2.2. a) In the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, a common perpendicular $M N$ is drawn to the lines $A_{1} B$ and $B_{1} C$ (point $M$ lies on the line $A_{1} B$). Find the ratio $A_{1} M: M B$.

b) Given the cube $A B C D A_{1} B_{1} C_{1} D_{1}$. Points $M$ and $N$ are taken on the segments $A A_{1}$ and $B C_{1}$ such that the lines $M N$ and $B_{1} D$ intersect. Find the difference of the ratios $B C_{1}: B N$ and $A M: A A_{1}$.
2.2. a) The first solution. Consider the projection of this cube onto a plane perpendicular to the line $B_{1} C$ (Fig. 18, a). On this drawing, the line $B_{1} C$ is represented by one point, and the segment $M N$ is a perpendicular dropped from this point to the line $A_{1} B$. It is also clear that on this drawing, $A_{1} B_{1} : B_{1} B = \sqrt{2} : 1$. Since $A_{1} M : M N = A_{1} B_{1} : B_{1} B$ and $M N : M B = A_{1} B_{1} : B_{1} B$, then $A_{1} M : M B = A_{1} B_{1}^2 : B_{1} B^2 = 2 : 1$.

The second solution. Consider the projection of the given cube onto a plane perpendicular to the line $A C_{1}$ (Fig. 18, b). The line $A C_{1}$ is perpendicular to the planes of the triangles $A_{1} B D$ and $B_{1} C D_{1}$, so it is perpendicular to the lines $A_{1} B$ and $B_{1} C$, i.e., the segment $M N$ is parallel to it. Thus, the segment $M N$ on this drawing is represented by the point of intersection of the segments $A_{1} B$ and $B_{1} C$. Therefore, on this drawing, $A_{1} M : M B = A_{1} C : B B_{1} = 2 : 1$.

b) Consider the projection of the cube onto a plane perpendicular to the diagonal $B_{1} D$ (Fig. 19). On this drawing, the hexagon $A B C C_{1} D_{1} A_{i}$ is regular, and the line $M N$ passes through its center; let $L$ be the point of intersection of the lines $M N$ and $A D_{1}$, and $P$ be the point of intersection of the line $A A_{1}$ and the line passing through the point $D_{1}$ parallel to $M N$. It is easy to verify that $\triangle A D M = \triangle A_{1} D_{1} P$, and therefore $A M = A_{1} P$. Hence, $B C_{1} : B N = A D_{1} : D_{1} L = A P : P M = (A A_{1} + A M) : A A_{1} = 1 + A M : A A_{1}$, i.e., the desired difference of the ratios is 1.

2.3: Let $A_{1}, B_{1}$, and $C_{1}$ be the projections of the vertices of the given equilateral triangle $A B C$ onto a line perpendicular to the given plane. If the angles between the given plane and the lines $A B, B C$, and $C A$ are $\gamma, \alpha$, and $\beta$, then $A_{1} B_{1} = a \sin \gamma, B_{1} C_{1} = a \sin \alpha$, and $C_{1} A_{1} = a \sin \beta$, where $a$ is the side of the triangle $A B C$. Let, for definiteness, the point $C_{1}$ lie on the segment $A_{1} B_{1}$. Then $A_{1} B_{1} = A_{1} C_{1} + C_{1} B_{1}$, i.e., $\sin \gamma = \sin \alpha + \sin \beta$.