Orthogonal Projection (Miscellaneous). Cube

Given a unit cube $A B C D A_1 B_1 C_1 D_1$, $M$ is the midpoint of $B B_1$. Find the angle and the distance between the lines $A B_1$ and $D M$. In what ratio does the common perpendicular of these lines divide the segments $D M$ and $A B_1$?
## ПІерьвъ这 :тим:

Let $M^{\prime}$ be the orthogonal projection of point $M$ onto the plane $B C D 1 A 1$, which is perpendicular to the line $A B 1$ and intersects it at point $K$ - the center of the square $A A 1 B 1 B$. Let $D^{\prime}$ be the orthogonal projection of point $D$ onto this plane (Fig.1). Then $D^{\prime}$ is the center of the square $C C 1 D 1 D$, and $D^{\prime} M^{\prime}$ is the orthogonal projection of the line $D M$ onto this plane, with $M^{\prime}$ lying on $A 1 B$ and $M^{\prime}$ being the midpoint of $B K$. The distance between the lines $A B 1$ and $D M$ is equal to the distance from point $K$ to the line $D^{\prime} M^{\prime}$. Let $P$ be the foot of the perpendicular dropped from point $K$ to the line $D^{\prime} M^{\prime}$. Then $K P$ is the height of the right triangle $K M^{\prime} D^{\prime}$, drawn from the vertex of the right angle (Fig.2). We further find:

$$
\begin{gathered}
D^{\prime} M^{\prime}=\sqrt{D^{\prime} K^{2}+M^{\prime} K^{2}}=\sqrt{1+\left(\frac{\sqrt{2}}{1}\right)^{2}}=\frac{3}{2 \sqrt{2}} \\
K P=K M^{\prime} \cdot \frac{K D^{\prime}}{D M^{\prime} M^{\prime}}=\frac{\sqrt{2}}{1} \cdot \frac{\frac{1}{2}}{2 \sqrt{2}}=\frac{1}{3}
\end{gathered}
$$

The angle $\alpha$ between the lines $A B 1$ and $D M$ complements to $90^{\circ}$ the angle between the intersecting lines $D M$ and $D^{\prime} M^{\prime}$ (Fig.3). Therefore,

$$
\sin \alpha=\cos \left(90^{\circ}-\alpha\right)=\frac{\frac{D N^{\prime}}{D M}}{\frac{\frac{3}{\Gamma^{2}}}{2}}=\frac{\sqrt{2}}{2}
$$

Thus, $\alpha=45^{\circ}$. Let $X Y$ be the common perpendicular of the lines $A B 1$ and $D M$ (point $X$ lies on $A B 1$, and $Y$ lies on $D M$). Then $K H Y P$ is a rectangle, $P Y \parallel M M^{\prime} \parallel D D^{\prime} \parallel A B 1$. Therefore,

$$
\frac{D Y}{Y M}=\frac{D P}{P M^{\prime}}=\frac{D K^{2}}{M^{\prime} K^{2}}=\frac{1}{\left.\left(\frac{1}{4}\right)^{2}\right)}=8
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_db48a70f246fb4a097d0g-21.jpg?height=43&width=274&top_left_y=2540&top_left_x=0)

Let $\overrightarrow{A B}=\vec{z}, \overrightarrow{A D}=\overrightarrow{y^{\prime}}, \overrightarrow{A_{A_{1}}}=\vec{z}, A B=x, A D=y, A A 1=z$, where $x=y=z=1$. Then

$$
\overrightarrow{A B_{1}}=\vec{z}+\vec{z}, \overrightarrow{D M}=\vec{z}-\vec{y}+\frac{1}{2} \vec{z}
$$

Let $\alpha$ be the angle between the lines $A B 1$ and $D M$. Then

$$
\cos \alpha=\frac{\left|\overrightarrow{A B_{1}^{2}} \cdot \overrightarrow{D M}\right|}{A B_{1} \cdot D M}=\frac{\left|(\vec{x}+\vec{Z}) \cdot\left(\vec{x}-\vec{y}+\frac{1}{2} \vec{Z}\right)\right|}{\sqrt{\pi^{2}+z^{2}} \cdot \sqrt{x^{2}+y^{2}+\frac{x^{2}}{4}}}=
$$

$$
=\frac{\left.p^{2}+\frac{1}{z^{2} \mid} \right\rvert\,}{\sqrt{\sqrt{x}^{2}+z^{2}} \sqrt{x^{2}+y^{2}+\frac{2}{4}}}=\frac{\sqrt[3]{\sqrt{2} \cdot \frac{2}{2}}}{=\frac{1}{\sqrt{2}}}
$$

