1. Given the positional relationship between two intersecting lines A, B and A∑ plane α, B and α ().

A. b∑αb. b intersects α.

CB？ αd. b∑α or b intersects α.

[answer] d

[Analysis] ∫a and B intersect, ∴a and B determine a plane as β, if β ∑ α, then b∧α, if β is not parallel to α, then B intersects α.

2. The following proposition is correct ()

(1) After passing a point, there must be a plane parallel to two straight lines in a different plane.

② If the straight line L and the plane α are parallel to the same straight line M, then L∧α

(3) If there is no common point between two straight lines, when one of them intersects, there must be a plane parallel to the other straight line.

A.①b .③

C.①③d .①②③

[answer] b

[Analysis] For example, it is the special case method.

(1) When a point is on a straight line, it does not exist;

②l？ α, m∑l, ② wrong;

(3) The two straight lines A and B have nothing in common, and there are two cases: I)A∑B ii)A and B are different planes, both of which have a plane α passing through the straight line B, α ∑ A.

So choose B.

3. In the space quadrilateral ABCD, M∈AB, N∈AD, if =, then the positional relationship between MN and plane BDC is ().

A.MN？ Plane BDC

B.Mn intersects with plane BDC

C.Mn∑ plane BDC

The positional relationship between D. Mn and plane BDC is uncertain.

[answer] c

[Resolution ]∫=∴Mn∨BD

MN again? Facing BDC ∴MN∥ Facing BDC.

4. Give the following conclusions.

(1) At a point outside the known plane, there is one and only one straight line parallel to the plane.

(2) When crossing a point outside the straight line, one and only one plane is parallel to the known straight line.

(3) If A and B are straight lines in different planes, only one plane parallel to A passes through B. 。

The correct one is ()

A. 1

C.3 D.4

[Answer] A.

[Resolution] (1) Error (2) Error

(3) correctly select point a on b. There is only one A' on the straight line parallel to A beyond this point. B and A' determine the unique plane α, A∧α.

5. Straight lines A and B in different planes are within α and β respectively, and α ∩ β = L, then the positional relationship between straight line L and A and B must be ().

A.l intersects at least one of a and b.

B.l intersects at most one of a and b.

C.l is parallel to at least one of a and b.

D.l intersects with both a and B.

[Answer] A.

[Analysis] According to the conditions, both L and A are in plane α, and both L and B are in plane β. If L does not intersect with A and B, then l∨a and l∨b, so a∨b contradicts A and B, and L intersects at least one of A and B. 。

6. Give the following conclusions:

(1) Two lines parallel to the same line are parallel;

(2) Two planes parallel to the same straight line are parallel;

(3) Two straight lines parallel to the same plane are parallel;

(4) Two planes parallel to the same plane are parallel.

The correct number is ()

A. 1

C.3 D.4

[answer] b

[Analysis] Axiom 4 shows that (1) is correct. In the cube ABCD-a1b1c1d1,the plane ABB 1A 1, DD 65438. Similarly, in the cube ABCD-a1b1d1,a1b1is parallel to the plane ABCD, so (3) is wrong. 4 correct, so choose B.

7. Give the following propositions:

(1) If there is no common point between the straight line and the plane, the straight line is parallel to the plane;

② If a straight line has no common point with any straight line in the plane, it is parallel to the plane;

③ If the straight line does not intersect with countless straight lines in the plane, the straight line is parallel to the plane;

④ If one of two parallel straight lines is parallel to a plane, the other is also parallel to the plane.

The serial number of the correct proposition is ()

A.①②b .③④

C.①③d .②④

[Answer] A.

[Resolution] By definition, ① is correct; If a straight line and any straight line in the plane have no common point, then this straight line and the plane have no common point, ② is correct; As shown in figure (1), the straight line A ∩ α = A, and any straight line in A and α that does not pass through point A does not intersect, so ③ is wrong; As shown in figure (2), a∨b, b? α satisfies a∨b and a∨α, so ④ is wrong.

8. straight line a'? Plane alpha, straight line b' Plane α, and A '∨B', where a' and b' are orthogonal projections of straight lines A and B on plane α, then the positional relationship between straight lines A and B is ().

A. Parallel or different planes

B. Intersection or nonplanarity

C. intersecting, parallel or different planes

D. None of the above answers are correct.

[Answer] A.

As shown in the figure, A and B can be parallel or different.

9. At △ABC, ∠ ACB = 90, BC = 3°, AC = 4°, and P is the point on AB, then the maximum value of the distance product from point P to AC and BC is ().

a . 0 b . 3

C.12 d. does not exist.

[answer] b

[Analysis] From the meaning of the question ab = 5, let pa = x, then 0≤x≤5, Pb = 5-x,

=,=,

∴PM PN=x (5-x)=x(5-x)，

When x =, the maximum value is 3.

10. as shown in the figure, in the cube ABCD-a1b1c1d, where e and f are the midpoints of the sides BC and C 1D 1D respectively, then EF and.

A.ef∑ Σ plane BB 1D 1D

B.ef intersects the plane bb1d1d.

C.ef? Plane BB 1D 1D

The positional relationship between D.EF and plane BB 1D 1D cannot be judged.

[Answer] A.

