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Chng I
PHP CHIU
Ngi ta dng php chiu xuyn tm, php chiu song song ( m trng hpc bit l php chiu vung gc ) biu din cc vt thtrong khng gian.
I.PHP CHIU XUYN TM
1.1: Chiu mt im A ttm chiu S ln mt phng hnh chiu P
Trong khng gian ly mt mt phng P v mt im S khng thuc P. Chiumt im A bt kca khng gian t tmS ln mt phng P l:
1. Vng thng SA.2. Xc nh giao im A Ca
ng thng SA vi mt phng P (H 1.1 )
Khi ngi ta gi A l hnh chiuca im A ttm S ln mt phng P.
Ta c cc tn gi: Hnh 1.1
S : Tm chiu
SA : Tia chiu ( ng thngchiu)
P : Mt phng hnh chiu
Vy hnh chiu ca im A l imA. Ddng thy rng nu A thuc mt
phng P th A trng vi A. Hin nhinA khng ch l hnh chiu ca im Am n cn l hnh chiu ca mt im bt kthuc ca ng thng SA. V dAcng l hnh chiu ca cc im B, C .....(H 1.2 ).
1.2:Chiu mt ng thng ttmchiu S ln mt phng hnh chiu P
Nhta bit ng thng l tp hpca v sim nn tm hnh chiu cang thng ta i tm hnh chiu ca hai
im thuc ng thng (H 1.3)
AB C
C
A
S
B
P
Hnh 1.2
B
P
A'
A
S
Hnh 1.3
P
A
S
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1.3:Chiu mt hnh ttm chiu S ln mt phng hnh chiu P
Hnh l mt tp hp
im. Chiu mt hnh ttm S ln mt phng P tc lchiu mi im ca hnh ttm S ln mt phng P.
Vy hnh chiu camt hnh l tp hp cchnh chiu ca mi imca hnh . Nhng nh ta
sthy, mun vhnh chiuca hnh ta ch cn v hnhchiu ca cc yu t xcnh hnh . ( H 1.4 )
1.4: Tnh cht ca php chiu xuyn tm
Tnh cht 1. Hnh
chiu ca mt ng thng
khng i qua tm chiu lmt ng thng. ( H 1.3 )
Tnh cht 2. Php
chiu xuyn tm bo ton ts kp ca 4 im thnghng.
Gi sA, B, C, D l 4im thuc ng thng k,hnh chiu ca chng lnlt l A, B, C, D thuc ng thng k ( H 1.5 )
Ta c:''
'':
''
'':
BD
DA
BC
CA
DB
AD
CB
AC=
S
A
AP
Hnh 1.4
AB
CD
AB
C
S
D
P
k
k
Hnh 1.5
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II. PHP CHIU SONG SONG
2.1: nh ngha
Trong khng gian ly 1 mt phng Pv 1 ng thng s ct mt phng P ( s gil hng chiu song song ) hnh chiusong song ca im A ln mt phng Ptheo hng chiu s l giao im A cang thng i qua A song song vi s vmt phng P. ( H 1.6 )
2.2: Tnh cht
Tnh cht 1. Trong php chiu song song hai ng thng song song chiuthnh hai ng thng song song. ( H 1.7 ) AB // CD AB // CD
Tnh cht 2.Trong php chiu song song t sn ca 3 im thng hng
bng tsn ca 3 im hnh chiu ca chng. ( H 1.8 ).''
''
CB
BA
BC
AB=
Hnh 1.7 Hnh 1.8
Hqu: Trong php chiu song song tsca hai on thng song song bngtsca hai on thng hnh chiu ca chng. ( H 1.7 )
GisAB // CD, AB v CD l hai hnh chiu ca chng khi ta c:
''
''
DC
BA
CD
AB=
A
S
AP
Hnh 1.6
D
C
B
A
B'A'
C'P
S
D'
B'C'
CBA
P
A'
S
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III. PHP CHIU VUNG GC
3.1: nh ngha.
Php chiu thng gc ( php chiu vung gc ) l php chiu song song chng chiu s vung gc vi mt phng hnh chiu P.
3.2: Tnh cht.
Php chiu thng gc l trng hp c bit ca php chiu song song do nc mi tnh cht ca php chiu song song, ngoi ra n cn c tnh cht sau:
* iu kin t c v mt gcvung chiu thnh mt gc vung
l, c mt cnh song song vi mtphng hnh chiu, cnh cn likhng vung gc vi mt phnghnh chiu. ( H 1.9 ) c AB songsong vi mt phng P cn BCkhng vung gc vi mt phng P.
Khi nu gc ABC = 90 th
gc ABC = 90
VI. MT SKIN THC VSONG SONG V VUNG GC
4.1: Mt squan hsong song trong khng gian.
a. Nu ng thng a khng thuc mt phng Q v a song song vi ngthng b thuc Q th ng thng a song song vi mt phng Q.
b.Nu ng thng a song song vi mt phng Q, mt phng R cha a v ctQ theo giao tuyn b th a song song vi b.
c.Nu 3 mt phng ct nhau theo 3 giao tuyn th 3 giao tuyn hoc songsong hoc ng quy.
d.Nu 2 mt phng P v Q cng song song vi ng thng a v ct nhau theogiao tuyn c th a song song vi c.
4.2: Mt squan hvung gc trong khng gian.
a.Mt ng thng vung gc vi mt mt phng th ng thng vunggc vi mi ng thng thuc mt phng .
A
C
C'
B'
A'P
B
Hnh 1.9
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b. Nu ng thng k vung gc vi hai ng thng ct nhau thuc mtphng Q th k vung gc vi Q.
c.Qua mt im bao gicng chdng c mt ng thng vung gc vimt mt phng cho trc.
d. Nu ng thng k vung gc vi mt phng Q th mi mt phng chang thng k u vung gc vi Q.
4.3: Khong cch.
a.Khong cch tmt im n mt phng l di on vung gc tnh tim n mt phng.
b.Khong cch tmt im n ng thng l di on vung gc tnh tim ti ng thng.
c. Khong cch gia hai mt phng song song l khong cch t mt imthuc mt phng ny n mt phng kia.
CU HI V BI TP
1. Nu nh ngha php chiu xuyn tm, php chiu song song, php chiuvung gc?
2.Nu cc tnh cht ca php chiu song song?
3. Cho hnh chiu song song trn mt mt phng ca tm O v hai nh A, Bca hnh bnh hnh ABCD l O, A, B, vhnh chiu ca hnh bnh hnh trn mt
phng ? ( H 1.11 )
Hnh 1.10 Hnh 1.11
4.Cho hnh chiu song song trn mt mt phng ca trng tm G v ca hai
nh A, B ca tam gic ABC ln lt l G, A, B, vhnh chiu ca nh th3 C
ca tam gic. ( H 1.11 )
A
B
O
A
B
G
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Chng II
IM
I. HNH CHIU CAIM TRONG HHAI MT PHNG HNH CHIU.
Phng php hai hnh chiu thng gc c dng rng ri trong kthut nhtl trong cc bn vckh v xy dng.
1.1. Hthng hai mt phng hnh chiu trong khng gian.
Trong khng gian ly hai mt phng vung gc P1v P2, ct nhau theo ng
thng x.
Ta c cc tn gi sau:
+ P1thng ng gi l mt phng hnh chiu ng.
+ P2nm ngang gi l mt phng hnh chiu bng.
+ Hng chiu s1vung gc vi P1gi l hng chiu ng.
+ Hng chiu s2vung gc vi P2gi l hng chiu bng.
+ ng thng x gi l trc hnh chiu.
Hnh 2.1 Hnh 2.2
S2
P1
X
Ax A2
A1A
A2
P2
S1
X
A1
Ax
A2
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Hai mt phng hnh chiu chia khng gian ra lm bn phn c nh sthtl I, II, III, VI. ( H 2.1 )
Mt phng P1sc chn lm mt phng bn vtc l mt phng trn svhnh biu din ca khng gian.
