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Daring Matematika 10 September 2020
XI TKJ SMK WALISONGO
Catat materi dan kerjakan latihan soalnya!
๏ถ Materi
1. Vektor Pada Bidang Datar (๐‘น๐Ÿ)
A. Vektor satuan ( ๐‘’
โƒ— )
๐‘’
โƒ— =
๐‘Ž
โƒ—
|๐‘Ž
โƒ—|
Contoh :
1. Tentukan vektor satuan dari ๐‘
โƒ—โƒ— =[
โˆ’3
4
]!
Jawab: ๐‘’
โƒ— =
๐‘
โƒ—โƒ—
|๐‘
โƒ—โƒ—|
==
[
โˆ’3
4
]
โˆš(โˆ’3)2
+42
=
[
โˆ’3
4
]
โˆš9+16
=
[
โˆ’3
4
]
โˆš25
=
[
โˆ’3
4
]
5
= [
โˆ’
3
5
4
5
]
2. Tentukan vektor satuan dari ๐‘
โƒ— =[
2
0
]!
Jawab: ๐‘’
โƒ— =
๐‘
โƒ—
|๐‘
โƒ—|
==
[2
0
]
โˆš22+02
=
[
2
0
]
โˆš4+0
=
[2
0
]
โˆš4
=
[2
0
]
2
= [
2
2
0
2
] = [
1
0
]
B. Perkalian vektor dengan skalar
Contoh:
Diketahui ๐‘Ž
โƒ— = [
2
โˆ’4
] ๐‘‘๐‘Ž๐‘› ๐‘
โƒ—โƒ— = [
1
โˆ’5
]
Tentukan ;
1. 2๐‘Ž
โƒ—
2. โˆ’3๐‘
โƒ—โƒ—
Jawab:
1. 2๐‘Ž
โƒ— =2. [
2
โˆ’4
] = [
4
โˆ’8
]
2. โˆ’3๐‘
โƒ—โƒ—=(-3). [
1
โˆ’5
] = [
โˆ’3
15
]
C. Penjumalahan Vektor
1. Cara Segitiga 2. Cara Jajar genjang
Contoh:
Diketahui ๐‘Ž
โƒ— = [
2
4
] ๐‘‘๐‘Ž๐‘› ๐‘
โƒ—โƒ— = [
1
โˆ’5
] , ๐‘š๐‘Ž๐‘˜๐‘Ž:
1. ๐‘Ž
โƒ— + ๐‘
โƒ—โƒ— =[
2
4
] + [
1
โˆ’5
] = [
3
โˆ’1
]
2. ๐‘Ž
โƒ—โƒ— + 2๐‘
โƒ—โƒ— =[
2
4
] + 2. [
1
โˆ’5
] = [
2
4
] + [
2
โˆ’10
] = [
4
โˆ’6
]
D. Selisih/pengurangan Vektor
Contoh:
Diketahui ๐‘Ž
โƒ— = [
2
4
] ๐‘‘๐‘Ž๐‘› ๐‘
โƒ—โƒ— = [
1
โˆ’5
] , ๐‘š๐‘Ž๐‘˜๐‘Ž:
1. ๐‘Ž
โƒ— โˆ’ ๐‘
โƒ—โƒ— =[
2
4
] โˆ’ [
1
โˆ’5
] = [
2 โˆ’ 1
4 โˆ’ (โˆ’5)
] = [
1
9
]
2. ๐‘Ž
โƒ—โƒ— โˆ’ 2๐‘
โƒ—โƒ— =[
2
4
] โˆ’ 2. [
1
โˆ’5
] = [
2
4
] โˆ’ [
2
โˆ’10
] = [
0
14
]
2. Vektor Pada Bangun Ruang (๐‘น๐Ÿ‘)
A. Vektor satuan ( ๐‘’
โƒ— )
๐‘’
โƒ— =
๐‘Ž
โƒ—
|๐‘Ž
โƒ—|
Contoh :
1. Tentukan vektor satuan dari ๐‘
โƒ—โƒ— =[
0
3
โˆ’4
]!
