1. Given $a^{2}+b^{2}=1, b^{2}+c^{2}=2, c^{2}+a^{2}=$ 2. Then the minimum value of $a b+b c+c a$ is ( ).
(A) $\sqrt{3}-\frac{1}{2}$
(B) $-\sqrt{3}+\frac{1}{2}$
(C) $-\sqrt{3}-\frac{1}{2}$
(D) $\sqrt{3}+\frac{1}{2}$
-、1.B.
From the problem, we have
$$
c= \pm \sqrt{\frac{3}{2}}, a= \pm \sqrt{\frac{1}{2}}, b= \pm \sqrt{\frac{1}{2}} .
$$

Notice that
$$
a b+b c+c a=\frac{(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)}{2},
$$

so we only need to consider the minimum value of $|a+b+c|$.
To make $|a+b+c|$ as small as possible, we can take
$$
a=b=-\sqrt{\frac{1}{2}}, c=\sqrt{\frac{3}{2}} \text {. }
$$

Thus, the minimum value of $a b+b c+c a$ is $-\sqrt{3}+\frac{1}{2}$.