1.2. Hnh chiu ca im.
Gisc mt im A thuc gc phn tthI. ( H 2.1 )
+ Chiu vung gc im A ln mt phng hnh chiu ng P1 ta c
im A1, A1gi l hnh chiu ngca im A
+ Chiu vung gc im A ln mt phng hnh chiu bng P2 ta c
im A2, A2gi l hnh chiu bngca im AMt phng P1c chn lm mt phng bn v, xoay mt phng P2quanh trc
x vtrng vi P1sao cho na trc ca P2vtrng vi na di ca P1( H2.1 ) .Khi hnh biu din ca im A l cp im ( A1, A2)
+ A1A2thuc ng thng vung gc vi trc x ti Ax
+ ng thng ni A1A2gi l ng dng ng ca im A ( H 2.2 )
1.3. Quy c vcao v xa ca mt im.
1.3.1 cao.
+ nh ngha: cao ca im A l khong cch ca im ti mt phnghnh chiu bng P2.
+ Quy c:
- Nu im A pha trn P2th c cao dng ( A1nm pha trn trc x) A1Ax > 0. ( A thuc gc phn tthI v thII )
- Nu im pha di P2th c cao m ( A1nm pha di trc x )A1Ax < 0. ( A thuc gc phn tthIII v thVI )
1.3.2 xa.
+ nh ngha: xa ca mt im l khong cch ca im ti mt phnghnh chiu ng P1.
+ Quy c:
- Nu im pha trc P1th c xa dng , A2nm pha di trc
x , A2Ax > 0. ( A thuc gc phn tthI v thVI )
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- Nu im pha sau P1th c cao m ( A2nm pha trn trc x )A2Ax < 0. ( A thuc gc phn tthII v thIII )
III. HNH CHIU CAIM TRONG H3 MT PHNG HNH CHIU.
Trong k thut c nhng vt th nu ch biu din trn hai mt phng hnhchiu th khng thhin c ht hnh dng v kt cu ca vt th, v vy i khicn biu din vt thtrong hba mt phng hnh chiu.
2.1: Hba mt phng hnh chiu trong khng gian.
Trong khng gian cng vi P1v P2ta ly thm mt phng P3vung gc vi
trc x tc l vung gc vi chai mt phng hnh chiu P1v P2( H 2.3 )
+ P3gi l mt phng hnh chiu cnh+ Hng chiu vung gc vi P3gi l hng chiu cnh
+ P1giao vi P2ti trc x, P1giao vi P3tai trc z, P2 giao vi P3ti trcy, to ln htrc ta oxyz ( H2.3 )
2.2. Hnh biu din im
Gi s c im Atrong khng gian (H2.3 )
+ Hnh chiu vunggc ca im A ln mt
phng hnh chiu ngP1 l A1. A1 gi l hnhchiu ng ca im A.
+ Hnh chiu vunggc ca im A ln mt
phng hnh chiu bng P2l A2. A2 gi l hnhchiu bng ca im A.
+ Hnh chiu vunggc ca im A ln mt
phng hnh chiu cnh P3l A3. A3 gi l hnhchiu cnh ca im A.
P1
XAx
A2
A1
A
A2
P2
P3
y
z
A3
Az
Ay
A3
o
Hnh 2.3
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im A c cc khong cch v ta sau:
+ A1Ax = AA2= ZA ( gi l cao ca im A )
+ A2Ax = AA1 = YA ( gi l xa ca im A )+ A1Az = AA3= XA ( gi l xa cnh ca im A )
* Xoay P2 v trng vi mtphng P1nhtrn, xoay P3vtrngvi P1nhhnh ( H 2.3). khi ta c
thc ca im A ( H 2.4 ).
Tthc ca im A ta c:
+ A1A2 nm trn ngdng vung gc vi trc x ti Ax.
+ A1A3 nm trn ngdng vung gc vi trc z ti Az.
+ AxA2= AzA3
CU HI V BI TP
1.Trnh by cch biu din mt im theo phng php hai hnh chiu thnggc ?
2. Xc nh vtr ca cc im so vi mt phng hnh chiu ca cc im A, BC, D, E, ( H 2.5 )
Hnh 2.5 Hnh 2.6 Hnh 2.7
3.Tm im A i xng vi im A qua mt phng hnh chiu ng P1, tm
im B i xng vi im B qua mt phng hnh chiu bng P2? ( H 2.6 )
z
A1 A3
A2
Ax
Ay
Az
X
y
o Ay Y
E1 E2
C2C1B2
B B1D2
D1
xA2
A1A
x
A1
A2
B1
B2A2
B2
A1
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4. Cho im A ( A1, A2) v hnh chiu bng ca im B l B2, xc nh vtrtrc x v hnh chiu ng ca im B? ( H 2.7 )
5. Vhnh chiu cnh ca im A (A1, A2), vhnh chiu bng ca im B( B1, B3) ? ( H 2.8 )
Hnh 2.8
X Ax
A1
A2
Y o X
B1 B3Bz
o
y
z
Y
z
y
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Chng III
NG THNG
I. HNH BIU DIN CANG THNG.
biu din mt ng thng bt k, ngi ta cng chiu ng thng y lncc mt phng hnh chiu P1, P2nhkhi biu din cc im. V hnh chiu ca mtng thng l mt ng thng v c xc nh bi hai im, nn mun vhnhchiu ca ng thng ta i vhnh chiu ca hai im thuc ng thng. ( H 3.1 )
Hnh 3.1
Gista c ng thng d, v A, B l hai im bt kthuc ng thng .+ im A c hnh chiu ng A1, hnh chiu bng A2.+ im B c hnh chiu ng B1, Hnh chiu bng B2.
Khi ng thng d1xc nh bi hai im A1, B1l ng thng hnh chiu
ng ca ng thng d, d2xc nh bi hai im A2, B2l ng thng hnh chiubng ca ng thng d.
II. NHNGNG THNG C VTRC BIT TRONG HTHNG MTPHNG CHIU.
2.1: ng thng ng mc.
ng thng ng mc l ng thng c vtr song song vi mt trong ccmt phng hnh chiu.
b
Ax
A1
A2
X
bx
b1
b2
a
d
d1
d2
P2
P1b1
d1A1
AxX bx
b2
A2
d2
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a. ng bng.
+ nh ngha: ng bng l ng thng song song vi mt phng hnh chiubng P2. ( H 3.3 )
Hnh 3.3
+ Tnh cht:
1. Hnh chiu ng d1song song vi trc x ( y l du hiu c trng nhn bit ng bng )
2. on thng thuc ng bng ( AB )song song v bng hnh chiubng (A2B2) ( AB // = A2B2)
3.Gc to bi ng thng hnh chiu bng d2vi trc x chnh l gc tobi gia ng bng v mt phng hnh chiu ng P1, gc .
b. ng mt.+ nh ngha: ng mt l ng thng song song vi mt phng hnh chiu
ng P1. ( H 3.4 )
+ Tnh cht:
1. Hnh chiu bng d2song song vi trc x ( y l du hiu c trng nhn bit ng mt )
b
A1
A2
b1
b2
a d
d1
d2
P1
P2
x
A1 B1
b2
d1
d2
x
A2
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Hnh 3.4
2. on thng thuc ng mt (AB) song song v bng hnh chiu ng(A1B1) ( AB // = A1B1)
3. Gc to bi ng thng hnh chiu ng d1vi trc x chnh l gc to
bi gia ng mt v mt phng hnh chiu bng P2, gc .c. ng cnh.+ nh ngha: ng cnh l ng thng song song vi mt phng hnh chiu
cnh P3. ( H 3.5 )+ Tnh cht:
1. Hnh chiu ng ( d1 ), hnh chiu bng ( d2 ) nm trn mt ngdng vung gc vi trc x.( chng cnh ta cn chr hai im thuc ngcnh ).
2. Hnh chiu cnh ( A3B3 ) song song v bng AB.
3. Gc to bi gia d v P1 bng gc to bi d3 v trc z.( gc )4. Gc to bi gia d v P2 bng gc to bi d3 v trc y.
(gc )2.2: ng thng chiu.L ng thng vung gc vi mt phng hnh chiu.
a. ng thng chiu ng+ nh ngha: ng thng chiu ng l ng thng vung gc vi mt
phng hnh chiu ng P1.
b
A1
A2
b1
b2
a
d
d1
d2
P1
P2
x
A1
B1
b2
d1
d2
x
A2
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Hnh 3.5
+ Tnh cht:1.Hnh chiu ng suy bin thnh mt im A1B1 d1.
