已知PD垂直平面ABCD,AD垂直DC,AD平行BC,PD:DC:BG=1:1:根号2,求(1)PB与平面PDC所成已知PD垂直平面ABCD,AD垂直DC,AD平行BC,PD:DC:BG=1:1:根号2,求二面角D-PB-C的正切值.
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![已知PD垂直平面ABCD,AD垂直DC,AD平行BC,PD:DC:BG=1:1:根号2,求(1)PB与平面PDC所成已知PD垂直平面ABCD,AD垂直DC,AD平行BC,PD:DC:BG=1:1:根号2,求二面角D-PB-C的正切值.](/uploads/image/z/1818424-64-4.jpg?t=%E5%B7%B2%E7%9F%A5PD%E5%9E%82%E7%9B%B4%E5%B9%B3%E9%9D%A2ABCD%2CAD%E5%9E%82%E7%9B%B4DC%2CAD%E5%B9%B3%E8%A1%8CBC%2CPD%3ADC%3ABG%3D1%3A1%3A%E6%A0%B9%E5%8F%B72%2C%E6%B1%82%281%29PB%E4%B8%8E%E5%B9%B3%E9%9D%A2PDC%E6%89%80%E6%88%90%E5%B7%B2%E7%9F%A5PD%E5%9E%82%E7%9B%B4%E5%B9%B3%E9%9D%A2ABCD%2CAD%E5%9E%82%E7%9B%B4DC%2CAD%E5%B9%B3%E8%A1%8CBC%2CPD%EF%BC%9ADC%EF%BC%9ABG%EF%BC%9D1%EF%BC%9A1%EF%BC%9A%E6%A0%B9%E5%8F%B72%2C%E6%B1%82%E4%BA%8C%E9%9D%A2%E8%A7%92D-PB-C%E7%9A%84%E6%AD%A3%E5%88%87%E5%80%BC.)
已知PD垂直平面ABCD,AD垂直DC,AD平行BC,PD:DC:BG=1:1:根号2,求(1)PB与平面PDC所成已知PD垂直平面ABCD,AD垂直DC,AD平行BC,PD:DC:BG=1:1:根号2,求二面角D-PB-C的正切值.
已知PD垂直平面ABCD,AD垂直DC,AD平行BC,PD:DC:BG=1:1:根号2,求(1)PB与平面PDC所成
已知PD垂直平面ABCD,AD垂直DC,AD平行BC,PD:DC:BG=1:1:根号2,求二面角D-PB-C的正切值.
已知PD垂直平面ABCD,AD垂直DC,AD平行BC,PD:DC:BG=1:1:根号2,求(1)PB与平面PDC所成已知PD垂直平面ABCD,AD垂直DC,AD平行BC,PD:DC:BG=1:1:根号2,求二面角D-PB-C的正切值.
应是PD:DC:BC=1:1:√2吧?
以D为原点,DA为X轴,DC为Y轴,DP为Z轴建立空间直角坐标系,
D(0,0,0),A(√2,0,0),
B(√2,1,0),C(0,1,0),
P(0,0,1),
设平面PDB的法向量n1(1,y,z),
向量DB(√2,1,0),向量DP(0,0,1),
向量n1⊥DP,n1⊥DB,n1•DP=0,n1•DB=0,
√2+y=0,y=-√2,z=0,n1=(1,- √2,0),
同理,设平面PBC的法向量n2(x,y,1), 向量PC=(0,1,-1),
向量PB(√2,1,-1),n2•PC=0,n2•PB=0,
y=1,x=0,n2=(0,1,1),
n1•n2=-√2,
|n1|=√3,|n2|=√2,
cos<n1,n2>= n1•n2/(|n1|*|n2|=-√2/(√3*√2)=- √3/3,
二法向量成角为钝角,则二面角D-PB-C为arccos(√3/3),
设二面角D-PB-C平面角为θ,cosθ=√3/3,secθ=1/ cosθ=√3,
tanθ=√[(secθ)^2-1]= √2. 二面角D-PB-C的正切值为√2.
是BC吧
哪来的G啊?题出错了吧?