0.0.1 ↑ 77. Hausaufgabe
0.0.1.1 ↑ Geometrie-Buch Seite 117, Aufgabe 2
Im Dreieck ABC ist \overrightarrow{BD} = \frac{3}{4} \overrightarrow{BD} und \overrightarrow{AS} = \frac{1}{2} \overrightarrow{AD}. BS schneidet AC in T.
In welchem Verhältnis teilt T die Strecke \overrightarrow{AC} beziehungsweise S die Strecke \overrightarrow{BT}?
\overrightarrow{AB}, \overrightarrow{AC} linear unabhängig.
\overrightarrow{AS} + \overrightarrow{ST} + \overrightarrow{TA} =\\{\quad}= \frac{1}{2} \underbrace{\left(\overrightarrow{AB} + \frac{3}{4} \overrightarrow{BC}\right)}_{\overrightarrow{AD}} + \underbrace{\lambda \overrightarrow{BT}}_{\overrightarrow{ST}} + \underbrace{\mu \overrightarrow{CA}}_{\overrightarrow{TA}} =\\{\quad}= \frac{1}{2} \overrightarrow{AB} + \frac{3}{8} \underbrace{\left(\overrightarrow{BA} + \overrightarrow{AC}\right)}_{\overrightarrow{BC}} + \lambda \underbrace{\left(\overrightarrow{BA} - \mu \overrightarrow{CA}\right)}_{\overrightarrow{BT}} + \mu \overrightarrow{CA} =\\{\quad}= \overrightarrow{AB} \left(\frac{1}{2} - \frac{3}{8} - \lambda\right) + \overrightarrow{AC} \left(\frac{3}{8} + \lambda\mu - \mu\right) =\\{\quad}= \vec 0;
\lambda = \frac{1}{2} - \frac{3}{8} = \frac{1}{8};
\mu = \frac{\frac{3}{8}}{1 - \lambda} = \frac{3}{7};
\overrightarrow{BS} = \beta \overrightarrow{ST}; ⇔ \beta = \frac{\overrightarrow{BS}}{\overrightarrow{ST}} = \frac{\overrightarrow{BT} + \overrightarrow{TS}}{\lambda \overrightarrow{BT}} = \frac{\overrightarrow{BT} - \lambda \overrightarrow{BT}}{\lambda \overrightarrow{BT}} = \frac{1 - \lambda}{\lambda} = 7;
\overrightarrow{AT} = \alpha \overrightarrow{TC}; ⇔ \alpha = \frac{\overrightarrow{AT}}{\overrightarrow{TC}} = \frac{\overrightarrow{AT}}{\overrightarrow{TA} + \overrightarrow{AC}} = \frac{\mu \overrightarrow{AC}}{-\mu \overrightarrow{AC} + \overrightarrow{AC}} = \frac{\mu}{1 - \mu} = \frac{3}{4};