( Du hiu c trng cho ng thng chiu ng )2.Hnh chiu bng d2 vung gc vi trc x.3.Hnh chiu bng A2B2 song song v bng AB.( H 3.6 )
Hnh 3.6
b.ng thng chiu bng.+ nh ngha: ng thng chiu bng l ng thng vung gc vi mt
phng hnh chiu bng P2.+ Tnh cht:
1. Hnh chiu bng suy bin thnh mt im A2 B2 d2.
b
A1
A2
b1
b2
a
d
d1
d2
P3
P2
x
A1
b1
Z
Y
P1
d3
Ob1
A1
d1
A2
b2
d2
B3
A3
d3
Z
Y
Yx o
d1
d
d2
p2
aa1
p1
a2
x
b2
b
b1b1a 1 d1
d2
b2
a2
x
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( Du hiu c trng cho ng thng chiu bng )2. Hnh chiu ng d1vung gc vi trc x.3. Hnh chiu ng A1B1 song song v bng AB.( H 3.7 )
Hnh 3.7
c. ng thng chiu cnh.+ nh ngha: ng thng chiu cnh l ng thng vung gc vi mt
phng hnh chiu cnh P3.+ Tnh cht:
1. Hnh chiu ng d1 v hnh chiu bng d2 song song vi trc x.( Duhiu c trng cho ng thng chiu cnh )
2. Hnh chiu cnh d3 suy bin thnh mt im A3 B3 d33. AB //= A1B1 //= A2B2. ( H 3.8 )
Hnh 3.8
B
x
P1
A2B2d2
A
P2
d
d1
A1
B1
d1
B1
A1
x
A2B2d2
y
y
z
xA3B3d3
A1
P3
y
z
x
d1
d
d2
P2
A
P1
A2B2
B
B1A1d1
B2
B1
d2
A2
A3B3d3
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III. SLIN THUC GIAIM VNG THNG.nh l: iu kin t c v , im C thuc ng thng AB ( AB khng
l ng cnh ) l,cc hnh ch iu ca im C thuc cc hnh chiu cng tn ca AB
( C1 A1B1, C2 A2B2 ) ( H 3.9 ).
Hnh 3.9
Nu thc ca im v mt ng thng ( khng phi l ng cnh) thamn: hnh chu ng ca im thuc hnh chiu ng ca ng thng, hnh chiu
bng ca im thuc hnh chiu bng ca ng thng, v chng nm trn mtng dng ng th thc ny biu din im thuc ng thng trong khnggian.
VI. VT CANG THNG.4.1: Vt ng ca ng thng
+ nh ngha: Vt ng ( N ) l giao im ca ng thng vi mt phnghnh chiu ng P1.
+ Tnh cht: N thuc P1 nn c xa bng khngN2 thuc trc x.
Vy mun xc nh vt ng N ca ng thng d ta i tm hnh chiu bng N2ca im N, N2 l giao ca d2 v trc x, sau tN2N1. ( H 3.10 )
Hnh 3.10
C2
C1
C2
C1
C
x
B2
A2
A1
B1B1B
B2
x
A2
p1
A1
A
p2
MM2
P1
M1
d
N2
NN1
P2
d1x
d2
NN1
d1
MM2
M1
N2
x
d2
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4.2: Vt bng ca ng thng
+ nh ngha: Vt bng ( M ) l giao im ca ng thng vi mt phnghnh chiu bng P2.
+ Tnh cht: M thuc P2 nn c cao bng khngM1 thuc trc x.
Vy mun xc nh vt bng M ca ng thng d ta i tm hnh chiu ngM1 ca im M, M1 l giao ca d1 v trc x, sau tM1M2. ( H 3.10 ) .
V. VTR TNGI CA HAING THNG.
5.1: Hai ng thng ct nhau.
- Hai ng thng thng:iu kin t c v 2 ng thng thng ctnhau l cc cp hnh chiu cng tn ca chng ct nhau ti nhng im trn cngmt ng dng. ( H 3.11 )
a ) b ) c )
Hnh 3.11
Hnh 3.11b: l chai ng thng a, b cng nm trn mt phng chiu ng.
Hnh 3.11c: l chai ng thng a, b cng nm trn mt phng chiu ng,nhng y a l ng thng chiu ng.
-Mt trong hai ng l ng cnh: Th nu hnh chiu ng ca hai ng
thng l hai ng thng ct nhau, hnh chiu cnh l hai ng thng ct nhau, vhai giao im ny nm trn mt ng dng ngang, th hai ng thng tng ngl ct nhau trong khng gian. ( H 3.12 a )
Trn hnh 3.12 b: AB l ng cnh, qua hnh chiu ng v hnh chiu bngchng ta khng kt lun ngay c chng c ct nhau hay khng. Chng ta c thitm hnh chiu cnh ca chng nu chng tho mn trng hp hnh (3.12a ) thchng l hai ng thng ct nhau trong khng gian.
x
b1
x
M2
M1
x
a1
a2
b2
M2
M1
a1 b1
a2
b2
M2
M1
a1 b1
a2
b2
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a) b)
Hnh 3.12
5.2: Hai ng thng song song.
-Hai ng thng thng: Nu trn thc, hnh chiu ng ca chng l haing thng song song, hnh chiu bng ca chng l hai ng thng song song,th thc biu din hai ng thng song song trong khng gian. (H 3.13 a )
a) b) c)
Hnh 3.13
Hnh 3.13 b: chai ng thng a, b cng nm trn mt phng chiu ng.
Hnh 3.13 c: chai ng thng a, b cng l ng thng chiu ng.
-Hai ng thng l ng cnh: Nu trn thc, hnh chiu cnh ca chngl hai ng thng song song th thc biu din hai ng thng song songtrong khng gian. ( H 3.14 a )
M3M1
x
a1a3
o
A1
B1
A3
B3
z
y
y
A1
A2
x
a1
a2
B1
a2
a1
x x
a1b1
b1
b2
xa2
b2
a2
b1
b2
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a) b)
Hnh 3.14
Trn hnh 3.14 b, AB v CD l hai ng cnh, qua hnh chiu ng v hnhchiu bng, chng ta khng kt lun ngay c chng c ct nhau hay khng.Chng ta c thi tm hnh chiu cnh ca chng nu chng thomn trng hphnh ( 3.14a ) th chng l hai ng thng ct nhau trong khng gian.
5.3: Hai ng thng cho nhau.
Nu hai ng thng cho nhau trong khng gian, th hnh chiu ca chngkhng thomn iu kin ct nhau ( song song nhau ) ni trn th hai ng thngl cho nhau. ( H 3.15 )
Hnh 3.15
xa1 b1
b3
z
y
ya3 x
B1
A1
A2
B2
D1
C1
C2
D2
x
b2
a2
a1b1
x
b2
a2
a1b1
x
b2a2
a1b1
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VI. XCNHLN THT CAON THNG
Gisc on thng AB c biu din bng cc hnh chiu A1B1v A2B2.Bi ton t ra l ta i xc nh di ca AB theo cc hnh chiu y.
Quay li cch lp thc caon thng AB ( H 3.16 ). Ta thyc rng AB l cnh huyn ca tamgic vung ABA* vung A*, v ccnh AA* song song v bng A1B1,cnh BA* bng hiu xa ca haiim A, B.
Do vic xc nh ln caon thng AB a v vic xc nhcnh huyn ca tam gic vung nitrn.
Tng t ta thy AB cng lcnh huyn ca tam gic vung ABB* vung B*, v c cnh BB* song song v
bng A2B2, cnh AB* bng hiu cao ca hai im A, B.
Hnh 3.17 Hnh 3.18
Trn hnh 3.17, di ca on thng AB l A2B*, l cnh huyn ca tam gicvung m mt cnh l hnh chiu bng A2B2, v cnh cn li l B*B2bng hiu cao ca hai im A v B l A1B0.
Trn hnh 3.18, di ca on thng AB l A2B* l cnh huyn ca tam gicvung m mt cnh l hnh chiu bng A2B2, v cnh cn li l B*B2bng hiu cao ca hai im A v B l A1B0.
x
P2
P1
B*
B A*
A
B2
A2
B1
A1
Hnh 3.16
x
A*
A2 A0
B2
A1
B1
B*
B0 B1
A1
B2
A2
x
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CU HI V BI TP
1. Trnh by cch biu din mt ng thng?
2.Pht biu
iu ki
n c
n v
m
tim thu
c
ng th
ng ?
