2.11. Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$, where $A(1,2,1), B(3,-1,7), C(7,4,-2)$.
Solution. From the definition of the scalar product (formula (2.30)), it follows that the cosine of the angle between the vectors is determined by the formula

$$
\cos \varphi=\frac{\vec{A} \vec{B} \cdot \overrightarrow{A C}}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}
$$

To find $\cos \varphi$ using this formula, it is necessary to find the coordinates of the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$. We have:

$$
\overrightarrow{A B}=\{2,-3,6\}, \overrightarrow{A C}=\{6,2,-3\}
$$

By calculating $\overline{A B} \cdot \overrightarrow{A C},|\overrightarrow{A B}|$ and $|\overrightarrow{A C}|$, we get

$$
\cos \varphi=\frac{2 \cdot 6+(-3) \cdot 2+6 \cdot(-3)}{\sqrt{2^{2}+(-3)^{2}+6^{2}} \sqrt{6^{2}+2^{2}+(-3)^{2}}}=-\frac{12}{49}
$$