vektor.ppt

X

1.GEOMETRI IN SPACE

Y

Z

VECTORS

233

222

211

321321

321321

)()()(

),,(),,(

:tan.1

),,(

QPQPQPPQ

QQQQandPPPP

formulacedisThe

PPPkPjPiPPPVector

3327)5()1()1(

)14()45()34(

)()()(

)4,5,4()1,4,3(

:1

222

222

2233

222

211

PQPQPQPQ

QP

Example

),,(

)()()(

)0,0,0(

:

222

2222

cbaiscentralthe

czbyax

iscentralthe

rzyx

equationtheandSphere

(0,0,0)

(a,b.c)

0

:tan

222 DCzByAxzyx

shpereofequationdartS

5)1,2,4(

25)1()2()4(

461416)1()2()4(

46...)2(....)4(...)8(

046248:

:2

222

222

222

222

radiusandcenterThe

zyx

zyx

zzyyxx

Solution

zyxzyxequationwith

sphereofradiusandcentertheFind

Example

4,-2,-1

x

y

z

DCzByAx

planeofequationThe

:

NormalplaneN(A,B,C)

trace

trace

trace

30,0

40,0

60,0

12432

:3

zyx

yzx

xzy

zyxofgraphthesketch

Example

6

3

4

trac

e

trace

trace

x

z

z

axiszthetoparalelplaneethso

axiszcrossneverplaneTheSolution

yxequationlinierthe

ofgraphthesketch

Example

:

63:

:4

1052:

.3

02522

:

.2

)1,2,3()2,1,1()1,0,3(int.1

:

222

yxequationliniertheof

graphthesketch

zyxzyx

equationwith

sphereofradiusandcentertheFind

CBApotheplot

Exercise

4

127

)2/1,1,1(

0254

111)2/1()1(1)(x

2..

222

r

isshereofCentre

zy

50

20

3

xy

yx

nojwb

2

5

originThec

axisyb

planexya

thetofromcedisFind

..

.

.

)1,3,2(tan.4

P(2.3.1)0,0,0

spacethreein

Vector

23

22

21

321 ),,(

.1

aaaa

bydenotedaoflength

aaaa

aoflength

VEKTOR

1. GAMBAR VEKTOR

2. PANJANG VEKTOR

3. PERKALIAN TITIK=DOT=HSL SKALAR

4. SUDUT 2 BUAH VEKTOR

5. PERKALIAN SILANG=a x b

HSLVEKTOR

6.Jumlah 2 buah vektor

7Persamaan garis dalam vektor

8.Pers bidang

9jarak 2 buahbid

cos

)(

),,(),,(

..2

332211

321321

baba

scalarbabababa

bbbbandaaaa

productdotThe

ba

baarc

ba

ba

bandabetweenangleThe

cos

cos

.3

2738

13cos

2738

13

2738

5126cos

)1(43532

)1,4,3()5,3,2(cos

cos:

)1,4,3()5,3,2(

:1

222222

arc

ba

baSolution

bandaangletheFind

Example

))(),(),((

),,(),,,(

.4

332211

321321

babababa

bbbbaaaa

vectorstwoofSum

a

b

ba

a

x

y

z

A

B

a

b

),,(),,(

),,(),,(

),,(int),,,(int

321321

321321

321321

bbbaaaBABAvector

aaabbbABABvector

bbbBpoaaaAPo

5.

taconsk

kakakaaaaaa

aaparalelarevectorstwoThe

tan

),,(),,(

//:.6

32123211

21

a1

a2

)()()(

)(

),,(),,,(

.7

122113312332

321

321

321321

babakbabajbabai

bbb

aaa

kji

bcandaccbxa

bbbbaaaa

productcrossThe

bdanaantarasudut

cxba

bxa

ba

c

b

a

soal

.4

)(.3

.2

.1

)5,3,3(

)2,6,4(

)7,4,3(

:

)2,34,50()2(3450

264

743.3

58)28(30

80

18

126

84

24

90

533

264

743

)(.2

22142412.1

)2,6,4(

)5,3,3(),7,4,3(

:

kji

kji

bxa

cxba

ba

b

ca

soal

)13,11,6()13()11()6(

)49()29()612(

331

243

)3,3,1(),2,4,3(:2

kji

kji

kji

bxa

baExample

pedparalelepiofareathebabxa

bxaa

bandabetweenangleis

spacethreeinvectorbebandaLet

ATheorem

sin.2

0)(.1

,

:

sina

a

b

bb sina AREA=

)5,1,3(,)1,2,3(),4.2.1(

3

CBA

triangleofareatheFind

Example

772

1

46492

1)2()8()3(

2

1

)2()62()30(2

1

112

3022

1

2

1

)1,1,2()4,2,1()5,1,3(

)3,0,2()4,2,1()1,2,3(3

kji

kji

kji

ACxABArea

ACAC

ABABSolution

baifonlyifparalelare

spacethreetheinbandavectorsTwo

BTheorema

//

cbacbacxbxa

bxcacbxa

axaaxxa

vxkabxak

cxabxacbxa

axbbxa

spacethreetheinvectorsarecandbandaIf

CTheorema

).().()(.6

).(.).(5

0,000.4

)()(.3

)()()(.2

)(.1

,

:

LINES

),,(

,,

)(,)(,)(

),,(),,(

),,(),,(),,(

000

000

000

00000

0000

zyxthroughequationparametricareThese

tczzbtyytaxx

tczzbtyyatxx

tctbtazzyyxxPPVtPP

cbaVandzyxPandzyxPLet

0P

P

V

),,(),,(

),,(),,(),,(

00000

0000

tctbtazzyyxxPPVtPP

cbaVandzyxPandzyxPLet

V

tz

ty

tx

Bmelgrsvektorpers

atau

tz

ty

tx

Amelgrsvektorpers

CBAABABvgrsarahjawab

xybidangmemotongtersebutgaristitikDimanab

BandA

titikmelaluigarisvektorpersCaria

Example

62

86

25

)2,6,5(

64

82

23

)4,2,3(

),,()6,8,2(

)(.

)2,6,5()4,2,3(

.

.1

tz

ty

tx

V

andthroughlinethefor

equationparametrictheFind

Example

64

82

23

)6,8,2()42,26,35(

)2,6,5()4,2,3(:

.1

?.

.

21

5

34

82

2

51

45

41

1

nberpotongab

sejajara

ldanlapakah

tz

ty

tx

l

tz

ty

tx

l

grspersdiketahui

?.

21

24

43

2

2

3

32

2

1

nberpotonga

ldanlapakah

tz

ty

tx

l

tz

ty

tx

l

grspersdiketahui

?.

21

32

61

1213

2

1

3

41

1

.36

nberpotonga

ldanlapakah

tz

ty

tx

l

z

ty

tx

l

grspersdiketahui

?.

21

27

6

4

2

35

3

71

1

nberpotonga

ldanlapakah

tz

y

tx

l

tz

ty

tx

l

grspersdiketahui

Normal

),,( 0000 zyxP

),,( zyxP

DCzByAx

zCyBxACzByAx

zzCyyBxxA

NPP

CBAnormalN

zzyyxxPPPP

zyxPzyxP

Planes

)()()(

0)()()(

)(0

),,()(

),,(

),,(),,,(

:

000

000

0

00000

0000

26642

0)2(6)3(4)1(2

0)()()(

)2,3,1(),,(:

