12.2. Consider the right rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, where $A B=a, B C=2 a, A A_{1}=3 a$. On the edges $C C_{1}$ and $A D$, points $M$ and $N$ are taken, respectively, such that $A N=C_{1} M=a$. Determine the measure of the angle between the lines $A M$ and $N B_{1}$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_06_06_6f54227717af4fe8c7a3g-1.jpg?height=975&width=1373&top_left_y=1483&top_left_x=430)

On the right support of the edge $B C$, consider the point $Q$, such that $N Q \| A C$.

Consider the point $Q_{1} \in B_{1} C_{1}$, such that $Q_{1} Q \| C_{1} C, Q_{1} Q=C_{1} C$, and the point $M_{1} \in Q_{1} Q$, such that $Q_{1} M_{1}=a$.

Then $A N\|C Q, C Q\| M M_{1}$ implies $A N \| M M_{1}$ and respectively $A M \| N M_{1}$. The measure of the angle between the lines $A M$ and $N B_{1}$ is equal to the measure of the angle $B_{1} N M_{1}$.

We determine

$$
\begin{gathered}
B_{1} N^{2}=A N^{2}+A B^{2}+B B_{1}^{2}=a^{2}+a^{2}+9 a^{2}=11 a^{2} \\
N M_{1}^{2}=A M^{2}=A D^{2}+D C^{2}+C M^{2}=4 a^{2}+a^{2}+4 a^{2}=9 a^{2} \\
B_{1} M_{1}^{2}=B_{1} Q_{1}^{2}+M_{1} Q_{1}^{2}=9 a^{2}+a^{2}=10 a^{2}
\end{gathered}
$$

According to the law of cosines in triangle $B_{1} N M_{1}$ we obtain

$B_{1} M_{1}^{2}=B_{1} N^{2}+N M_{1}^{2}-2 B_{1} N \cdot N M_{1} \cos \varphi$, where $\varphi=m\left(\angle B_{1} N M_{1}\right)$.

Then $\cos \varphi=\frac{11 a^{2}+9 a^{2}-10 a^{2}}{2 a \sqrt{11} \cdot 3 a}=\frac{5 \sqrt{11}}{33}$ and $m\left(\angle B_{1} N M_{1}\right)=\arccos \frac{5 \sqrt{11}}{33}$.