a
2
v
1
w
2
v
1
−
w
1
v
2
{\displaystyle t_{2}={\frac {a_{1}v_{2}-a_{2}v_{1}}{w_{2}v_{1}-w_{1}v_{2}}}}
Bedingung:
w
2
v
1
−
w
1
v
2
≠
0
{\displaystyle w_{2}v_{1}-w_{1}v_{2}\neq 0}
⇒
{\displaystyle \Rightarrow }
x
=
a
+
t
2
w
{\displaystyle \mathbf {x} =\mathbf {a} +t_{2}\mathbf {w} }
x
c
o
g
=
p
1
+
x
{\displaystyle \mathbf {x} _{cog}=\mathbf {p} _{1}+\mathbf {x} }
