11.3. On one main diagonal and all edges of a cube, directions are chosen. What is the smallest length that the sum of the resulting 13 vectors can have, if the edge length is 1, and the length of the main diagonal is $\sqrt{3}$.
Answer: $\sqrt{3}$

Solution: Let's choose a basis of three vectors along the edges of the cube, such that the vector of the diagonal equals $\overrightarrow{e_{1}}+\overrightarrow{e_{2}}+\overrightarrow{e_{3}}$. Then, four vectors will be equal to $\pm \overrightarrow{e_{1}}$, another four to $\pm \overrightarrow{e_{2}}$, and another four to $\pm \overrightarrow{e_{3}}$. Thus, the sum $\vec{S}$ of the vectors has the form $k \cdot \overrightarrow{e_{1}}+m \cdot \overrightarrow{e_{2}}+n \cdot \overrightarrow{e_{3}}$, where $k, m$, and $n$ are odd integers, and $|\vec{S}|=\sqrt{k^{2}+m^{2}+n^{2}} \geq \sqrt{3}$, since $k^{2} \geq 1, m^{2} \geq 1, n^{2} \geq 1$. Clearly, by pairing the parallel edges of the cube into oppositely directed vectors, we get that $k=m=n=1$ and $|\vec{S}|=\sqrt{3}$.