Question

高二数学题。与几何向量相关的

Answer 1

(1)以A为原点AB,AD,AA1为轴建系,设AB=2k>0
则D1(0,1,1),B1(2k,0,1),E(k,1,0)
∴AD1→=(0,1,1),B1E→=(-k,1,-1)
∵AD1→·B1E→=0+1-1=0,∴AD1⊥B1E
(2)设P(0,0,t)(0≤t≤1),D(0,1,0),∴DP→=(0,-1,t)
AE→=(k,1,0)
设面AEB1法向量为n→=(x,y,2),则
kx+y=0,-kx+y-2=0,解得x=-1/k,y=1,∴n→=(-1/k,1,2)
要使DP∥面AEB1,则DP→⊥n→,即0-1+2t=0
t=1/2,即P(0,0,1/2),∴AP=1/2
(3)易证面A1B1E法向量m→=(0,1,1),设二面角A-B1E-A1为∠B1E
则cosB1E=cos30°=|cos<m→,n→>|=|0+1+2|/[√(0+1+1)*√(1/k^2+1+4)=√3/2
解得k=1,∴AB=2k=2