3.Cho im A ( A1, A2 ), im C ( C1, C2 ), qua A von thng AB songsong vi mt phng hnh chiu bng bit AB c di bng 25mm, v nghing vimt phng hnh chiu ng gc 30. ( H 3.16 a).Qua C von thng CD song songvi mt phng hnh chiu ng bit CD c di bng 30mm, v nghing vi mt
phng hnh chiu ng gc 45. (H 3.16 b)
a) b)
Hnh 3.16
4.Cho hai ng thng bt ka ( a1
, a2
) v b ( b1
, b2
) hy vng thnghnh chiu ng c, ng thng hnh chiu bng d, ct chai ng thng a v b?
( H 3.17 )
Hnh 3.17 Hnh 3.18
5.Cho ng thng a tm trn a cc im : ( H 3.18 )
+A c cao bng 0 +C c hai hnh chiu trng nhau
+B c xa bng 0 +D c hai hnh chiu i xng nhau qua trc x
6. Xc nh di cnh ca on thng AB? ( H 3.19 )
x
A1
A2
x
A1
A2
x
b2
a2
a1
b1x
a2
a1
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6.Cho hai ng thng bt k a ( a1, a2) v b ( b1, b2) v hnh chiu ng M1ca im M. Tm hnh chiu bng M2bit M thuc ng thng c song song vi av ct b. ( H 3.20 )
Hnh 3.19 Hnh 3.20
7.Qua im N vng bng b ( b1, b2) ct ng thng a? ( H 3.20 )
8.Cho hnh chiu ng ca hai im A v B, hnh chiu bng ca hai im Cv D, hy tm cc hnh chiu cn li ca A, B, C, D bit rng cbn im uthuc mt ng thng? ( H 3.21 )
9.Cho im A ( A1, A2), ng thng b ( b1, b2) v c ( c1, c2). Hy vqua Ang thng ct cb v c? (H 3.22 )
Hnh 3.20 Hnh 3.21 Hnh 3.22
x
b2a2
a1
b1
M1
M1
a1a2
M2
x
D2
A1
B1
C2
A1
A2
b1
b2
c1
c2
xx
x
B1
A1
A2
B2
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Chng IV
MT PHNGI. HNH CHIU CA MT PHNGHnh chiu ca mt phng l hnh chiu ca cc yu txc nh mt phng y,
v thhnh chiu ca mt phng thng cho bi:
+ Ba im khng thng hng. ( H 4.1 a )
+ Mt ng thng v im khng thuc ng thng ( H 4.1 b )
+ Hai ng thng song song ( H 4.1 c )
+ Hai ng thng ct nhau ( H 4.1 d )
a) b)
b) d)
Hnh 4.1
x
A1
B1
C1
A2
C2
B2
A2
A1
x
b2
b1
b2
b1
x
a1
x
b2
b1
a2
a1
a2
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II VT CA MT PHNG
L giao tuyn ca mt phng vi mt phng hnh chiu.( H 4.2 )
2.1. Vt ng
Vt ng ca mt phng l giaotuyn ca mt phng vi mt phnghnh chiu ng P1. l vt ( m)
2.2. Vt bng
Vt bng ca mt phng l giaotuyn ca mt phng vi mt phnghnh chiu bng P2. l vt ( n)
2.3. Vt cnhVt cnh ca mt phng l giao
tuyn ca mt phng vi mt phnghnh chiu cnh P3. l vt ( p)
Nhvy ta c hnh biu din cacc vt nhsau:
+ Vt ng v vt bng ct nhau ( H 4.3 a )
+ Vt ng v vt bng khng ct nhau ( H 4.3 b )
a) b)
Hnh 4.3
x
p
o
y
P1
m
n
z
P3
Hnh 4.2
x
p
y
m
n
z
y
x
p
y
m
n
z
y
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III. NHNG MT PHNGC BIT
3.1: Mt phng chiu.
Mt phng chiu l mt phng vung gc vi mt phng hnh chiu.
1. Mt phng chiu ng.
+ Mt phng chiu ng l mt phng vung gc vi mt phng hnhchiu ng P1Hnh chiu ng suy bin thnh mt ng thng. ( H 4.4 )
Hnh 4.4
+ Nu mt phng chiu ng c biu din bng hai vt ( m, n). Th vtbng nvung gc vi trc x. ( H 4.5 )
Hnh 4.5
+ Gc ( mt phng , mt phng P2) = Gc ( vt ng m, trc x )
2. Mt phng chiu bng.
+ Mt phng chiu bng l mt phng vung gc vi mt phng hnh chiubng P1Hnh chiu bng suy bin thnh mt ng thng. ( H 4.6 )
+ Nu mt phng chiu bng c biu din bng hai vt ( m, n). Th vtng mvung gc vi trc x. ( H 4.7 )
C
C2
C1
P2
A
A1P1
A2
x
B2
B
B1
C2
C1
A1
A2
x
B2
B1
n
m
x
P2
P1
n
m
x
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Hnh 4.6
Hnh 4.7
+ Gc ( mt phng , mt phng P1) = Gc ( vt bng n, trc x )
3. Mt phng chiu cnh
+Mt phng chiu cnh l mt phng vung gc vi mt phng hnh chiu cnhP3Hnh chiu cnh suy bin thnh ng thng. ( H 4.8 )
Hnh 4.8
C
C2
C1
P2
A
A1P1
A2
x
B2
B
B1
C2
C1
A1
A2
x
B2
B1
n
m
x
P2
P1
n
m
x
n
y
p
x
z
y
m
o
n
P1
p
x
z
y
m
o
P3
P2
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+ Gc ( mt phng , mt phng P1) = Gc ( vt cnh p, trc z )
+ Gc ( mt phng , mt phng P2) = Gc ( vt cnh p, trc x )
+ Nu mt phng chiu cnh c biu din bng hai vt ( m, n). Th vtbng n. vt ng mu song song vi trc x. ( H 4.8 )
3.2 Mt phng ng mc
Mt phng ng mc l mt phng song song vi mt phng hnh chiu. Tngng vi ba mt phng hnh chiu, c ba loi mt phng ng mc, chng c nhngha v tnh cht tng tnhau.
1. Mt phng bng
+ Mt phng bng l mt phng song song vi mt phng hnh chiu bng P2.Mt phng ( ) ( H 4.9 )
Hnh 4.9
+ Hnh chiu ng suy bin thnh ng thng song song vi trc x.
( A1B1C1m// x )
+ Hnh chiu bng ca mt hnh phng trn mt phng bng l bng hnh tht
( ABC = A2B2C2)
2. Mt phng mt
+ Mt phng bng l mt phng song song vi mt phng hnh chiu ng P1.Mt phng ( ) ( H 4.10 )
+ Hnh chiu bng suy bin thnh ng thng song song vi trc x.
( A2B2C2n// x )
C
C2
C1
P2
A
A1
P1
A2x
B2
BB1
m
C1
B2
C2
A1 mB1
A2
x
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+ Hnh chiu ng ca mt hnh phng trn mt phng mt l bng hnh tht
( ABC = A1B1C1)
Hnh 4.10
1.
Mt phng cnh
+ Mt phng cnh l mt phng song song vi mt phng hnh chiu cnh P3,mt phng ( ) ( H 4.10 )
Hnh 4.11
+ Vt ng v vt bng nm trn ng thng vung gc vi trc x
+ Hnh chiu cnh ca hnh phng nm trn mt phng cnh l bng hnh tht
( ABC = A3B3C3)
C
C2
C1
P2
A
A1
P1
A2x
B2
B
n
B1
C2
C1
A1
A2
x
n
B1
B2
C
C2
C3
A
A3
P1
A2
x
Bm
B3
P3
P2
B2n
z
B1
C1
C2
C3
A3
A2
x
m
B3
B2
n
A1z
B1
C1
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IV. SLIN THUC CAIM VNG THNG VI MT PHNG
Ta c hai mnh lm cscho stng quan lin thuc gia im v ngthng vi mt phng:
Mnh 1:
ng thng d thuc mt phng P nu d c hai im thuc mt phng P.
