直三棱柱abc-a1b1c1中,ab=ac=1 ∠BAC=90° 且异面直线a1b与b1c1所成角为60° 且AA1=1直三棱柱abc-a1b1c1中,ab=ac=1 ∠BAC=90° 且异面直线a1b与b1c1所成角为60° 且AA1=1 D是b1c1上任意一点,求D到平面A1BC的距离
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![直三棱柱abc-a1b1c1中,ab=ac=1 ∠BAC=90° 且异面直线a1b与b1c1所成角为60° 且AA1=1直三棱柱abc-a1b1c1中,ab=ac=1 ∠BAC=90° 且异面直线a1b与b1c1所成角为60° 且AA1=1 D是b1c1上任意一点,求D到平面A1BC的距离](/uploads/image/z/3996394-34-4.jpg?t=%E7%9B%B4%E4%B8%89%E6%A3%B1%E6%9F%B1abc-a1b1c1%E4%B8%AD%2Cab%3Dac%3D1+%E2%88%A0BAC%3D90%C2%B0+%E4%B8%94%E5%BC%82%E9%9D%A2%E7%9B%B4%E7%BA%BFa1b%E4%B8%8Eb1c1%E6%89%80%E6%88%90%E8%A7%92%E4%B8%BA60%C2%B0+%E4%B8%94AA1%3D1%E7%9B%B4%E4%B8%89%E6%A3%B1%E6%9F%B1abc-a1b1c1%E4%B8%AD%2Cab%3Dac%3D1+%E2%88%A0BAC%3D90%C2%B0+%E4%B8%94%E5%BC%82%E9%9D%A2%E7%9B%B4%E7%BA%BFa1b%E4%B8%8Eb1c1%E6%89%80%E6%88%90%E8%A7%92%E4%B8%BA60%C2%B0+%E4%B8%94AA1%3D1+D%E6%98%AFb1c1%E4%B8%8A%E4%BB%BB%E6%84%8F%E4%B8%80%E7%82%B9%2C%E6%B1%82D%E5%88%B0%E5%B9%B3%E9%9D%A2A1BC%E7%9A%84%E8%B7%9D%E7%A6%BB)
直三棱柱abc-a1b1c1中,ab=ac=1 ∠BAC=90° 且异面直线a1b与b1c1所成角为60° 且AA1=1直三棱柱abc-a1b1c1中,ab=ac=1 ∠BAC=90° 且异面直线a1b与b1c1所成角为60° 且AA1=1 D是b1c1上任意一点,求D到平面A1BC的距离
直三棱柱abc-a1b1c1中,ab=ac=1 ∠BAC=90° 且异面直线a1b与b1c1所成角为60° 且AA1=1
直三棱柱abc-a1b1c1中,ab=ac=1 ∠BAC=90° 且异面直线a1b与b1c1所成角为60° 且AA1=1 D是b1c1上任意一点,求D到平面A1BC的距离
直三棱柱abc-a1b1c1中,ab=ac=1 ∠BAC=90° 且异面直线a1b与b1c1所成角为60° 且AA1=1直三棱柱abc-a1b1c1中,ab=ac=1 ∠BAC=90° 且异面直线a1b与b1c1所成角为60° 且AA1=1 D是b1c1上任意一点,求D到平面A1BC的距离
作BC、B1C1的中点E、E1,连结EE1、A1E1、A1E,
∵直棱柱中B1C1∥BC,∴B1C1∥平面A1BC,则B1C1上任一点到平面A1BC的距离相等,
作E1H⊥A1E于H,∵A1B=A1C、E是BC中点,A1E⊥BC、又E1E⊥BC,BC⊥平面A1E1E,
∴BC⊥E1H,∵BC交A1E,即E1H是直线B1C1到平面A1BC的距离,
∵AB=AC=1、∠BAC=90°,则A1B1=A1C1=1、∠B1A1C1=90°,
∴A1E1=√2/2,又E1E=1,Rt△A1E1E中,A1E=√(A1E1^2+E1E^2)=√(√2/2)^2+1)=√6/2,
∴E1H=A1E1*E1E/A1E=(√2/2*1)/(√6/2)=√3/3,点D到平面A1BC的距离也是√3/3,解毕.
∴AA1=a=√[(√2)²-1²]=1.
设E为B1C1中点,则A1E⊥平面B1BC1.作EF⊥BC1.F∈BC1,
∠A1FE为平面A1BC1与平面B1BC1所成的锐二面角的平面角。
....
A1E=√2/2.BC1=√3.EF/EC1=B1B/BC1.算得EF=√6/6
AB=AC=1,角ABC=90度.AC是斜边=直角边?。打错了吧!∠BAC=90°
如图,BC‖B1C1,∴∠A1BC=60°,从对称性,⊿A1BC为正三角形,A1B=√2
∴AA1=a=√[(√2)²-1²]=1.
设E为B1C1中点,则A1E⊥平面B1BC1.作EF⊥BC1.F∈BC1,
∠A1FE为平面A1BC1与平面B1BC1所成的锐二面...
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AB=AC=1,角ABC=90度.AC是斜边=直角边?。打错了吧!∠BAC=90°
如图,BC‖B1C1,∴∠A1BC=60°,从对称性,⊿A1BC为正三角形,A1B=√2
∴AA1=a=√[(√2)²-1²]=1.
设E为B1C1中点,则A1E⊥平面B1BC1.作EF⊥BC1.F∈BC1,
∠A1FE为平面A1BC1与平面B1BC1所成的锐二面角的平面角。
A1E=√2/2.BC1=√3.EF/EC1=B1B/BC1.算得EF=√6/6
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