[Prove] Take the midpoint o of D 1B 1, even, OB,

B 1C 1，BE B 1C 1，∴OF BE，

∴ quadrilateral OFEB is a parallelogram, ∴EF∥BO.

∵EF？ Plane BB 1D 1D, BO? Plane BB 1D 1D,

∴EF∥ plane BB 1D 1D, so choose a.

Second, fill in the blanks

1 1. as shown in the figure, in the cube ABCD-a1b1c1d1,m is the midpoint of a1,then a straight line.

[Answer] Intersection

[Analysis] Because m is the midpoint of A 1D 1 and the straight line DM intersects with the straight line AA 1, DM has a common point with the plane A 1CC 1, so DM and the plane A 1CC 1.

Third, answer questions.

12. In the geometry shown in the figure, △ABC is an arbitrary triangle, AE∨CD, AE = AB = 2A, CD = A, and F is the midpoint of BE.

Verification: DF∑ plane ABC.

【 Proof 】 As shown in the figure, take the midpoint G of AB and connect FG and CG, where ∵F and G are the midpoints of BE and AB respectively.

∴FG 綊 AE,

Ae = 2a，CD = a，

∴CD=AE，

And AE∑CD, ∴CD ∴ FG,

∴ Quadrilateral CDFG is a parallelogram,

∴DF∥CG.

CG again? Plane ABC, DF? ABC aircraft, df, ABC aircraft

13. As shown in the figure, it is known that E, F, G and M are the midpoint of the sides AD, CD, BD and BC of a tetrahedron, and it is proved that the AM∑ plane EFG.

[Proof] As shown in the figure, connect DM, cross GF at point O, and connect OE.

In △BCD, G and F are the midpoint of BD and CD, ∴GF∥BC, respectively.

∫G is the midpoint of BD,

∴O is the midpoint of MD.

In △AMD, ∫e and O are the midpoint between AD and MD, ∴EO∥AM.

∵AM？ Aircraft EFG, EO aircraft EFG. ∴AM∥ airplane EFG.

14. As shown in the figure, P is a point out of the plane of the parallelogram ABCD, and M and N are the midpoint of AB and PC respectively.

(1) verification: MN∑ plane PAD；;

(2) If MN = BC = 4 and PA = 4, find the angle formed by the straight line PA and MN.

[Analysis] (1) Take the midpoint H of PD, connect AH, NH, ∫n as the midpoint of PC, ∴NH ∶ DC. Where m is the midpoint of AB,

∴NH ∴ that is, quadrilateral AMNH is a parallelogram.

∴MN∥AH.

By MN? Flat mat, huh? Flat pad,

∴MN∥ airport

(2) Connect AC, take the middle point O, connect OM, ON,

∴OM 綊 BC, Moon 綊 Dad.

∴∠ONM is the angle formed by straight lines PA and MN.

From Mn = BC = 4, PA = 4, OM = 2, ON = 2.

∴MO2+ON2=MN2,∴∠ONM=30，

That is, the non-planar straight line PA forms an angle of 30 with MN.

15. As shown in the figure, square ABCD and square ADEF intersect in AD, where m and n are points on BD and AE respectively, and an = BM. Proof: MN∑ plane EDC (proved by two proofs).

[Proof] Proof 1: If NP∨AD crosses DE in P and MQ∨AD crosses DC in Q, then NP∨MQ. an = bm，∴ NE = DM，

= =, again =, =,

∴NP = MQ，∴ NP MQ

∴MNPQ is a parallelogram, ∴MN∥PQ.

PQ and aircraft EDC MN? Aircraft deck,

Mn ∥ Graphic Design Center

Proof 2: Connect AM, extend the intersection DC to H, and connect EH.

∫ab∨CD ∴=

And BM = an, BD = AE, ∴ =, ∴NM∥EH.

∵MN？ Aircraft EDC, huh? Plane EDC

∴MN∥ Aircraft Design Center.

16.(09 Shandong) As shown in the figure, in the square prism ABCD-A1B1D1,the bottom ABCD is an isosceles trapezoid, AB∑CD, AB = 4, BC = CD = 2, AA6538.

[resolution] take the midpoint F 1 of A 1B 1 and connect FF 1, C 1F 1.

Since ff1∨ bb1∨ cc1,

So F 1∈ plane FCC 1,

So the plane FCC 1 is the plane C 1CFF 1, which connects A 1D and f1C. Starting with A 1F 1D 1CD,

So quadrilateral A 1DCF 1 is a parallelogram,

Therefore, a1d ∨ f1c.

And ee 1∑a 1D and ee 1∑f 1C,

And EE 1? Aircraft FCC 1, F 1C? Aircraft FCC 1,

So ee1Σ plane FCC 1.

[Comment] After learning the next lesson, it can be proved that the following are parallel to each other:

Because F is the midpoint of AB, CD = 2, AB = 4, AB∨CD, so CD∨AF, so the quadrilateral AFCD is a parallelogram, so AD∨FC.

And cc 1∨DD 1, FC ∩ cc 1 = c, FC? Aircraft FCC 1, CC 1? Plane FCC 1, so plane add 1a 1∑ plane FCC 1,

EE 1 again? Plane ADD 1A 1,

So ee1Σ plane FCC 1.