)6,4,2(

)2,3,1(

:1

000

00000

zyx

zyx

zzCyyBxxA

PzyxPSolution

Ntolarperpendicu

throughplaneequationtheFind

Example

BatanglethecheckH

BAif

triangle

rightaisABCtranglethatShow

Exercise

int

)5.2,10,1)(1,1,3(),3,3,6(

:1

tz

ty

tx

land

tz

ty

tx

l

linesparalelthe

containsthatplaneequationtheFind

Exercise

1

43

22

:2

2

41

22

:1

2

planetheofequationthefind andIntersect6

2

1

1

1

22

4

3

2

4

1

:.3

zyx

andzyx

linesthethatShow

)3,2,6()5,1,2(

linethefor

equationparametrictheFind.4

andthrough

tz

ty

tx

garislurustegakdan

ikmelaluititbidangpersamaanCari

23

2

)3,4,1(.5

42

0)3(1)4(2)1(1

)0()0()0(

),,()1,2,1(

23

2

)0,0,0()3,4,1(.5

zyx

zyx

zzcyyBxxAbidpers

CBAnormalbidgarisarah

tz

ty

tx

garislurustegakdan

zyxikmelaluititbidangpersamaanCari

234

)3,0,1(

)4,1,2(.5

zyxbidlurustegakharus

dan

ikmelaluititbidangpersamaanCari

1134

0)3()0(13)1(4

)3,0,1(

1134

0)4()1(13)2(4

)4,1,2(

)1,13,4(

314

113

1

1

)3,1,4()1(

)1,1,3(

234

)3,0,1(

)4,1,2(.5

zyx

zyx

melbidpers

zyx

zyx

melbidpers

kj

xNABdicariygbidN

NdiketahuiygNbid

ABAB

zyxbidlurustegakharus

danB

ikAmelaluititbidangpersamaanCari

tz

ty

tx

garisdanmelalui

ikAmelaluititbidangpersamaanCari

24

1

)3,0,2(.22

537

0)4(3)(1)1(7

537

0)3(3)(1)2(7

)3,1,7(

211

703

1

)7,0,3(

)2,1,1(

)4,0,1(

24

1

)3,0,2(.22

zyx

zyxbidpers

zyx

zyxbidpers

kj

VxABbidNormal

ABABvektorBentuk

Vgarisarah

dilaluiBygtitik

tz

ty

tx

garisdanmelalui

ikAmelaluititbidangpersamaanCari

)(.)(

061812)(.)(

)5.1,9,4(

)4,2,3(

:1

BCBA

BCBA

BCCB

BAAB

Solution

116

0)2(12)1(2)2(2

)12,2,2()12()2()2(

142

124

)1,4,2()1,2,4(

),1,2,4(

)1,4,2(),1,3,2(:2

)1,4,2(),2,1,2(:1:2

121

1221

22

11

zyx

zyx

kji

kji

VPPNormal

PPPP

VPl

VPlSolutions

)4,2,1(intsecint

106t-22t-43

10t13t22

24t-11

6t-2z,t1y,t-2x:2 line

2t-4z,3t2y,4t-1x:1line

3

2121

2121

21

222

111

potheatterarelinesthethen

ttif

ttt

Sulution

562216

0)4(1)2(22)1(16

0)()()()4,2,1(

),,()1,22,16(

)1(31.4())1(26)4(()1.26.3(

611

234

)6,1,1(6t-2z,t1y,t-2x:2 line

)2,3,4(2t-4z,3t2y,4t-1x:1line

:isplanetheofequationThe

000

21

2222

1111

zyx

zyx

zzCyyBxxAPlet

CBAN

kji

kji

xVVLinenormalThe

V

V

tztytx

PQV

zyxthroughequationparametricareThese

tczzbtyytaxx

andQthroughPequationline

no

85,1,82

)8,1,8()53,12,26()(

),,(

,,

)3,2,6()5,1,2(

:4

000

000

tz

ty

txl

tz

ty

txl

sejajar

grsbuahbelaluibidpersamaancari

24

32

4

23

21

:

2

937

0)(1)4(3)3(7

)0,4,3(

937

0)4(1)3(3)2(7

)4,3,2(

),,()1,3,7(1,2,1

)4,1,1(

)4,1,1(

)0,4,3(24

3

2

)4,3,2(

4

23

2

1

:

2

21

zyx

zyx

Btitikmelalui

zyx

zyx

Atitikmelalui

NCBAx

xVABatauxVABNormal

ABAB

Btitik

tz

ty

tx

l

Atitik

tz

ty

tx

l

sejajar

grsbuahbelaluibidpersamaancari

tz

ty

txl

z

ty

txl

grsbuahbelaluibidpersamaancari

32

61

2132

1

3

411

2.28

2712123

0)1(12)3(12)1(3)1,3,1(

),,()12,12,3()3,6,12(

)0,1,4(

)3,6,12(

32

61

1213

2

)0,1,4(

1

3

41

1

2.28

21

2

1

zyx

zyxmelbidpers

CBAxxVVNbid

V

tz

ty

tx

l

V

z

ty

tx

l

grsbuahbelaluibidpersamaancari

soal2

mbarkanzbidangdibawahterletakyang

yxzparaboloidbagianluasTentukan

zyxdanyxz

olehdibatasiygbendavolTentukan

polarbentukkedalamubahdydxyx

dxdyyxfmenjadidydxyxfUbah

dxdyyxfmenjadidydxyxfUbah

x

y

y

x

,2

.5

:.4

....)(.3

),(),(.2

),(),(.1

22

22222

21

0

1

0

2

1

0

2

0 0

2

2

SUDUT DARI DUA BUAH BIDANG

21

2.1cosNN

NNarc

linesNormalthebetweenangletheis

planestwobetweenangleThe

N1

N2

1056101734

:

:

zyxandzyx

planesbetweenangletheFind

Example

16174

93cos

253610049916

)5,6,10).(7,3,4(cos

)5,6,10(

)7,3,4(

:

2

1

arc

arcN

N

Solution

Garis perpotongan dari dua buah bidang

)sec(int

6,505,274

)6,50,27(

5610

734

)0,5,4()0,.....,4(40,892

....105610

1734

21

planesoflinetheoftioner

tztytx

kji

NNV

xzzx

zyx

andzyx

Contoh

I

II

N2

N1

bidduatgarisperpoperscari

zyx

andzyx

soal

....32

22

21

tz

ty

tx

Vtitik

zxyambilyx

zyx

zyx

xNNVNzyx

Nzyx

bidduatgarisperpoperscari

zyx

andzyx

soal

34

0

1

)3,1,1(),4,0,1(

4,1,0,1

32

22

)3,1,1(21)1,2,1(232

)1,1,2(122

....32

22

21

diatasperpotgrsmemuat

danmelaluibidpersCarib

bidduatgarisperpoperscari

zyx

andzyx

soal

)2,4,1(.