Mnh 2:
im A thuc mt phng P, nu A thuc ng thng no ca P.
T ta thy rng vic biu din slin thuc ca im vi mt phng hay cang thng vi mt phng a vvic biu din slin thuc ca mt im vi
ng thng m ta nghin cu trn.
l sng tta xt cc v dsau:
V d1 : Cho mt phng ABC, vmt ng thng bt k thuc mt phng
Gii : Ly hai im bt k thuc mt phng, chng hn im M thuc AB,
im N thuc BC, ng thng b m hnh chiu ng b1M1N1v hnh chiu bng
b2M2N2l ng thng phi v( H 4.12 )
Hnh 4.12 Hnh 4.13
V d2: Cho mt phng xc nh bi hai ng thng ct nhau p v q. Vmtng bng bt kthuc mt phng y? ( H 4.13 )
Gii : ng bng ca mt phng l ng thng thuc mt phng v songsong vi mt phng hnh chiu bng P2, gisb l ng bng phi v, Th hnh
b2
M1
x
N1
M2
p2
p1
b1
q1
q2
b2
C2
N2
M1
x
N1
M2
B2
b1
B1
A2
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chiu ng b1ca n phi song song vi trc x. Gi M, N l giao im ca b vi p
v q, khi ta c M1l giao ca b1vi p1, N1l giao ca b1vi q1, tM1M2
thuc p2, tN1N2thuc q2. ng thng b2i qua M2v N2l hnh chiu bngca ng thng phi v.
V d3: Cho mt phng ABC. Qua A vmt ng mt thuc mt phng ?
Gii: ng mt ca mtphng ABC l ng thng ca mtphng y v song song vi mtphng hnh chiu ng P1. Gi sgi m l ng mt phi vhnhchiu bng m2ca n phi i qua A2v song song vi trc x. Gi E lgiao im ca m vi BC, ta tmc E2 l giao ca m2v B2C2 tE2 E1 thuc B1C1 ng thngm1 i qua A1 v E1 l hnh chiung ca ng mt phi v.( H 4.14 )
CU HI V BI TP1. Trnh by cch biu din mt mt phng ?
2. Th no gi l mt phng chiu, c my loi mt phng chiu, nu nhngha v cc tnh cht ca chng?
3. Thno gi l mt phng ng mc, c my loi mt phng ng mc, nunh ngha v cc tnh cht ca chng?
4. nh ngha vt ca mt phng?
5. Cho mt phng xc nh bi 3 im A, B, C v im D thuc mt phng .Bit hnh chiu ng ca im D l D1tm D2? ( H 4. 15 )
6. Cho im A ( A1, A2) v ng thng a ( a1, a2). ( H 4.16 )
+ Qua im A dng mt phng chiu ng ( ) nghing vi mt phng
hnh chiu bng mt gc 30?
+ Qua ng thng a dng mt phng chiu bng ( )?
m2
C2
E1
x
C1
E2
m1
Hnh 4.14
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Hnh 4.15 Hnh 4.16
7. Vhnh chiu cn thiu ca on thng AB bit AB song song Vi mtphng chiu ng ( ) ? ( H 4.17 )
8. Bit im A thuc mt phng ( ), bit hnh chiu bng A2tm A1?
( H 4.18 )
Hnh 4.17 Hnh 4.18
C2
E1
x
C1
B2
B1
a2
x
a1
A1
x
m
n
x
m
n
A1
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Chng VNHNG BI TON VVTR
Trong chng ny ta nghin cu cc bi ton vvtr gia cc yu thnh hccbn. Giao im ca ng thng vi mt phng, giao tuyn ca hai mt phng .
I. GIAOIM CA MT PHNG CHIU VING THNG
1.1: Giao im ca ng thng vi mt phng chiu
Bi ton 1: Cho mt phng chiu bng ABC v mt ng thng d. Xc nhgiao im ca ng thng d vi mt phng ABC ( H 5.1 )
Gii: gi I l giao im ca ng thng d v mt phng ABC. V ABC l mtphng chiu bng nn mi im thuc ABC u c hnh chiu bng thuc A2B2C2
I2thuc A2B2C2ng thi I2phi thuc d2I2l giao ca A2B2C2v d2, tI2
I1thuc d1. ( H 5.1 )
Hnh 5.1 Hnh 5.2
Bi ton 2: Cho mt phng chiu ng ABC v mt ng thng d. Xc nhgiao im ca ng thng d vi mt phng ABC ( H 5.1 )
Gii: gi I l giao im ca ng thng d v mt phng ABC. V ABC l mtphng chiu ng nn mi im thuc ABC u c hnh chiu ng thuc A1B1C1
I1thuc A1B1C1ng thi I1phi thuc d1I1l giao ca A1B1C1v d1, tI1
I2thuc d2. ( H 5.2 )
C2
I1
xC1
B2
A1B1
A2
d1
d2
I2
C2
d1
xC1
B2
A1
B1
A2
d2
I1
I2
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II. GIAO TUYN CA MT PHNG CHIU VI MT PHNG
Bi ton :Cho mt phng R xc nh bi hai ng thng l v d ct nhau ti
im O v mt phng chiu ng ABC. Hy vgiao tuyn ca hai mt phng y?Gii : Nh ta bit
giao tuyn ca hai mt phngl mt ng thng. Munxc nh ng thng ny tai xc nh hai im chungca hai mt phng cho. y hai im chung y l hai
giao im N, M ca ccng thng d, l vi mt
phng ABC m cch tm ta
bit trn NM l giao
tuyn cn tm. ( H 5.3 )
III. GIAO IM CA
NG THNG VI MTPHNG
trn khi tm giao im ca ng thng vi mt phng chiu v giao tuynca mt phng vi mt phng chiu, ta thy mt hnh chiu ca giao im hay giaotuyn bit c ngay. Trong trnghp ng thng v mt phng iul mt phng thng. Giscn xcnh giao im I ca ng thng dv mt phng P, ta lm nhsau:
+ Qua d dng mt phng chiuR.
+ Xc nh giao tuyn MN caP v R.
+ Xc nh giao im I ca d
v MN I chnh l giao im cn
tm.( H 5.4 )
C2
N1
x C1
B2
A1
B1
A2
d1
d2N2
M1
l1
l2M2
O1
O2
Hnh 5.3
N
P
R
d
M
I
Oa
b
Hnh 5.4
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Bi ton: Cho ngthng d v mt phng Pxc nh bi hai ngthng ct nhau a v b tiim O. Xc nh giaoim I ca d v P?
Gii : Qua d dngmt phng chiu ng R
D1 R1, ddng thyrng mt phng R ct b ti
N, ct a ti M N1, M1ln lt l giao ca d1 vi
b1 v a1 khi ta tm
c N2, M2 thuc b2, a2
M2N2 ct d2 ti I2I1 . ( H 5.5 )IV. GIAO TUYN CA HAI MT PHNGxc nh giao tuyn ca hai mt phng, thng chcn xc nh hai im
chung thuc mt phng . Ta c mt im chung nhthbng cch tm giao im
ca mt ng thng bt kca mt phng ny vi mt phng kia. Do vy vic tmgiao tuyn ca hai mt phng c tha vvic gii lin tip hai ln bi ton xcnh giao im ca ng thng vi mt phng m ta va nghin cu trn.
Di y ta trnh by mt phng php khc tm im chung ca hai mtphng .
Cho mt phng P v Q,trong mt phng P c xcnh bi hai ng thng song
song e v g, mt phng Q cxc nh bi hai ng thngct nhau h v f. ( H 5.6 )
c mt im chung cahai mt phng P v Q ta ct Pv Q bng mt mt phng R.
+ R ct P theo giao tuynAB ( Trong A l giao ca ng thng e v mt phng R, B l giao ca ng
thng g vi mt phng R )
I2
b2
d2
d1 R1
I1
x
M2
N2
N1M1
O1
O2a2
a1b1
Hnh 5.5
T
Uk
N
M
S
ICB
D
ARQ
Pe
g h
f
Hnh 5.6
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+ R ct Q theo giao tuyn CD ( Trong C l giao ca ng thng h v mtphng R, D l giao ca ng thng f vi mt R )
Hai ng thng AB v CD cng thuc mt phng R nn phi ct nhau ti mt
im I, tt nhin I l im chung ca P v Q. c c im chung thhai K talm tng t. ( H 5.6 ) Ct P v Q bng mt phng thhai S.
