5.26. In a regular quadrilateral pyramid $S A B C D$, the length of each edge is $a$. Point $M \in S C$ and $S M: M C=2: 1$. Find the angle between vectors $\overline{D C}$ and $\overline{A M}$.
5.26. By the condition, $S A B C D-$ is a regular quadrilateral pyramid, $S A=A B=a, M \in S C$, $S M: M C=2: 1$ (Fig. 5.16). Let $\overline{A D}=\bar{m}, \overline{A B}=\bar{n}, \overline{A S}=\bar{p}$ and $(D \widehat{C ; A M})=\varphi$. The angle $\varphi$ we are looking for will be found using the formula

$$
\cos \varphi=\frac{\overline{D C} \cdot \overline{A M}}{|\overline{D C}| \overline{A M} \mid}
$$

We will decompose the vectors $\overline{D C}$ and $\overline{A M}$ into the vectors $\bar{m}, \bar{n}$ and $\bar{p} . \quad$ We have $\quad \overline{D C}=\overline{A B}=\bar{n}, \overline{A M}=$ $\overline{A D}+\overline{D C}+\overline{C M}$. From the condition $S M: M C=2: 1$ it follows that $\overline{C M}=\frac{1}{3} \overline{C S}$, where $\overline{C S}=\overline{A S}-\overline{A C}=\bar{p}-(\bar{m}+\bar{n})$. Thus,

$$
\begin{aligned}
\overline{A M} & =\bar{m}+\bar{n}+\frac{1}{3} \bar{p}-\frac{1}{3}(\bar{m}+\bar{n})=\frac{2}{3} \bar{m}+\frac{2}{3} \bar{n}+\frac{1}{3} \bar{p} \\
\overline{D C} \cdot \overline{A M} & =\bar{n}\left(\frac{2}{3} \bar{m}+\frac{2}{3} \bar{n}+\frac{1}{3} \bar{p}\right)=\frac{1}{3}\left(2 \bar{n} \bar{m}+2 \bar{n}^{2}+\bar{n} \bar{p}\right)
\end{aligned}
$$

But $\bar{n} \bar{m}=0$ (since $\bar{n} \perp \bar{m}$ ), $\bar{n}^{2}=a^{2}$ and $\bar{n} \bar{p}=a^{2} \cos 60^{\circ}=\frac{a^{2}}{2} ;$ therefore, $\overline{D C} \cdot \overline{A M}=\frac{1}{3}\left(2 a^{2}+\frac{a^{2}}{2}\right)=\frac{5}{6} a^{2}$ Next, we find $|\overline{D C}|=a$,

$$
\begin{gathered}
|A M|=\sqrt{\frac{4}{9} a^{2}+\frac{4}{9} a^{2}+\frac{1}{9} a^{2}+\frac{8}{9} \bar{m} \bar{n}+\frac{4}{9} \bar{m} \bar{p}+\frac{4}{9} \bar{n} \bar{p}}= \\
=\sqrt{a^{2}+\frac{8}{9} a^{2} \cdot \frac{1}{2}}=\frac{a \sqrt{13}}{3}
\end{gathered}
$$

Substituting the found expressions into equation (1), we finally get $\cos \varphi=\frac{5 \sqrt{13}}{26}$.

Answer: $\arccos \frac{5 \sqrt{13}}{26}$.