Question 2. $E$ is the midpoint of edge $A' D'$ of a cube $ABCD$ $\qquad$ $A'B'C'D'$ with edge length $a$. Find the lengths of the two segments into which $EB$ and $EC$ are divided by the plane $AB'D'$.
To solve as shown in Figure 3, to discuss the two segments of $\mathrm{EB}$ divided by the plane $\mathrm{AB}^{\prime} \mathrm{D}^{\prime}$, we must study the intersection point of $\mathrm{EB}$ and the plane $\mathrm{AB}^{\prime} \mathrm{D}^{\prime}$. For this, we look at the plane $\mathrm{EBD}^{\prime}$, i.e., the intersection line of the planes $\mathrm{BCD}^{\prime} \mathrm{A}^{\prime}$ and $\mathrm{AB}^{\prime} \mathrm{D}^{\prime}$. Connecting $\mathrm{A}^{\prime} \mathrm{B}$ and $\mathrm{AB}^{\prime}$ gives the intersection point $S$, so $D^{\prime} S$ is the intersection line of these two planes, and it intersects $\mathrm{EB}$ at $\mathrm{O}$. Then $\mathrm{OE}$ and $\mathrm{OR}$ are the two segments we are looking for.

Clearly, $\mathrm{S}$ is the midpoint of $\mathrm{A}^{\prime} \mathrm{B}$, and thus, $\mathrm{O}$ is the centroid of $\triangle \mathrm{A}^{\prime} \mathrm{BD}^{\prime}$, $\therefore|\mathrm{OE}|:|\mathrm{OB}|=1: 2$.
$$
\begin{aligned}
\text{Also, } |\mathrm{BE}| & =\sqrt{\left|\mathrm{A}^{\prime} \mathrm{B}\right|^{2}+\left|\mathrm{A}^{\prime} \mathrm{E}\right|^{2}} \\
& =\frac{3}{2} \mathrm{a} . \\
\therefore|\mathrm{OE}| & =\frac{1}{3}|\mathrm{BE}|=\frac{1}{2} \mathrm{a} . \\
|\mathrm{OB}| & =\frac{2}{3}|\mathrm{BE}|=\mathrm{a} .
\end{aligned}
$$

Next, we find the lengths of the two segments of $\mathrm{EC}$ divided by the plane $\mathrm{AB}^{\prime} \mathrm{D}^{\prime}$.

Let $\mathrm{EC}$ intersect $\mathrm{D}^{\prime} \mathrm{S}$ at point $\mathrm{Q}$, i.e., we need to find the lengths of $\mathrm{E}$ and $\mathrm{QC}$. To facilitate the calculation, we draw the front view of $\mathrm{BCD}^{\prime} \mathrm{A}^{\prime}$ (as shown in Figure 4).
[Method 1] Connect $\mathrm{A}^{\prime} \mathrm{C}$ and $\mathrm{A}^{\prime} \mathrm{H}$ (where $\mathrm{H}$ is the midpoint of $\mathrm{BC}$) with $\mathrm{SD}^{\prime}$ intersecting at $\mathrm{O}$ and $\mathrm{T}$. It is easy to see that $\mathrm{A}^{\prime} \mathrm{T} \parallel \mathrm{EC}$, and $O$ is the centroid of $\triangle A^{\prime} B D^{\prime}$.
$$
\begin{array}{l}
\therefore\left|A^{\prime} \mathrm{O}\right|:|\mathrm{OC}|=1: 2, \\
\text { Also, }\left|\mathrm{A}^{\prime} \mathrm{T}\right|:|\mathrm{QC}| \\
\quad=\left|\mathrm{A}^{\prime} \mathrm{O}\right|:|\mathrm{OC}|=1: 2, \\
\text { Also, }|\mathrm{EQ}|:\left|\mathrm{A}^{\prime} \mathrm{T}\right|=1: 2, \\
\therefore|\mathrm{EQ}|:|\mathrm{QC}|=1: 4 . \\
\therefore|\mathrm{EQ}|=\frac{1}{5}|\mathrm{EC}| \\
=\frac{1}{5} \sqrt{\left|\mathrm{D}^{\prime} \mathrm{C}\right|^{2}+\left|\mathrm{D}^{\prime} \mathrm{E}\right|^{2}}=\frac{3}{10} \mathrm{a} . \\
\therefore|Q C|=\frac{4}{5}|\mathrm{EC}|=1 \frac{1}{5} \mathrm{a} .
\end{array}
$$
[Method 2] Establish a coordinate system, with $\mathrm{A}^{\prime} \mathrm{D}^{\prime}$ and $\mathrm{CD}^{\prime}$ as the $\mathrm{X}$-axis and $Y$-axis, respectively. Thus, we have
$$
\begin{array}{l}
E\left(-\frac{a}{2}, 0\right), C(0,-\sqrt{2} a), \\
S\left(-a,-\frac{\sqrt{2}}{2} a\right) .
\end{array}
$$
The equation of $\mathrm{SD}^{\prime}$: $\mathrm{y}=\frac{1}{\sqrt{2}} \mathrm{x}$,
$$
\text { The equation of EC }: \frac{x}{-\frac{a}{2}}+\frac{y}{-\sqrt{2} a}=1 .
$$

The $x$-coordinate of the intersection point $Q$ is $+\frac{2}{5} a$.
$$
\begin{array}{l}
\therefore|C Q|:|C E|=|x|:\left|E^{\prime}\right| \\
\quad=\frac{2}{5} a: \frac{a}{2}=4: 5 . \\
\therefore|E Q|=\frac{1}{5}|E C|=\frac{3}{10} . \\
\quad|Q C|=\frac{4}{5}|E C|=1 \frac{1}{5} a .
\end{array}
$$

Additionally, in solving for the two segments on $\mathrm{EB}$ in problem 2, the key lies in selecting the plane $\mathrm{EBD}^{\prime}$. Similarly, we can also choose the plane $\mathrm{EBA}$ or $\mathrm{EBB}^{\prime}$, i.e., converting the three-dimensional problem into a two-dimensional problem based on the intrinsic relationship between the known and unknown elements of the problem. Specifically, we need to select a plane that includes $\mathrm{EB}$ and intersects the plane $\mathrm{AB}^{\prime} \mathrm{D}^{\prime}$, yielding a special intersection line to find the two segments on $\mathrm{EB}$. It is important to note that this method can be used under specific conditions.