In the cube abcd-a'b'c'd ', P and Q are the midpoint of a'B', BB ', respectively Find the angle between AP and CQ and the angle between AP and BD Let the edge length of cube be 2 (1) Take the midpoint m of AB and the midpoint n of CC 'to connect B'm and B'n Then: the angle mb'n is the angle formed by the line AP and CQ B'M=B'N=√5,MN=√6 From the cosine theorem, it is concluded that cos(MB'N)=2/5 Angle mb'n = arccos (2 / 5) (2) Connect b'd ', then the angle mb'd' is the angle formed by the line AP and BD B'D'=2√2,D'M=3 From the cosine theorem, it is concluded that: Cos (MB'd ') = from cosine theorem: cos(MB'D')=√10/10 Angle MB'd '= arccos (√ 10 / 10) Why is d'm equal to 3?

In the cube abcd-a'b'c'd ', P and Q are the midpoint of a'B', BB ', respectively Find the angle between AP and CQ and the angle between AP and BD Let the edge length of cube be 2 (1) Take the midpoint m of AB and the midpoint n of CC 'to connect B'm and B'n Then: the angle mb'n is the angle formed by the line AP and CQ B'M=B'N=√5,MN=√6 From the cosine theorem, it is concluded that cos(MB'N)=2/5 Angle mb'n = arccos (2 / 5) (2) Connect b'd ', then the angle mb'd' is the angle formed by the line AP and BD B'D'=2√2,D'M=3 From the cosine theorem, it is concluded that: Cos (MB'd ') = from cosine theorem: cos(MB'D')=√10/10 Angle MB'd '= arccos (√ 10 / 10) Why is d'm equal to 3?

Knowing ad = 2, am = 1 can calculate DM
We also know that DD '= 2, MDD' = 90 degrees
D'm can be calculated

Given that E and F are the midpoint of BB1 and CC1 in the cube abcd-a1b1c1d1, the cosine of the angle between AE and d1f is ()
A. -45B. 35C. 34D. -35

Let the edge length of the cube abcd-a1b1c1d1 be 2, take DA as the X axis, DC as the Y axis, dd1 as the Z axis, and establish the space rectangular coordinate system, then a (2, 0, 0), e (2, 2, 1) D1 (0, 0, 2), f (0, 2, 1) ∧ AE = (0, 2, 1), d1f = (0, 2, - 1), let the angle between AE and d1f be θ

As shown in the figure, ABCD is a square, O is the center of the square, Po ⊥ is the bottom, ABCD, and E is the midpoint of PC. this paper proves: (1) Pa ⊥ plane BDE; (2) plane PAC ⊥ plane BDE. (3) if Po = 1, ab = 2, then the cosine of the angle between the out of plane line OE and AD

It is proved that: (1) connecting AC and OE, AC ∩ BD = O, in △ PAC, ∩ e is the midpoint of PC, O is the midpoint of ac.. PA ⊂ EO, and ∞ EO ⊂ plane EBD, PA ⊄ plane EBD, and ⊂ PA ⊂ plane BDE. (2) ∨ Po ⊥ bottom ABCD, ∨ Po ⊥ BD. and ∨ BD ⊥ AC, and ⊂ BD ⊥ plane PAC

As shown in the figure, ABCD is a square, O is the center of the square, Po ⊥ bottom ABCD, e is the midpoint of PC. Po = 2, ab = 2, prove: (1) Pa ∥ plane BDE; (2) plane PAC ⊥ plane BDE

It is proved that (1) ∩ o is the midpoint of AC, e is the midpoint of PC, ∩ OE ⊂ AP, and ⊂ OE ⊂ plane BDE, PA ⊄ plane BDE, PA ⊄ plane BDE (2) ∵ Po ⊥ bottom ABCD, ∩ Po ⊥ BD, and ≁ AC ⊥ BD, and AC ∩ Po = O ∩ BD ⊥ plane PAC, and BD ⊂ plane BDE, ⊥ plane PAC ⊥ plane BDE