Gaokao Math Practice Question
已知正四棱柱 $A B C D-A_{1} B_{1} C_{1} D_{1}$ 中, $A B=2, C C_{1}=2 \sqrt{2}, E$ 为 $C C_{1}$ 的中 点, 则直线 $A C_{1}$ 与平面 $\mathrm{BED}$ 的距离为 ($\qquad$)
- A.2
- B.$\sqrt{3}$
- C.$\sqrt{2}$
- D.1
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