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19. (12 分) 如图, 在四棱雉 P−ABCDP-A B C D 中, $A B=B D=B P=\sqrt{5}, P A=P D=\sqrt{2}, \...
Mar 22, 2024
19. (12 分) 如图, 在四棱雉 P−ABCDP-A B C D 中, AB=BD=BP=5,PA=PD=2,∠APD=90∘,EA B=B D=B P=\sqrt{5}, P A=P D=\sqrt{2}, \angle A P D=90^{\circ}, E 足棱 PAP A 的中点, 且 BE//B E / / 平面 PCDP C D. (1) 证明: CD⊥C D \perp 平面 PADP A D; (2) 若 CD=1C D=1, 求二面角 A−PB−CA-P B-C 的余弦值.
Solution by Steps
step 1
To prove CD⊥ CD \perp plane PAD PAD , we need to show that CD CD is perpendicular to two distinct lines that lie on plane PAD PAD
step 2
Since ∠APD=90∘ \angle APD = 90^{\circ} , line PD PD is already perpendicular to PA PA which lies on plane PAD PAD
step 3
If CD CD is perpendicular to PA PA , then CD CD will be perpendicular to the plane PAD PAD because it would be perpendicular to two intersecting lines on the plane
step 4
To show CD CD is perpendicular to PA PA , we can demonstrate that triangle PCD PCD is a right triangle with ∠CPD=90∘ \angle CPD = 90^{\circ} by using the Pythagorean theorem, given PD=2 PD = \sqrt{2} and CD=1 CD = 1
step 5
Calculate PC2+PD2 PC^2 + PD^2 and if it equals CD2 CD^2 , then ∠CPD=90∘ \angle CPD = 90^{\circ} and CD⊥ CD \perp plane PAD PAD
step 6
PC2+PD2=12+(2)2=1+2=3 PC^2 + PD^2 = 1^2 + (\sqrt{2})^2 = 1 + 2 = 3
step 7
Since CD2=12=1 CD^2 = 1^2 = 1 and PC2+PD2=3 PC^2 + PD^2 = 3 , CD CD is not perpendicular to PA PA , thus CD CD is not perpendicular to plane PAD PAD . There must be an error in the given information or the question's setup
Answer
The statement CD⊥ CD \perp plane PAD PAD cannot be proven with the given information as it is inconsistent.
Key Concept
Perpendicularity in 3D Geometry
Explanation
To prove a line is perpendicular to a plane, it must be shown to be perpendicular to two distinct lines on that plane. The Pythagorean theorem is used to verify perpendicularity in triangles.
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Solution by Steps
step 1
To find the cosine of the dihedral angle A−PB−C A-PB-C , we can use the Law of Cosines in triangle ABP ABP and BPC BPC
step 2
The Law of Cosines states AB2+BP2−2⋅AB⋅BP⋅cos⁡(θ)=PC2 AB^2 + BP^2 - 2 \cdot AB \cdot BP \cdot \cos(\theta) = PC^2 , where θ \theta is the angle ABP ABP
step 3
Substitute AB=5 AB = \sqrt{5} , BP=5 BP = \sqrt{5} , and PC=1 PC = 1 into the Law of Cosines
step 4
Solve for cos⁡(θ) \cos(\theta) : (5)2+(5)2−2⋅5⋅5⋅cos⁡(θ)=12 (\sqrt{5})^2 + (\sqrt{5})^2 - 2 \cdot \sqrt{5} \cdot \sqrt{5} \cdot \cos(\theta) = 1^2
step 5
Simplify the equation: 5+5−10⋅cos⁡(θ)=1 5 + 5 - 10 \cdot \cos(\theta) = 1
step 6
Solve for cos⁡(θ) \cos(\theta) : 10−10⋅cos⁡(θ)=1 10 - 10 \cdot \cos(\theta) = 1
step 7
Subtract 10 from both sides: −10⋅cos⁡(θ)=−9 -10 \cdot \cos(\theta) = -9
step 8
Divide by -10: cos⁡(θ)=910 \cos(\theta) = \frac{9}{10}
Answer
The cosine of the dihedral angle A−PB−C A-PB-C is 910 \frac{9}{10} .
Key Concept
Law of Cosines
Explanation
The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles, which can be used to find the cosine of the dihedral angle in this context.
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