长方体ABCD-A1B1C1D1中,AB=8,BC=6,BB1=6,求BD1与平面ABCD,

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  • 令B1D1的中点为E.利用赋值法,令AA1=1.

    ∵ABCD-A1B1C1D1是正方体,

    ∴AA1=A1B1=A1D1=1、∠AA1B1=∠AA1D1=∠B1A1D1=90°,

    ∴AB1=AD1=B1D1=√2.

    ∵∠B1A1D1=90°、E∈B1D1且B1E=D1E,∴A1E=B1D1/2=√2/2.

    ∵A1B1=A1D1、E∈B1D1且B1E=D1E,∴A1E⊥B1D1.

    ∵AB1=AD1、E∈B1D1且B1E=D1E,∴AE⊥B1D1.

    由A1E⊥B1D1、AE⊥B1D1,得:∠AEA1=二面角A-B1D1-A1的平面角.

    ∵ABCD-A1B1C1D1是正方体,∴AA1⊥平面A1B1D1,∴AA1⊥A1E,

    ∴tan∠AEA1=AA1/A1E=1/(√2/2)=√2.

    ∴∠AEA1=arctan√2.

    ∴二面角A-B1D1-A1的大小为arctan√2.