高一數(shù)學(xué)空間向量在立體幾何中的應(yīng)用(理)

空間向量在立體幾何的應(yīng)用例題解析考點(diǎn)1.利用空間向證明空間垂直問題例1PAABCABAC

AB

NABAB=4AN,M,S..AABACAPxy0,0,11,0,

11,0,0,022111CM),,0)222

11CM2

.【點(diǎn)評(píng)】0例2BC1

F

,

CC

=AB=2CE,

AD:=1:4

ED.A,D,(0,0,4),E

,

AF2,

EA

,4

,

ED

AF

EA

=0

AFED

AFEA

,

AF

,

EDEAFAED【點(diǎn)評(píng)1

33例3BCD是MAPD//MAEFMB

ADPD2MA.EFG..A

AM

z

AM=1AD=AB=PD=2,B(0,2,0),C2,2,0),D(2,0,0),P(2,0,2),E(0,1,1,1,1),F(2,1,1)

EG

),

GF

=(1,0,0),EFG

=

z

EG

=

12

=0

=

=0

=1

x

=

z

=0

m

=01,0,PDC

DA

=(2,0,0),

=

2

=0,

DA

,

EFG

PDC【點(diǎn)評(píng)】.考點(diǎn)2.利用空間向處理空間平行關(guān)系例4

E

1

FABE1.A2

A

B

∴

(－

BA

=，02

BEA

m=ym

=

xy

=0

m

=

2x

=0

x

=1z

=1

=

m=1,22

BFCDBF

FBF(22)02,00

22

BF

=

1

32

=0

=1

D1

B

A

【點(diǎn)評(píng)】;例5ABCABC

=

90

BC

B

D

D

BC

ABD

.BABC=2

BBcAa2b)B)A(a0)(0bc)y2AD=

0)

1

=(

b

)

BA1

=(

0

)

BD1

=0

ACD

=(

AD

=

=0

m

=

=0y=x=y10

ay0

=

ayab

B

D

m

=

ay0

ay0

=

ma

BD=(1

,

,

),

n

=

ax=0n2

=

by

=0z

=1=

cc=n=(1)ba3

cFG11,cFG11,

n

AACD.11【點(diǎn)評(píng)】.考點(diǎn)3利用空間向處理異面直線夾角、線面角、二面角等空間角問題.3.1求異面直線所成的角例6ABD11

BCCB1

,1

E

DCC1

平面FEE

GE1

E

1

1

1

1

1

CC

1

DD

1

1

ED

DE1

EFG

DE1

S

FGRtFGRtDG

11222

DEFG1

EE

EDEFG

12S33

2

DA

E

(0,0,1)

G(2,0,1)

F(0,1,2)

(1,2,1)

FG1

1

1

(

FE11

FGFE1

4

(0,00(0,00

FE

FG

FEE

G11

1

1EA1

GE1

sin

3

3.2求直線與平面所成的角例7BDHPHEAD12APB=PAPEHH,,yz

C(,0,0),P)(0,n0)1,,,0).21PECm(,1,022PE22

BC33,故C(,0,0)33D(

33,326

0)(0,0,1)

nxx)

PEH

5

o

12z

n3,0)PA

coPA

PA

PEH

3.2求二面角例如圖在四面體ABOC，O⊥。⊥OB∠°，且OA=OB=OC=1（Ⅰ）設(shè)PAC的中點(diǎn)，Q在且AB=3AQ證明：PQOA；（Ⅱ）求二面角O-AC-B的平面角的余弦值。6

例

ABEF111

,

CC

,

:ADAA

1

EF

AD

ED

：，系A(chǔ)AB,依D

,

F(1,2,1)

,

(0,0,4)

,

E1,,0(1)

,1AD(0,2,,EF,D

AD1A1

35

,

EFA

(2)

=

(xz)

EF

=

=0

=

1y2

=0

x

=1,

=(1,2,1),AED

=

m

mED

=

12

n

=0

mDA

=

p

=0p=1,=2=1=1,2,1cosn,m

n=nm|

sin,

A-ED-F

【點(diǎn)評(píng)（1）

m

m

cos=cosm,n

mmnAB

nm

=

AB

m,n

.例10

ABCD-

CD中B與的1117

BCDBCDA

326333.

D(0,0,0),C(0,1,0)B(1,1,0),A(1,0,0),D(0,0,1),=1,0=1,0,1),=(0,0,1),111ACDxzy0==x=1=1

=1n

sin

=

||n

=

=D.【點(diǎn)評(píng)】n，n

|m|=n||n|||n|

考點(diǎn)4利用空間向處理空間距離問題例111,F

GC

解析：

G(24)

(4),n,y,

xF

D

2xy4xy

E

2EBd|例12如圖，在三棱柱A中，AB11

C棱CC上異于1

C

、的一點(diǎn)，EB，已知2,BCBCC11（1異面直線AB與EB的距離（2二面角EB的平面角的正切值118

3

,求

11解：取如圖所示空間直角坐標(biāo)系，由AB2,BBBCBCC111133(0,0,2),2,0),((,0)AZ22233,,2B22

3

,

ABYEB330EB,2a2413a或a（舍去）E(,0)222設(shè),z)異面直線、EB方向量，yz)2)2zz0(1)由30x,y,)1,xy3(2)2（1AB(0,2,1設(shè)AB的距離為，則

CX

d

n

((3103（2設(shè)nx,z平面的法向量，則011

33,x)xyx3y2B2),