[ Intersecting lines, angle between them ] Cube

Given a cube $A B C D A 1 B 1 C 1 D 1$. Find the angles between the lines: a) $A A 1$ and $B D 1$; b) $B D 1$ and $D C 1$; c) $A D 1$ and $D C 1$.
## First Method.

a) Let the edge of the cube be $a$. Since the line $A A 1$ is parallel to the line $B B 1$ (Fig.1), the angle between the lines $A A 1$ and $B D 1$ is equal to the angle between the lines $B B 1$ and $B D 1$, i.e., the angle $B 1 B D 1$. From the right triangle $B 1 B D 1$, we find that

$$
\cos \angle B 1 B D 1=\frac{B B_{1}}{B D_{1}}=\frac{a}{a \sqrt{3}}=\frac{1}{\sqrt{3}}
$$

b) Since $B C$ is perpendicular to the plane $D D 1 C 1 C$ (Fig.2), the line $C D 1$ is the orthogonal projection of the inclined line $B D 1$ onto the plane $D D 1 C 1 C$, and since $C D 1 \perp D C 1$ (as the diagonals of the square $D D 1 C 1 C$), by the theorem of three perpendiculars, $B D 1 \perp D C 1$, i.e., the angle between the lines $B D 1$ and $D C 1$ is $90^{\circ}$. c) Since the line $A B 1$ is parallel to the line $D C 1$ (Fig.3), the angle between the lines $A D 1$ and $D C 1$ is equal to the angle between the lines $A B 1$ and $A D 1$, i.e., the angle $B 1 A D 1$, and since the triangle $B 1 A D 1$ is equilateral, $\angle B 1 A D 1=60^{\circ}$.

Second Method.

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

$$
\overrightarrow{B D_{1}}=-\vec{x}+\vec{y}+\vec{z}, \overrightarrow{D C_{1}}=\vec{x}+\vec{z}, \overrightarrow{A D_{1}}=\vec{y}+\vec{z}
$$

If $\alpha, \beta$ and $\gamma-$ are the desired angles, then

$$
\cos \alpha=\frac{\overrightarrow{A A_{1}} \cdot \overrightarrow{B D_{1}}}{A A_{1} \cdot B D_{1}}=\frac{\vec{z}(-\vec{x}+\vec{y}+\vec{z})}{a \sqrt{a^{2}+a^{2}+a^{2}}}=
$$

$$
\begin{gathered}
=\frac{-\vec{z} \cdot \vec{z}+\vec{r} \cdot \vec{z}+\vec{z} \cdot \vec{z}}{a^{2} \sqrt{3}}=\frac{0+0+a^{2}}{a^{2} \sqrt{3}}=\frac{1}{\sqrt{3}} \\
\cos \beta=\frac{\overrightarrow{B D_{1}} \cdot \overrightarrow{D C_{1}}}{B D_{1} \cdot D C_{1}}=\frac{(-\vec{x}+\vec{y}+\vec{z})(\vec{x}+\vec{z})}{\sqrt{\alpha^{2}+a^{2}+a^{2}} \sqrt{\alpha^{2}+a^{2}}}= \\
=\frac{(-\vec{x} \cdot \vec{x}+\vec{z} \cdot \vec{z})}{a \sqrt{3} \cdot a \sqrt{2}}=\frac{-a^{2}+a^{2}}{a^{2} \sqrt{6}}=0 \\
\cos \gamma=\frac{\overrightarrow{A D_{1}} \cdot \overrightarrow{A C_{1}}}{A D_{1} \cdot D C_{1}}=\frac{(\vec{y}+\vec{z})(\vec{z}+\vec{z})}{\sqrt{a^{2}+a^{2}} \cdot \sqrt{a^{2}+a^{2}}}=\frac{\vec{z} \cdot \vec{z}}{2 a^{2}}=\frac{a^{2}}{2 a^{2}}=\frac{1}{2}
\end{gathered}
$$

Therefore,

$$
\alpha=\arccos \frac{1}{\sqrt{3}}, \beta=90^{\circ}, \gamma=60^{\circ} .
$$

## Answer

a) $\arccos \frac{1}{\sqrt{3}}$; b) $90^{\circ}$; c) $60^{\circ}$.