在直角坐标系xOy中，设A（2，2），B（-2，-3），沿y轴把坐标平面折成120°的二面角后，AB的长是（ ）
|
A（2，2），B（-2，-3），作AC垂直x轴，BD垂直x轴，BM平行等于CD，
连接AB，MC，则|CD|=4，|BD|=3，|AC|=2，
∵BD⊥x轴，MC⊥x轴（MC∥BD），∴∠ACM就是二面角的平面角，即∠ACM=120°
∴|AM|=
=9+4-2×2×3cos120°
，19
∵|BM|=4
∴|AB|=
=19+16
．35
故选：A．
| 用方框内的词的适当形式填空. | |
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