Dreieck/Umfang und Flächeninhalt/1/Aufgabe/Lösung

${\displaystyle {}{\begin{aligned}&\,\,\,\,\,\,\,\Vert {{\begin{pmatrix}5\\2\end{pmatrix}}-{\begin{pmatrix}-1\\1\end{pmatrix}}}\Vert +\Vert {{\begin{pmatrix}5\\2\end{pmatrix}}-{\begin{pmatrix}2\\-3\end{pmatrix}}}\Vert +\Vert {{\begin{pmatrix}2\\-3\end{pmatrix}}-{\begin{pmatrix}-1\\1\end{pmatrix}}}\Vert \\&=\Vert {\begin{pmatrix}6\\1\end{pmatrix}}\Vert +\Vert {\begin{pmatrix}3\\5\end{pmatrix}}\Vert +\Vert {\begin{pmatrix}3\\-4\end{pmatrix}}\Vert \\&={\sqrt {37}}+{\sqrt {34}}+{\sqrt {25}}\\&={\sqrt {37}}+{\sqrt {34}}+5\,\end{aligned}}}$

${\displaystyle {}{\sqrt {37}}\geq 6\,}$

und

${\displaystyle {}{\sqrt {34}}\geq 5\,}$

ist der Umfang nach unten durch

${\displaystyle {}{\sqrt {37}}+{\sqrt {34}}+5\geq 16\,}$

beschränkt. Wir zeigen

${\displaystyle {}{\sqrt {37}}+{\sqrt {34}}+5\leq 17\,,}$

was zu

${\displaystyle {}{\sqrt {37}}+{\sqrt {34}}\leq 12\,}$

äquivalent ist. Quadrieren ergibt

${\displaystyle {}37+34+2{\sqrt {37\cdot 34}}=71+2{\sqrt {1258}}\leq 144\,}$

bzw.

${\displaystyle {}2{\sqrt {1258}}\leq 73\,.}$

Erneutes Quadrieren ergibt

${\displaystyle {}4\cdot 1258=5032\leq 5329\,,}$

was wahr ist.

${\displaystyle {}{\begin{pmatrix}-1\\1\end{pmatrix}}}$

${\displaystyle {\begin{pmatrix}3\\-4\end{pmatrix}}\,\,{\text{ und }}\,\,{\begin{pmatrix}6\\1\end{pmatrix}}}$

Daher ist der Kosinus des eingeschlossenen Winkels ${\displaystyle {}\alpha }$ nach Bemerkung gleich

${\displaystyle {}{\begin{aligned}\cos \alpha &={\frac {\left\langle {\begin{pmatrix}3\\-4\end{pmatrix}},{\begin{pmatrix}6\\1\end{pmatrix}}\right\rangle }{\Vert {\begin{pmatrix}3\\-4\end{pmatrix}}\Vert \cdot \Vert {\begin{pmatrix}6\\1\end{pmatrix}}\Vert }}\\&={\frac {18-4}{5{\sqrt {37}}}}\\&={\frac {14}{5{\sqrt {37}}}}.\end{aligned}}}$

${\displaystyle {}{\begin{pmatrix}5\\2\end{pmatrix}}}$

${\displaystyle {}{\begin{pmatrix}2\\-3\end{pmatrix}}}$

${\displaystyle {}{\begin{pmatrix}-1\\1\end{pmatrix}}}$

${\displaystyle {}{\begin{pmatrix}3\\-4\end{pmatrix}}}$

${\displaystyle {}{\begin{pmatrix}4\\3\end{pmatrix}}}$

${\displaystyle {\left\{{\begin{pmatrix}5\\2\end{pmatrix}}+s{\begin{pmatrix}4\\3\end{pmatrix}}\mid s\in \mathbb {R} \right\}}.}$

${\displaystyle {}{\begin{pmatrix}5\\2\end{pmatrix}}+s{\begin{pmatrix}4\\3\end{pmatrix}}={\begin{pmatrix}-1\\1\end{pmatrix}}+t{\begin{pmatrix}3\\-4\end{pmatrix}}\,}$

bzw.

${\displaystyle {}{\begin{pmatrix}6\\1\end{pmatrix}}=-s{\begin{pmatrix}4\\3\end{pmatrix}}+t{\begin{pmatrix}3\\-4\end{pmatrix}}\,.}$

Dies führt auf ${\displaystyle {}t={\frac {14}{25}}}$ und ${\displaystyle {}s=-{\frac {27}{25}}}$, der Höhenfußpunkt ist also

${\displaystyle {}{\begin{pmatrix}5\\2\end{pmatrix}}-{\frac {27}{25}}{\begin{pmatrix}4\\3\end{pmatrix}}={\begin{pmatrix}{\frac {17}{25}}\\-{\frac {31}{25}}\end{pmatrix}}\,.}$

${\displaystyle {}{\frac {1}{2}}\det {\begin{pmatrix}3&6\\-4&1\end{pmatrix}}={\frac {27}{2}}\,.}$