3. Given point $A(a, \sqrt{2})$ is the intersection point of the graphs of the two functions $y=k x-2$ and $y=(\sqrt{2}-1) x$. Then the real number $k$ equals ( ).
(A) $-\sqrt{2}$
(B) $1-\sqrt{2}$
(C) $\sqrt{2}-1$
(D) 1
3.D.

From the condition, we know $\sqrt{2}=(\sqrt{2}-1) a$, then
$$
\begin{array}{l}
a=\frac{\sqrt{2}}{\sqrt{2}-1}=2+\sqrt{2} . \\
\text { Also } \sqrt{2}=k a-2 \text {, i.e., } \sqrt{2}=k(\sqrt{2}+2)-2 .
\end{array}
$$

Solving for $k$ gives $k=1$.