Problem 26. $A B C D A_{1} B_{1} C_{1} D_{1}$ is a cube (Fig. 13). $A B=$ $=1 ; \overline{D M}=\frac{8}{15} \overline{D B}_{1}$. The plane $\alpha$ passes through point $M$ and is perpendicular to the line $B_{1} D$. Find the ratio in which the plane $\alpha$ divides the edge $A_{1} D_{1}$.
Let the plane $\alpha$ and the line $A_{1} D_{1}$ intersect at point $X$. According to the problem,

$$
\overline{M X} \cdot \overline{D B_{1}}=0
$$

Consider the basis vectors $\overline{D A}, \overline{D C}$, and $\overline{D D}_{1}$.

$\overline{D B}_{1}=\overline{D D}_{1}+\overline{D A}+\overline{D C}$. Let $\overline{D_{1} X}=n \overline{D_{1} A_{1}}$. Then $\overline{M X}=\overline{D D}_{1}+\overline{D_{1} X}-\overline{D M}=\frac{7}{15} \overline{D D}_{1}+\left(n-\frac{8}{15}\right) \overline{D A}-\frac{8}{15} D C$. After this, equation (1) takes the form:

$$
\begin{gathered}
\left(\frac{7}{15} \overline{D D}_{1}+\left(n-\frac{8}{15}\right) \overline{D A}-\frac{8}{15} \overline{D C}\right) \times \\
\times\left(\overline{D D}_{1}+\overline{D A}+\overline{D C}\right)=0
\end{gathered}
$$

The vectors $\overline{D D}_{1}, \overline{D A}, \overline{D C}$ are mutually perpendicular, and the length of each is 1. Therefore, equation (2) simplifies to:

$$
\frac{7}{15}+\left(n-\frac{8}{15}\right)-\frac{8}{15}=0
$$

From this, $n=0.6$, i.e., $D_{1} X: X A_{1}=3: 2$.

## Problem 13. SECTIONS OF POLYHEDRONS. PERIMETERS AND AREAS