17. As shown in the figure, in $\triangle \mathrm{ABC}$, $\angle \mathrm{C}=90^{\circ}, \mathrm{AC}=\mathrm{BC}=\sqrt{2}$, $\triangle \mathrm{ABC}$ is rotated $60^{\circ}$ clockwise around point $\mathrm{A}$ to the position of $\triangle \mathrm{AB}^{\prime} \mathrm{C}^{\prime}$, and $\mathrm{C}^{\prime} \mathrm{B}$ is connected. Then the length of $\mathrm{C}^{\prime} \mathrm{B}$ is ( )
(A) $2-\sqrt{2}$
(B) $\frac{\sqrt{3}}{2}$
(C) $\sqrt{3}-1$
(D) 1
Answer: (C)
Solution: As shown in the figure, connect $\mathrm{BB}^{\prime}$.
Since $\triangle A B C$ is rotated $60^{\circ}$ clockwise around point $A$ to get $\triangle A B^{\prime} C^{\prime}$,
we have
$$
\mathrm{AB}=\mathrm{AB}^{\prime}, \angle \mathrm{BAB}^{\prime}=60^{\circ} \text {, }
$$

Therefore, $\triangle \mathrm{ABB}^{\prime}$ is an equilateral triangle,
which gives

In $\triangle \mathrm{ABC}^{\prime}$ and $\triangle \mathrm{B}^{\prime} \mathrm{BC}^{\prime}$,

we have
which gives
Extend $\mathrm{BC}^{\prime}$ to intersect $\mathrm{AB}^{\prime}$ at $\mathrm{D}$, then $\mathrm{BD} \perp \mathrm{AB}^{\prime}$,
Since
By the Pythagorean theorem, we get
Therefore
So
$$
\begin{array}{c}
\angle \mathrm{C}=90^{\circ}, \mathrm{AC}=\mathrm{BC}=\sqrt{2}, \\
\mathrm{AB}=\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}=2, \\
\mathrm{BD}=2 \times \frac{\sqrt{3}}{2}=\sqrt{3}, \quad \mathrm{C}^{\prime} \mathrm{D}=2 \times \frac{1}{2}=1, \\
\mathrm{BC}^{\prime}=\mathrm{BD}-\mathrm{C}^{\prime} \mathrm{D}=\sqrt{3}-1 .
\end{array}
$$