Seven. Line segment $\mathrm{AB}$ is parallel to plane $\mathrm{M}$. From $\mathrm{A}$ and $\mathrm{B}$, perpendicular lines $\mathrm{AA}_{1}$ and $\mathrm{BB}_{1}$ are drawn to this line segment, making angles with plane $\mathrm{M}$ of $\beta$ and $\alpha(\alpha>\beta)$, respectively. Given $\mathrm{AB}=\mathrm{a}, \mathrm{A}_{1} \mathrm{~B}_{1}=\mathrm{b}$, find the distance between $\mathrm{AB}$ and the plane.
Seven. Solution Let C, Y be the projections of points A, B on plane M, respectively.
We discuss in two cases.
(1) AA_1 and BB_1 are on the same side of AB,
\because AA_1 \perp AB,
BB_1 \perp AB, AB \parallel plane M,
AB \perp AC, AB \perp BD,
\therefore AB \perp plane A_1CA, AB \perp plane B_1DB,
\therefore plane A_1CA \parallel plane B_1DB,
\therefore A_1C \parallel B_1D.
Draw B_1E \perp A_1C at E,
then B_1E = CD = AB = a.
Let AC = x, then
\[
\begin{aligned}
A_1C & = A_1E + EC = A_1E + B_1D \\
& = \sqrt{b^2 - a^2} + x \cot \alpha,
\end{aligned}
\]
\[
\begin{array}{l}
\text{Also, } \because A_1C = x \cot \beta, \\
\therefore \quad x \cot \beta - x \cot \alpha = \sqrt{b^2 - a^2}, \\
\therefore \quad x = \frac{\sqrt{b^2 - a^2} \sin \alpha \cdot \sin \beta}{\sin (\alpha - \beta)}.
\end{array}
\]
(2) AA_1 and BB_1 are on opposite sides of AB.
Let AC = x, then,
\[
\begin{array}{l}
A_1E = A_1C + CE = x \cos \beta + 2 \cot \alpha, \\
\text{Also, } \quad A_1E = \sqrt{b^2 - a^2} \\
\therefore \sqrt{b^2 - a^2} = x \cot \beta + x \cot \alpha. \\
x = \frac{\sin \alpha \cdot \sin \beta \cdot \sqrt{b^2 - a^2}}{\sin (\alpha + \beta)}.
\end{array}
\]