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.