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Mathematical geometry ppt
1, let the center line AM be on the side of BC, and the vectors AB+AC=2AM, 2OA+2AM=0, so AM is the radius of the circumscribed circle, the center o is at the midpoint m of BC, the triangle ABC is an RT triangle, | BO | =| AO |+CO |, and |AB|=|OA|, then the triangle ABO is

|AB|=|OA|= 1, and the projection of BA in the direction of vector BC is | AB | * COS60 = 1/2.

2. If e and f are taken as planes AA 1D 1D, then this plane is a great circle passing through the center of the sphere with a diameter of =√3.

On the great circle, the chord center distance = 1/2, the radius R=√3/2, and the length of the segment cut by the sphere is MN. According to Pythagorean theorem, Mn/2 = √ 2, and MN=√2.

3. The triangular pyramid is a semi-cuboid, with side lengths of 1, 2 and √3 respectively. Its circumscribed sphere diameter D=√( 1+4+3)=2√2, radius R=√2, AB chord =2, and central angle AOB, sin.

AB arc length =2π√2/4=π√2/2

The spherical distance between AB is π√2/2.

5.( 1)F is the midpoint of BC, AB=AC, AF⊥BC, and because it is a right-angled prism.

Therefore, AF⊥ plane BCC 1B 1, B 1F∈ plane BCC 1B 1,

AF⊥B 1F, in triangle B 1FE, B 1F=√6, EF=√3.

B 1E=3。 According to the inverse theorem of Pythagorean theorem, triangle B 1FE is RT triangle, < B 1fe = 90 degrees, B 1F⊥EF, EF∩AF=F,

∴B 1F⊥ Aircraft Company.

(2) Finding the volume of pyramid E-AB 1F can be transformed into finding the volume of pyramid A-B 1FE.

SRT△b 1FE = b 1F * EF/2 =√6 *√3/2 = 3√2/2 .

AF=√2, V Pyramid A-B1Fe = (3 √ 2/2) * √ 2/3 =1.

6.( 1) In the original trapezoid, CD=AD=AE=CE, the quadrilateral AECD is a diamond, AC⊥DE, let AC and DE intersect at point O, and after folding along DE, CF⊥DE, P(A)F⊥DE,