Thus, $\alpha=45^{\circ}$. Let $X Y$ be the common perpendicular of the lines $A B 1$ and $D M$ (point $X$ lies on $A B 1$, and $Y$ lies on $D M$), and $\overrightarrow{A X}=\overrightarrow{A B_{1}}$ and $\overrightarrow{D Y}=\mu \overrightarrow{D M}$. Then

$$
\begin{aligned}
\overrightarrow{X Y}= & \overrightarrow{X A}+\overrightarrow{A D}+\overrightarrow{D Y}=(1-) \overrightarrow{B_{1} A}+\overrightarrow{A D}+\mu \overrightarrow{D M}= \\
= & -(1-)(\vec{x}+\vec{z})+\vec{y}+\mu\left(\vec{x} \cdot \vec{y}+\frac{1}{2} \vec{z}\right)= \\
& =(-1+\mu) \vec{z}+(1-\mu) \vec{y}+\left(-1+\frac{1}{2} \mu\right) \cdot \vec{z}
\end{aligned}
$$

Since $\overrightarrow{X Y} \perp \overrightarrow{A B_{1}}$ and $\overrightarrow{X Y} \perp \overrightarrow{D M}$, we have $\overrightarrow{X Y} \cdot \overrightarrow{A B_{1}}=0$ and $\overrightarrow{X Y} \cdot \overrightarrow{D M}=0$, or

$$
\begin{gathered}
\left((-1+\mu) \vec{z}+(1-\mu) \vec{y}+\left(-1+\frac{1}{2} \mu\right) \vec{z}\right)(\vec{x}+\vec{z})= \\
=(-1+\mu)-\left(-1+\frac{1}{2} \mu\right)=2+\frac{3}{2} \mu-2=0 \\
\left((-1+\mu) \vec{z}+(1-\mu) \vec{y}+\left(-1+\frac{1}{2} \mu\right) \vec{z}\right)\left(\vec{z}-\vec{y}+\frac{1}{2} \vec{z}\right)= \\
=(-1+\mu)+(\mu-1)+\frac{1}{2}\left(-1+\frac{1}{2} \mu\right)=\frac{3}{2}+\frac{9}{1} \mu-\frac{5}{2}=0
\end{gathered}
$$

From the system

$$
\left\{\begin{array}{l}
2 \lambda+\frac{3}{3} \mu-2=0 \\
\frac{3}{2} \lambda+\frac{3}{4} \mu-\frac{5}{2}=0
\end{array}\right.
$$

we find that $\lambda=\frac{1}{3}, \mu=\frac{8}{9}$. Therefore,

$$
\overrightarrow{X Y}=(-1+\mu) \vec{z}+(1-\mu) \vec{y}+\left(-1+\frac{1}{2} \mu\right) \vec{z}=\frac{2}{9} \vec{z}+\frac{1}{9} \vec{y}-\frac{2}{9} \bar{z}
$$

Thus,

$$
\begin{gathered}
X Y=\sqrt{\overline{X Y^{2}}}=\sqrt{\left(\frac{2}{9} \vec{X}+\frac{1}{9} \vec{y}-\frac{2}{9} \vec{Z}\right)^{2}}=\frac{\sqrt{1+1+1}}{9}=\frac{1}{3}, \\
\frac{A X}{A B_{1}}=\frac{1}{3}, \frac{A X}{X B_{1}}=\frac{1}{2}, \frac{D Y}{D M}=\mu=\frac{8}{9}, \frac{D Y}{Y M}=8 .
\end{gathered}
$$

## ТретънЧь :тчш:мак.

Assume that $\alpha=45^{\circ}$ is already known. Let $V$ be the volume of the tetrahedron $A D M B 1$ with vertex $D$. Then

$$
V=\frac{1}{3} S_{\triangle A M B} 1 \cdot D A=\frac{1}{3} \cdot \frac{1}{1} \cdot 1=\frac{1}{12} .
$$

On the other hand, if $d$ is the distance between the lines $A B 1$ and $D M$, then

$$
V=\frac{1}{6} A B 1 \cdot D M \cdot d \sin \alpha=\frac{1}{6} \cdot \sqrt{2} \cdot \frac{3}{2} \cdot d \cdot \frac{\sqrt{2}}{2}=\frac{d}{1} \text {. }
$$

From the equation $\frac{d}{\frac{d}{1}}=\frac{1}{12}$, we find that $d=\frac{1}{3}$.

## Answer

$4