解:∵直三棱柱ABC-A1B1C1中, ∠BAC=,AB=AC=AA1=1,D和E分别为棱AC、AB上的动点(不包括端点), ∴分别以AB,AC,AA1为x轴,y轴,z轴,作空间直角坐标系, 则B1(1,0,1),C1(0,1,1), 设E(t1,0,0),D(0,t2,0),t1,t2∈(0,1), 则= (t1,-1,-1), =(-1,t2,-1), ∵C1E⊥B1D, ∴-t1-t2+1=0, 即t1+t2=1. ∵=(t1,-t2,0), ∴|| =, ∵0<t1<1, ∴当t1=时,||min=, 当|| ==1. ∴线段DE长度的取值范围为[,1). 故选C. |