如图所示,正方体ABCD-A1B1C1D1的棱长为2,P,Q分别是BC,CD上的动点,且PQ=根号2.1)确定P,Q的位
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以A为原点,分别以AB、AD、AA1为X、Y、Z轴建立空间坐标系,

A(0,0,0),A1(0,0,2),

B(2,0,0),B1(2,0,2),

C(2,2,0),C1(2,2,2),

D(0,2,0),D1(0,2,2),

P(2,y0,0),Q(x0,2,0),

向量B1Q=(x0-2,2,-2),

向量D1P=(2,y0-2.-2),

∵向量B1Q⊥D1P,

∴向B1Q·D1P=2x0-4+2y0-4+4=2x0+2y0-4=0,

x0+y0=2,

x0=2-y0,

PQ^2=(x0-2)^2+(2-y0)^2=(√2)^2=2,

(x0-2)^2+x0^2=2,

2x0^2-4x0+2=0,

(x0-1)^2=0,

x0=1,

y0=1,

∴P(2,1,0),Q(1,2,0),

当P在BC中点,Q在DC中点时,使B1Q⊥D1P.