1. The mapping $f$ from the set $A$ of planar vectors to $A$ is defined by $f(x)=x-2(x \cdot a) a$, where $a$ is a constant vector. If the mapping $f$ satisfies $f(x) \cdot f(y)=x \cdot y$ for any $x, y \in A$, then the coordinates of $a$ could be ( ).
(A) $\left(\frac{\sqrt{5}}{2},-\frac{1}{2}\right)$
(B) $\left(\frac{\sqrt{2}}{4}, \frac{\sqrt{2}}{4}\right)$
(C) $\left(\frac{3}{4}, \frac{1}{4}\right)$
(D) $\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$
$$
\begin{array}{l}
\text {-,1.D. } \\
f(x) \cdot f(y)=[x-2(x \cdot a) a] \cdot[y-2(y \cdot a) a] \\
=x \cdot y+4\left(a^{2}-1\right)(x \cdot a)(y \cdot a)
\end{array}
$$

Therefore, from the condition, we should get $a^{2}-1=0$, i.e., $|a|=1$.