Jawab: ๐‘’
โƒ— =
๐‘
โƒ—โƒ—
|๐‘
โƒ—โƒ—|
==
[
0
3
โˆ’4
]
โˆš02+32(โˆ’4)
2
=
[
0
3
โˆ’4
]
โˆš0+9+16
=
[
0
3
โˆ’4
]
โˆš25
=
[
0
3
โˆ’4
]
5
=
[
0
5
3
5
โˆ’
4
5]
=[
0
3
5
โˆ’
4
5
]
2. Tentukan vektor satuan dari ๐‘
โƒ— =2๐‘–
โƒ—+ ๐‘—
โƒ— + ๐‘˜
โƒ—โƒ—
Jawab: ๐‘
โƒ— =2๐‘–
โƒ— + ๐‘—
โƒ— + ๐‘˜
โƒ—โƒ—=[
๐‘–
๐‘—
๐‘˜
] = [
2
1
1
], maka ๐‘’
โƒ— =
๐‘
โƒ—
|๐‘
โƒ—|
=
[
2
1
1
]
โˆš22
+12
+12
=
[
2
1
1
]
โˆš4+1+1
=
[
2
1
1
]
โˆš6
=
[
2
โˆš6
1
โˆš6
1
โˆš6]
B. Perkalian vektor dengan skalar
Contoh:
Diketahui ๐‘Ž
โƒ— = [
2
0
โˆ’4
] ๐‘‘๐‘Ž๐‘› ๐‘
โƒ—โƒ— = [
1
8
โˆ’5
]
Tentukan ;
1. 3๐‘Ž
โƒ—
2. โˆ’2๐‘
โƒ—โƒ—
Jawab:
1. 3๐‘Ž
โƒ— =3. [
2
0
โˆ’4
] = [
6
0
โˆ’12
]
2. โˆ’2๐‘
โƒ—โƒ—=(-2). [
1
8
โˆ’5
] = [
โˆ’2
โˆ’16
10
]
C. Penjumalahan Vektor
Contoh:
Diketahui ๐‘Ž
โƒ— = [
2
4
3
] ๐‘‘๐‘Ž๐‘› ๐‘
โƒ—โƒ— = [
1
โˆ’5
โˆ’4
] , ๐‘š๐‘Ž๐‘˜๐‘Ž:
1. ๐‘Ž
โƒ— + ๐‘
โƒ—โƒ— =[
2
4
3
] + [
1
โˆ’5
โˆ’4
] = [
3
โˆ’1
โˆ’1
]
2. 3๐‘Ž
โƒ—โƒ— + ๐‘
โƒ—โƒ— =3. [
2
4
3
] + [
1
โˆ’5
โˆ’4
] = [
6
12
9
] + [
1
โˆ’5
โˆ’4
] = [
7
7
5
]
D. Selisih/pengurangan Vektor
Contoh:
Diketahui ๐‘Ž
โƒ— = [
2
4
1
] ๐‘‘๐‘Ž๐‘› ๐‘
โƒ—โƒ— = [
1
โˆ’5
3
] , ๐‘š๐‘Ž๐‘˜๐‘Ž:
1. ๐‘Ž
โƒ— โˆ’ ๐‘
โƒ—โƒ— =[
2
4
1
] โˆ’ [
1
โˆ’5
3
] = [
1
10
โˆ’2
]
2. ๐‘Ž
โƒ—โƒ— โˆ’ 2๐‘
โƒ—โƒ— =[
2
1
1
] โˆ’ 2. [
1
โˆ’5
3
] = [
2
4
1
] โˆ’ [
2
โˆ’10
6
] = [
0
14
โˆ’5
]
E. Materi Tambahan
Contoh;
1. Tentukan vektor satuannya
a. ๐‘‘
โƒ— = 4๐‘–
โƒ—โˆ’ ๐‘—
โƒ— + ๐‘˜
โƒ—โƒ—
b. ๐‘
โƒ— = โˆ’4๐‘—
โƒ—โˆ’ 2๐‘˜
โƒ—โƒ—
c. ๐‘ž
โƒ— = ๐‘–
โƒ— โˆ’ 5๐‘˜
โƒ—โƒ—
Jawab.
a. ๐‘‘
โƒ— = 4๐‘–
โƒ—โˆ’ ๐‘—
โƒ— + ๐‘˜
โƒ—โƒ— = [
๐‘–
๐‘—
๐‘˜
] = [
4
โˆ’1
1
], maka ๐‘’
โƒ— =
๐‘
โƒ—
|๐‘‘
โƒ—|
=
[
4
โˆ’1
1
]
โˆš42+(โˆ’1)2
+12
=
[
4
โˆ’1
1
]
โˆš16+1+1
=
[
4
โˆ’1
1
]
โˆš18
=
[
4
โˆš18
โˆ’
1
โˆš18
1
โˆš18 ]
b. ๐‘
โƒ— = โˆ’4๐‘—
โƒ—โˆ’ 2๐‘˜
โƒ—โƒ—=[
๐‘–
๐‘—
๐‘˜
] = [
0
โˆ’4
โˆ’2
], maka ๐‘’
โƒ— =
๐‘
โƒ—
|๐‘
โƒ—|
=
[
0
โˆ’4
โˆ’2
]
โˆš02
+(โˆ’4)2
+(โˆ’2)2
=
[
0
โˆ’4
โˆ’2
]
โˆš0+16+4
=
[
0
โˆ’4
โˆ’2
]
โˆš20
=
[
0
โˆš20
โˆ’
4
โˆš20
โˆ’
2
โˆš20]
= [
0
โˆ’
4
