40. In an equilateral triangle $ABC$ with a side length of 6 (Fig. 42), find the scalar product of the vectors: a) $\overrightarrow{A B}$ and $\overrightarrow{A C} ;$ b) $\overrightarrow{A B}$ and $\overrightarrow{B C}$.
Solution. a) Since the angle $\varphi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$ (and their directions) is $60^{\circ}$, for the scalar product of these vectors we get

$$
\overrightarrow{A B} \cdot \overrightarrow{A C}=|\overrightarrow{A B}| \cdot|\overrightarrow{A C}| \cdot \cos B \widehat{A C}=6 \cdot 6 \cdot \cos 60^{\circ}=36 \cdot 0.5=18
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-126.jpg?height=194&width=386&top_left_y=295&top_left_x=91)

Fig. 41

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-126.jpg?height=311&width=311&top_left_y=181&top_left_x=829)

Fig. 42

b) The angle between the vectors $\overrightarrow{A B}$ and $\overrightarrow{B C}$ (i.e., the angle between their directions) is the angle $\varphi_{1}=120^{\circ}$, therefore

$\overrightarrow{A B}: \overrightarrow{B C}=|\overrightarrow{A B}| \cdot|\overrightarrow{B C}| \cdot \cos N \widehat{B C}=6 \cdot 6 \cdot \cos 120^{\circ}=6 \cdot 6 \cdot(-0.5)=-18$, since $\cos 120^{\circ}=\cos \left(180^{\circ}-60^{\circ}\right)=-\cos 60^{\circ}=-0.5$.