6. BC-BA=...

$$\overrightarrow{BC} \cdot \overrightarrow{BA} = |\overrightarrow{BC}| \cdot |\overrightarrow{BA}| \cdot cos(\angle ABC)$$

По теореме косинусов:

$$|\overrightarrow{BC}| = 6 \cdot sin(30^\circ) = 6 \cdot \frac{1}{2} = 3$$

$$|\overrightarrow{BA}| = 6$$

$$\angle ABC = 60^\circ$$

$$\overrightarrow{BC} \cdot \overrightarrow{BA} = 3 \cdot 6 \cdot cos(60^\circ) = 18 \cdot \frac{1}{2} = 9$$

Ответ: a) 9