....32

22

21

322

024

)2,4,1(.17

)1,2,1(.

....32

22

21

zyxdan

zyxperpotgrsmemuat

danmelaluibidpersCarib

diatasperpotgrsmemuat

danmelaluibidpersCarib

bidduatgarisperpoperscari

zyx

andzyx

soal

tersebutmelaluigrsbidpersCari

tz

ty

tx

l

z

ty

tx

l

potonganberldangarislapakah

soal

2

32

61

1213

2

01

3

41

1

???21

28

JARAK TITIK KE BIDANG

Langkah 2:

• Jarak titik P kke bidang

• Ambil smbarang titik pada bidang tsb(misal titik A)

• Buat vektor AP=P-A

• Cari N maka ketemu rumus dibawah ini

planetheinpoanyAwhichN

NAPL

orCBA

DCzByAxL

DCzByAxplanetheto

zyxPpoafromcedisThe

int

),,(inttan

222

000

000

L

A

Normal

P

N

NAPL

N

NAPL

APL

ALPAP

cos.

cos.

052

)7,1,2(tan

:

zyxplanetheto

PfromcedistheFind

Example

29

26

29

12184

1694

)4,3,2()3,6,2()4,3,2(

)3,6,2()0,0,0(

0432int

:

N

NAPN

APAPA

zyxplanetheonpoanyA

Solution

1543

0862

tan

:

222

zyxplanetheto

zyxzyx

spherethefromcedistheFind

Example

0lforethere

2626

41218)1,4,3(

)4,3,6(

)0,0,5(

)4,3,1(,26radiushassphereThe

26)4()3()1(

01691)4()3()1(

:Solution

222

222

N

NAPLN

APAP

A

centralthe

zyx

zyx

4

1)2sin)2/1((

4

1

)2cos1(2

1

2

1)(sin2/1

.

min.

,0,sin,0

:min.1

0

0

2

0

sin

00

xx

dxxdxx

dxdyyM

massapusatTitikb

alaMassaaTentukan

ymassarapatxxyy

olehdibatasiaLaDiketahui

x

9

4)

3

11()

3

11((

3

1

)cos)3/1((cos3

1

sin)cos1(3

1

)(sin3

1

3

1

0

3

2

0

3

0

sin

0

3

0

2sin

00

xx

dxxx

dxxdxy

dxdyyM

x

x

x

8)

4

1

4

1(

4

1

8

2cos2

1

2

12sin

2

1(

4

1

8

)2(sin2

12sin

2

1(

4

1

8

int)2cos2

1(

2

1

2

1

)2cos1(2

1

2

1

)(sin2

1

2

1

22

0

2

2

0

2

0

2

0

sin

0

2

0

sin

00

xxx

dxxxx

parsialdxxxx

dxxx

dxxxdxxy

dxdyyxM

x

x

y

3

16)10(

3

16

cos83

12sin8

3

12

sin3

12sin2

2

.

min.

,4

:min.2

2/

0

2/

0

2

0

3

0

2

0

2/

0

2

0

2/

0

2

d

drddrrr

dxdyyM

massapusatTitikb

alaMassaaTentukan

ymassarapatxy

olehdibatasiaLaDiketahui

4

2sin2

1(8)2cos1(

2

132

4

12

)2cos1(4

12sin2

2

.

min.

,4

:min.2

2/

00

2

0

4

0

222

0

2/

0

2

0

2/

0

2

d

drddrrr

dxdyyyM

massapusatTitikb

alaMassaaTentukan

ymassarapatxy

olehdibatasiaLaDiketahui

x

0sincos8

)sin(cos4

12sincos2

2

2/

0

2

0

4

0

22

0

2/

0

2

0

2/

0

d

drddrrr

dxdyxyM y

3

4

3

1

)0/1(

.

min.

2,0,1

,

:min3

2

1

1

0

3

2

1

21

0

1

0

2

10

1

0

xx

dxxxdxdyx

dxdyxdxdyxM

massapusatTitikb

alaMassaaTentukan

xmassarapatxyx

yxy

olehdibatasiaLaDiketahui

xy

y

xy

y

4

7)14(

2

1

4

1

2

1

4

1

)0/1()0(

2

1

21

0

4

22

1

21

0

2

1

0

2

1

2

0

1

0

xx

dxxxdxxx

dxdyxdxdyxMx

y

y

xy

y

y

2ln2

1

8

1ln

2

1

4

1

2

1

2

1

2

1

2

1

2

1

2

1

1

0

4

12

1

31

0

1

0

22

10

21

0

1

0

2

10

1

0

xx

dxxdxx

dxxydxxy

dxdyxydxdyxyM

xx

xy

y

xy

y

x