+ S ct P theo giao tuyn MN ( Trong M l giao ca ng thng e v mtphng R, N l giao ca ng thng g vi mt phng R )
+ S ct Q theo giao tuyn UT ( Trong U l giao ca ng thng f v mtphng R, T l giao ca ng thng h vi mt R )
IK l giao tuyn cn tm gia mt phng P v Q.
Nhng mt phng phtrR v S l nhng mt phng bt k. Tuy nhin ta phichn chng th no cthvc giao tuyn cachng vi mt phng P vQ, thng thng ta dngcc mt phng ph tr lcc mt phng chiu mvic v giao tuyn ca
chng vi mt phngthng ta bit.
Trn ( H 5.7 ) trnh
by cch vgiao tuyn IKtrn thc ca hai mt
phng P v Q ni trn, y ta chn mt ph trR l mt phng chiu ng mhnh chiu ng l ng thng R1, v S l mt phng hnh chiu bng m hnh
chiu bng l ng thng S2.
V. QUY C VTHY KHUT TRNTHC
Nh chung ta bit khi quan st cc hnh thc t, chng hn khi nh hai tmphng ct nhau, mt tm mu xanh v mt tm mu , ta chnhn thy mt phn
ca tm mu xanh v mt phn tm mu , cn nhng phn cn li th bkhut. hnh biu din cng gy cho ngi xem n tng ging nhn tng c c khiquan xt thc t v mt hnh dng, trn thc ngi ta cng biu din s thykhut theo nhng quy c sau.
e2g2
I2
f2 h2
C2
D2
B2A2
K2
K1
x
R1
B1D1
C1f1
h1 I1g1
e1
A1
S2
Hnh 5.7
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5.1 Xt thy khut trn hnh chiu ng
Ngi quan xt ng pha trc mt phng hnh chiu ng. Khi xt thy
khut ngi quan xt xem nh ng xa v tn trn hng thng gc vi mtphng y.
Hnh biu din u l vt thc, mt phng hnh chiu cng l mt phng c
Vy im no nm pha sau mt phng hnh chiu ng l im khut. Haiim cng tia chiu ng im no c xa ln hn sl im nh thy trn hnhchiu ng. im A l im nhn thy trn hnh chiu ng ( H 5.8 )
Hnh 5.8 Hnh 5.9
5.2 Xt thy khut trn hnh chiu bng
Ngi quan xt ng pha trn mt phng hnh chiu bng. Khi xt thykhut ngi quan xt xem nh ng xa v tn trn hng thng gc vi mt
phng y.
Hnh biu din u l vt thc, mt phng hnh chiu cng l mt phng c
Vy im no nm pha di mt phng hnh chiu bng l im khut. Hai
im cng tia chiu bng im no c cao ln hn sl im nh thy trn hnhchiu bng. im C l im nhn thy trn hnh chiu bng ( H 5.9 )
V d: Cho ng thng a v tm phng tam gic CDE. Vgiao im ca onthng vi tm phng v chphn thy, phn khut ca on thng
Vic xc nh giao im ca on thng a vi mt phng CDE ta bit trnhnh 5.10 giao im I c xc mh bng cch dng mt phng ph tr l mt
phng chiu ng cha ng thng a. xt thy, khut ca on thng ta ch rng im I l im thy trn cc hnh chiu.
x
B2
A1 B1
A2 x
C1
D1
C2D2
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Xt thy, khut trn hnhchiu ng:
Gi M, N l hai imcng tia chiu ng, y Mthuc ng thng a, N thucDE. V xa ca im M nhhn xa ca im N nn Ml im khut do phnon IM l phn khut, phncn li ca on
Xt thy, khut trn hnhchiu bng:
Gi T, S l hai imcng tia chiu bng, y Tthuc ng thng a, S thucCD. V cao ca imSnhhn cao ca im T nn S l im khut do phn on IT l phn khut,
phn cn li ca on thng l thy.
CU HI V BI TP
1. Trnh by cc bc gii bi ton xc nh giao im ca ng thng vimt phng ?
2. Trnh by cc bc gii bi tonxc nh giao tuyn ca mt phng vi mt
phng ?
3. Tm giao im ca ng thngchiu bng a v mt phng P xc nh bihai ng thng ct nhau d v b. ( H 5.11 )
4. Vgiao tuyn ca mt phng ABCvi mt phng chiu bng Q cho bng haivt mv n. ( H5.12 )
5. Tm giao im ca ng thng d
vi mt phng Q cho bi hai ng thng song song b v f. ( H 5.13 )
a2
I2
C2
D2
M2
E2
N2
T1
x
a1
D1
C1
I1S1
E1
M1N1
S2T2
Hnh 5.10
d2
a2
x
d1
a1
b2
b1
Hnh 5.11
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6. Tm giao im ca ng thng d vi mt phng Q cho bi ng thng bv im A.
7. Tm giao im ca ng thng d v mt phng Q cho bi im A v trc x.
A2
C2
n
m
A1
B1
B2
C1
x
Hnh 5.12
d2
x
d1
f1
b1
f2b2
Hnh 5.13
b2
d2A2
x
b1
d1
A1
Hnh 5.14
d2A2
x
d1
A1
Hnh 5.15
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Chng VI
CC PHP BIN I HNH CHIUVic tm li gii skh khng nu Hnh ca bi ton cho vtr bt k. Khi
y ngi ta dng cc php bin i hnh chiu, a Hnh cho vvtr cbit, khi vic tm li gii c ddng hn. Sau ngi ta thng bin ingc li a kt qu thnh bin i vhnh ban u. Di y xin giithiu phng php thay mt phng hnh chiu, v php di hnh song song vi ccmt phng hnh chiu.
I.PHNG PHP THAY MT PHNG HNH CHIU1.1. Thay mt trong hai mt phng hnh chiu
Phng php chung : Thay mt mt phng hnh chiu l trong h hai mtphng hnh chiu cho, ta gii nguyn mt mt phng v thay mt phng kia bngmt mt phng khc, vn vung gc vi mt phng hnh chiu cn li, hng chiulc ny thay i v vung gc vi mt phng hnh chiu mi. Vn quay mt phnghnh chiu quanh trc x vtrng vi mt phng hnh chiu kia trong h.
1.1.1.
Thay mt phng hnh chiu bngGista c hthng mt phng hnh chiu P1, P2( H 6.1 a ). Thay mt phng
hnh chiu bng P2, tc l ly mt mt phng P2 vung gc vi P1, lm mt phnghnh chiu bng mi v ly hng chiu vung gc vi P2 lm hng chiu mi.
Sau xoay P2 vtrng vi mt phng P1.
Ax
A'xA'x
Ax
A2
A2
X
x
A1
x'
P'2
A'2
A
A2
A1
P1
P2
x
Hnh 6.1
a ) b )
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Ta nhn thy :
+ Hnh chiu ng A1ca A khng thay i
+ xa ca im A trong h thng cbng xa ca im A trong hthng mi, tc l: A2Ax = A2Ax = AA1.
Da trn nhng nhn xt , ta thy vic thay hnh chiu bng c tin hnhtrc tip nhsau:
Gistrong hthng cim A c cc hnh chiu A1v A2.
+ Vtrc hnh chiu x, dnhin mi vtr xc nh ca P2 sc mt vtr catrc x tng ng. Vic ta chn n nhthno l ty theo yu cu ca tng bi ton
cth.
+ Vhnh chiu bng mi A2 ca A, y A1A2 phi vung gc vi trc xv A2Ax = A2Ax. ( H 6.1 b )
V d : S dng phngphp thay mt phng hnh chiubng, a on thng AB vng bng trong h thng mt
phng hnh chiu mi. ( H 6.2 )Gii: Nh ta bit iu
kin t c v AB l ngbng th A1B1phi song song vitrc x. Do gii bi tonny, ta chn trc x song songvi A1B1. Hnh chiu bng mica AB l A2B2 ( A2Ax =A2Ax, B2Bx = B2Bx ) V trong
h thng mt phng mi AB lng bng nn ta c di ca on thng AB chnh bng A2B2.
