正三棱柱ABC-A1B1C1的底面边长为a,侧棱长为根号2a,求AC1与侧面AB1所成的角(请以A为原点)
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正三棱柱ABC-A1B1C1的底面边长为a,侧棱长为根号2a,求AC1与侧面AB1所成的角(请以A为原点)
正三棱柱ABC-A1B1C1的底面边长为a,侧棱长为根号2a,求AC1与侧面AB1所成的角(请以A为原点)
正三棱柱ABC-A1B1C1的底面边长为a,侧棱长为根号2a,求AC1与侧面AB1所成的角(请以A为原点)
c以A为原点,分别以AB,底面AB的垂线和AA1为X轴、Y轴、Z轴建立空间直角坐标系,以a为1个单位,
A(0,0,0),B(1,0,0),C(1/2,√3/2,0),
A1(0,0,√2),B1(1,0,√2),C1(1/2,√3/2,√2),
向量AC1=(1/2,√3/2,√2),
∵平面ABB1A1在XOZ平面,
∴法向量n=(0,1,0),
向量AC1·n=√3/2,
|n|=1,
|AC1|=√(1/4+3/4+2)=√3,
设n和AC1所成角为α1,AC1与平面ABB1A1所成角为α,
α+α1=π/2,
cosα1=AC1·n/(|AC1|*|n|)=(√3/2)/(√3)=1/2,
cosα=sinα1=γ(1-1/4)=√3/2,
α=30°
∴AC1与侧面AB1所成的角30°
为什么一定要用向量解?
取A1B1中点M,
∵△A1B1C1是正△,
∴C1M⊥A1B1,
∵平面ABB1A1⊥平面A1B1C1,
∴C1M⊥平面ABB1A1,
∴〈C1AM就是AC1与平面ABB11所成角
C1M=√3/2,
AC1=√3,
AM=√(AA1^2+A1M^2)=3/2,
cos
AC1=AB1=a^2+(根号2a)^2=根号3a
AC1^2+AB1^2-2AC1*AB1cosA=B1C1^2
解得cosA=5/6,A=arccos5/6