1.1.2
Thay mt phng hnh chiu ng
Tng tnh trn, gi s ta c h thng mt phng hnh chiu P1, P2. Thaymt phng hnh chiu ng P1, tc l ly mt mt phng P1 vung gc vi P2, lm
A2
A1
B1
B2
x
Hnh 6.2
Bxx
Ax
Bx
Ax
B2
A2
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mt phng hnh chiu ng mi v ly hng chiu vung gc vi P1 lm hngchiu mi. Sau xoay P1 vtrng vi mt phng P2. ( H 6.3a )
Ta nhn thy :
+ Hnh chiu bng A2ca A khng thay i
+ cao ca im A trong hthng cbng cao ca im A trong hthng mi, tc l: A1Ax = A1Ax = AA2.
Da trn nhng nhn xt , ta thy vic thay hnh chiu ng c tin hnhtrc tip nhsau:
Gistrong hthng cim A c cc hnh chiu A1v A2.
+ Vtrc hnh chiu x, dnhin mi vtr xc nh ca P1 sc mt vtr catrc x tng ng. Vic ta chn n nhthno l ty theo yu cu ca tng bi toncth.
+ Vhnh chiu ng mi A1 ca A,
y A1A2phi vung gc vi trc x vA1Ax = A1Ax. ( H 6.3 b )
V d : Thay mt phng hnh chiung ng bng AB trthnh ngthng chiu ng trong h thng mt
phng hnh chiu mi. ( H 6.4 )
Gii :ng thng AB tr thnhng thng chiu ng ta phi chn trc
x'
A'1P'1 A'1
A'x x
P2
P1
A1
A2x'
A1
x
A2
AxA'x
Ax
A
Hnh 6.3
a ) b )
A2
A1 B1
B2
x
Hnh 6.4
BxAx
AxBx
A1 B1
x
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A1
B1
A2
B2
x
A2
B2
A1
B1
Hnh 6.7
song song vi A2B2 ( Php thay mt phng hnh chiu ng ). Sau thay mtphng hnh chiu bng AB l ng thng chiu bng, t x vung gc viA2B2. ( H 6. 6 )
II. PHP DI HNH
nh ngha :Php di hnh l php di hnh n mt vtr mi sao cho khongcch ca hai im bt kA v B bng khong cch ca hai im tng ng A, B.
Tnh cht :+ Php di hnh l php bin i mt i mt.
+ Php di hnh bo tn tnh lin thuc ca im v ng thng.
2.1. Php di hnh song song mt phng hnh chiu bng
nh ngha : Php di hnh song song mt phng hnh chiu bng l php dihnh m cc ng thng ni cc im tng ng u song song vi mt phng hnhchiu bng.
Tnh cht : + Hnh chiu ngca cc on thng ni cc cp imtng ng AA, BB......u song songvi trc x ( A1A1// B1B1// ..... x )
Thc vy, trong php di hnhsong song vi mt phng hnh chiu
bng nhng on thng AA, BB ...
u l nhng ng bng.
+ Hnh chiu bng ca hai hnh tng ng trong php di hnh lbng nhau. ( A2B2= A2B2 ). ( H 6.7 )
Do php di hnh song song vi mt phng hnh chiu bng P2c thc
hin trn thc nhsau:
+ Vhnh chiu bng 2 = 2 ( V tr 2 c xc nh theo yu cu ca
tng trng hp )
+ V1. Mi im A1 ca 1 c xc nh bng giao im ca hai ngging. ng ging qua A1 v song song vi trc x v ng thng qua A2 vvung gc vi trc x. ( H 6.7 )
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V d:Xc nh di ca on thng AB. ( H 6.8 )
Gii : xc nh di ca
on thng AB ta di AB n ABsong song vi mt phng hnh chiung P1( AB l ng mt ) vy tadi A2B2 n A2B2 song song vitrc x ( A2B2= A2B2 ). Theo A1B1v A2B2, xc nh A1B1. dica A1B1 chnh l di ca AB.
2.2. Php di hnh song song
mt phng hnh chiu ng
nh ngha : Php di hnh song song mt phng hnh chiu ng l php dihnh m cc ng thng ni cc im tng ng u song song vi mt phng hnhchiu ng. ( H 6.9 )
Tnh cht : + Hnh chiu bng cacc on thng ni cc cp im tngng AA, BB......u song song vi trc x
( A2A2// B2B2// ..... x )Thc vy, trong php di hnh song
song vi mt phng hnh chiu ngnhng on thng AA, BB ... u lnhng ng mt.
+ Hnh chiu ng cahai hnh tng ng trong php di hnh l
bng nhau. ( A1B
1= A
1B
1 ). ( H 6.9 )
Do php di hnh song song vi mt phng hnh chiu ng P1 c thc hintrn thc nhsau:
+ Vhnh chiu ng 1 = 1 ( Vtr 1 c xc nh theo yu cu ca
tng trng hp )
+ V2. Mi im A2 ca 2 c xc nh bng giao im ca hai ngging. ng ging qua A2 v song song vi trc x v ng thng qua A1 v
vung gc vi trc x. ( H 6.9 )
A1
B1
A2
B2
x
A2 B2
A1
B1
Hnh 6.8
A1
B1
A2
B2
x
A2
B2
A1
B1
Hnh 6.9
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V d:Xc nh hnh dng tht ca tam gic ABC. ( H 6.10 )
Gii :T hnh v ta thy A1B1C1 nm trn mt ng thng Mt phng
ABC l mt phng chiu ng. V vy php di hnh song song vi mt phng hnhchiu ng c tha mt phng ABC vmt mt phng bng. Khi hnh dngtht ca ABC chnh l A2B2C2. Mun vy di A1B1C1 vsong song vi trc x.( H 6.10 )
2.3. Thc hin lin tip cc php di hnh song song vi mt phng hnh
chiu
Cng nhtrong php thay mt phng hnh chiu, vi nhng bi ton phc tpta khng thchdng mt php di hnh song song vi mt phng hnh chiu m
phi dng lin tip hai hoc nhiu hn na cc php di y.
V d: Xc nh hnh dng tht ca tam gic ABC ( H 6.11 ).
Gii : xc nh c dng tht ca tam gic ABC ta a chng v mtphng bng. V vy lm c vic ny trc tin ta phi a ABC vmt phngchiu ng, mun vy ta phi di A2B2C2 n A2B2C2 sao cho v tr nyng bng AE l ng thng chiu ng. ( A2E2 vung gc vi trc x ) sau
A1
B1
A2
x
A2
B2
A1 B1
Hnh 6.10
C1
C2
B2
C2
C1
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tng tnhv dtrn ta a ABC vmt phng bng. Khi hnh dng tht ca
ABC chnh l A2B2C2. ( H 6.11 ). Lu tng tnh trn ta c thgii biton bng vic a mt phng ABC vmt phng mt.
CU HI V BI TP
1.nh ngha php thay mt phng hnh chiu bnmt phng, php thay mtphng hnh chiu ng, v nu cc tnh cht ?
2.nh ngha php di hnh song song mt phng hnh chiu bng, php dihnh song song mt phng hnh chiu ng, v nu cc tnh cht ?
3.Xcnh gc nghing ca mt phng P ( ABC )i vi mt phng hnh chiubng. ( H 6.12 )
4.Xc nh trng tm ca tam gic ABC. ( H 6.13 )
A1
C1
A2
x
E1
E2
B2
B1
A1E1
A2
A2
C1
C2
B2
C1 B1
C2
B2
A1
C2
B2
Hnh 6.11
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A1
B1
A2
x
C1
C2
B2
Hnh 6.12
B1
x
C2
B2
A2
A1
C1
Hnh 6.13
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Vng thng chiu ng I J ct SB I vct AC J. Theo hnh vtheo hnh vta thy rngim I c xa ln hn xa ca im J, do nhn tngoi vo th ng thng chiu ng I Js ct SB trc khi ct AC. Vy SB s l ngthng thy v AC l khut.
Tng t bng cch v ng thng chiubng GH ct SA v BC. Ta sxt c trong haicnh SA v BC cch no thy cnh no khut trnhnh chiu bng. Theo hnh v ta thy cnh SA l
thy cnh BC l khut trn hnh chiu bng.Vy trn mi hnh chiu th ng bao quanh
hnh chiu bao gi cng nhn thy c, vbngnt lin m. Cnh no c hnh chiu bn trongng bao quanh hnh chiu th phi xt thy khut cnh thy vbng nt lin m,cnh khut vbng nt t.
1.3 Biu din im thuc a din
+ vim thuc cnh a dinta p dng bi ton v im thucng thng. Nhim H ( H 7.3 )
+ vim thuc mt a din tap dng bi ton xc nh im thucmt phng. Nhim I ( H 7.3 )
Ch :
* Vim thuc mt bn ca mthnh chp th im thng c gnvo ng thng i qua im v
nh chp. im I ( H 7.3 )
* Vim thuc mt bn ca mthnh lng trth ta thng p dng gn
im vo ng thng song songvi cnh bn ca lng tr.
J1
G1
A1
B1
C1
S1
H1
I1
J1A1
C1
G1
I1
B1
H1
S1
Hnh 7.2
K2
K1
D2
D1
I2
I1
H2
H1
A1B1
C1
C2A2
B2
S2
S1
Hnh 7.3
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* Trn hnh chiu ang xt ca a din mt im thuc mt thy ca a din thim thy, mt im thuc mt khut th im khut.
1.4. Giao ca mt phng vi a dinGiao ca mt phng vi a din l mt
a gic. nh ca a gic ny l giao imca mt cnh a din vi mt phng ct.cnh ca a gic l giao tuyn ca mt mta din vi mt phng ct.
V d : Tm giao ca mt phng chiu
ng R vi khi a din SABC. ( H 7.4 )
Gii :Ta thy mt phng R giao vi adin SABC ti 4 mt phng l 3 mt bnSAC, SBC, SAB v mt y ABC.Do giao tuyn l a gic c bn cnh, cc cnh l giao ca cc mt phng trn vi mt
phng R. Cc nh ca a gic l giao cacc cnh SC, SB, AB, v AC vi mt phng
R. Nu ta gi cc nh ca a gic ln ltl I, N, H, K . Khi bi ton ch nn tm giao im ca ng thng vi mt
phng chiu bit trn.
1.5. Giao ca ng thng vi a din
Giao ca ng thng vi a din, thc cht l tp hp cc giao im ca
ng thng vi cc mt phng to thnh a din, m giao ca ng thng vi mtphng ta a bit cch gii trn.
Mun tm giao ca mt ng thng vi mt a din, ngi ta thng dng phngphp mt phng phtr. Ni dung ca phng php nhsau:
+ Qua ng thng cho dng mt phng chiu gi l mt phng phtr.
+ Tm giao ca mt phng phtrvi a din cho. Giao ny gi l giao ph.
+ Tm tp hp cc giao im ca ng thng cho vi giao ph. Tp hpcc im l giao phi tm.
R1
N2
N1
B2
C2
A2
H2
I2K2
I1
H1K1 C1B1A1
S1
S2
Hnh 7.4
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V d:Xc nh giao ca ng thng d vi khi chp SABC.
Gii : Dng mt phng ph tr lmt phng chiu ng R cha ngthng d. Sau tm cc giao tuyn phgia mt phng ph tr R vi cc mt
phng hnh chp SAB l GE, SBC l HE.Sau tm cc giao im ca cc giao
tuyn phvi ng thng d l I,J chnh l giao im ca ng thng d vi khichp SABC. ( H 7.5 ) v cch vc thhin trn thc ( H 7.6 )
II. MT CONG
2.1. Mt trtrn xoay
2.1.1. Hnh biu din.
hnh biu din ca mt trn ginngi ta thng t mt y ca mt trsongsong vi mt phng hnh chiu bng P2. Khi hnh chiu bng s l ng trn, hnhchiu ng l hnh chnht. (H 7.7 )
tm im thuc mt tr ta gn imvo ng sinh thuc mt tr. Khi hnhchiu bng ca im thuc ng trn hnhchiu bng. im A ( H 7.7 )
R1 d1
J1
B2
C2
A2
H2
I2
J2
I1 H1
C1B1
A1
S1
S2
G1
E1
E2
G2
Hnh 7.6
d2
R
G
H
E IJ C
B
A
S
d
Hnh 7.5
A1
A2
Hnh 7.7
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2.1.2. Giao ca ng thng vi mt tr
Nhta bit mt ng thng ct mt trti hai
im, tm hnh chiu ca hai giao im ny ta tmc ngy hnh chiu bng ca chng chnh l giao cang thng hnh chiu bng v ng trn hnh chiu
bng ca mt tr, trn ( H 7.8 ) tm giao im cang thng d vi mt tr ta tm c ngay hai hnhchiu bng ca giao im l A2, B2.Thnh chiu bngny ta tm c hnh chiu ng A1, B1.A, B chnh lgiao im ca ng thng d vi mt tr.
2.1.3. Giao ca mt phng vi mt tr
Khi mt phng giao vi mt tr, tu theo v trtng i gia mt phng v mt trm ta c cc dnggiao tuyn khc nhau:
+ Khi mt phng song song vi trc ca mt trtac giao tuyn l hnh chnht.
+ Khi mt phng vung gc vi trc ca trta c giao tuyn l ng trn.
+ Khi mt phng nghing vi trc ca trv ct tt c cc ng sinh ta cgiao tuyn l ng elp.
2.2 Mt nn trn xoay
2.2.1. Hnh biu din
hnh biu din ca mt nn ngin ngi ta thng t mt y ca mt
nn song song vi mt phng hnh chiubng P2. Khi hnh chiu bng s l
ng trn, hnh chiu ng lmttam giccn. (H 7.9)
vim thuc mt nn ta p dngbi ton gn im vo ng sinh ca nnin A ( H 7.9 ). Hoc gn vo ng trnsong song vi mt y. im B ( H 7.9 )
A2
A1
B2
B1
Hnh 7.9
A2
A1
B2
d1
B1
Hnh 7.8
d2
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2.2.2. Giao ca ng thng vi mt nn.
Trng hp ng thng l ng thng chiu th ta d dng tm c mt
hnh chiu ca giao im vic tm hnh chiu cn ta bit cch tm, ( H 7.10 )
Trng hp ng thng l ng thng thng ta phi s dng mt phngphtrl mt phng cha ng thng v i qua nh ca nn, ( H 7.11 )
Trn ( H 7.11 ) chn mt phng phtr l mt phng cha ng thng d v
nh S. Sau ta tm giao tuyn NT ca mt phng ph trvi mt phng y
S2I2, S2J2, chnh l giao tuyn phvi nn ty ta ddng tim c A2, B2 A1, B1.
2.2.3. Giao ca mt phng vi mt nn
Khi mt phng giao vi mt nn, tutheo v tr tng i gia mt phng vmt nn m ta c cc dng giao tuyn khc nhau:
+ Khi mt phng vung gc vi trc ca nn giao tuyn l ng trn.
+ Khi mt phng nghing vi trc ca nn v ct tt ccc ng sinh giaotuyn l ng elp.
d1 B1 A1
A2
B2
d2
Hnh 7.10
A2
d1
B2
Hnh 7.11
d2
A1
B1
I2J1 T2
N2
N1 T1
S2
S1
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+ Khi mt phng cha nh nn giao tuyn tam gic cn.
+
+
2.3 Mt cu
2.3.1 Hnh biu din
Hnh biu din ca mt cu lun lun l cc
ng trn c cng bn knh vi mt cu.
tm hnh chiu ca im thuc mt cu tagn im vo cc ng trn trn mt cu thuccc mt phng song song vi mt trong cc mt
phng hnh chiu. im A trn ( H 7.12 ) l ta gnvo ng trn song song vi mt phng hnhchiu bng.
2.3.2. Giao ca ng thng vi mt cu
Nu ng thng l ng thng chiu th vic tm l ddng ( H 7.13 )
Nu ng thng l ng thng thng, th ta khng thchdng mt phngph tr tm giao im c, m ta phi p dung phng php ny vi phngphp thay mt phng hnh chiu. ( H 7.14 )
A2
A1