โˆš20
โˆ’
2
โˆš20
]
c. ๐‘ž
โƒ— = ๐‘–
โƒ— โˆ’ 5๐‘˜
โƒ—โƒ— = [
๐‘–
๐‘—
๐‘˜
] = [
1
0
โˆ’5
], maka ๐‘’
โƒ— =
๐‘
โƒ—
|๐‘ž
โƒ—โƒ—|
=
[
1
0
โˆ’5
]
โˆš12
+02
+(โˆ’5)2
=
[
1
0
โˆ’5
]
โˆš1+0+25
=
[
1
0
โˆ’5
]
โˆš26
=
[
1
โˆš26
0
โˆš26
โˆ’
5
โˆš26]
= [
1
โˆš26
0
โˆ’
5
โˆš26
]
๏ถ Latihan Soal
1. Tentukan vektor satuannya
a. ๐‘Ÿ
โƒ— =[
โˆ’6
8
]
b. ๐‘
โƒ—โƒ— = [
0
4
]
c. ๐‘
โƒ— = 3๐‘–
โƒ— + ๐‘—
โƒ— + ๐‘˜
โƒ—โƒ—
d. ๐‘‘
โƒ— = ๐‘–
โƒ— โˆ’ ๐‘—
โƒ— + ๐‘˜
โƒ—โƒ—
e. ๐‘
โƒ— = โˆ’4๐‘—
โƒ— + 2๐‘˜
โƒ—โƒ—
---Selamat mengerjakan---

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8 Daring Matematika 10 September 2020 - XI.docx

  • 1. Daring Matematika 10 September 2020 XI TKJ SMK WALISONGO Catat materi dan kerjakan latihan soalnya! ๏ถ Materi 1. Vektor Pada Bidang Datar (๐‘น๐Ÿ) A. Vektor satuan ( ๐‘’ โƒ— ) ๐‘’ โƒ— = ๐‘Ž โƒ— |๐‘Ž โƒ—| Contoh : 1. Tentukan vektor satuan dari ๐‘ โƒ—โƒ— =[ โˆ’3 4 ]! Jawab: ๐‘’ โƒ— = ๐‘ โƒ—โƒ— |๐‘ โƒ—โƒ—| == [ โˆ’3 4 ] โˆš(โˆ’3)2 +42 = [ โˆ’3 4 ] โˆš9+16 = [ โˆ’3 4 ] โˆš25 = [ โˆ’3 4 ] 5 = [ โˆ’ 3 5 4 5 ] 2. Tentukan vektor satuan dari ๐‘ โƒ— =[ 2 0 ]! Jawab: ๐‘’ โƒ— = ๐‘ โƒ— |๐‘ โƒ—| == [2 0 ] โˆš22+02 = [ 2 0 ] โˆš4+0 = [2 0 ] โˆš4 = [2 0 ] 2 = [ 2 2 0 2 ] = [ 1 0 ] B. Perkalian vektor dengan skalar Contoh: Diketahui ๐‘Ž โƒ— = [ 2 โˆ’4 ] ๐‘‘๐‘Ž๐‘› ๐‘ โƒ—โƒ— = [ 1 โˆ’5 ] Tentukan ; 1. 2๐‘Ž โƒ— 2. โˆ’3๐‘ โƒ—โƒ— Jawab: 1. 2๐‘Ž โƒ— =2. [ 2 โˆ’4 ] = [ 4 โˆ’8 ] 2. โˆ’3๐‘ โƒ—โƒ—=(-3). [ 1 โˆ’5 ] = [ โˆ’3 15 ] C. Penjumalahan Vektor 1. Cara Segitiga 2. Cara Jajar genjang Contoh: Diketahui ๐‘Ž โƒ— = [ 2 4 ] ๐‘‘๐‘Ž๐‘› ๐‘ โƒ—โƒ— = [ 1 โˆ’5 ] , ๐‘š๐‘Ž๐‘˜๐‘Ž: 1. ๐‘Ž โƒ— + ๐‘ โƒ—โƒ— =[ 2 4 ] + [ 1 โˆ’5 ] = [ 3 โˆ’1 ] 2. ๐‘Ž โƒ—โƒ— + 2๐‘ โƒ—โƒ— =[ 2 4 ] + 2. [ 1 โˆ’5 ] = [ 2 4 ] + [ 2 โˆ’10 ] = [ 4 โˆ’6 ] D. Selisih/pengurangan Vektor Contoh: Diketahui ๐‘Ž โƒ— = [ 2 4 ] ๐‘‘๐‘Ž๐‘› ๐‘ โƒ—โƒ— = [ 1 โˆ’5 ] , ๐‘š๐‘Ž๐‘˜๐‘Ž: 1. ๐‘Ž โƒ— โˆ’ ๐‘ โƒ—โƒ— =[ 2 4 ] โˆ’ [ 1 โˆ’5 ] = [ 2 โˆ’ 1 4 โˆ’ (โˆ’5) ] = [ 1 9 ] 2. ๐‘Ž โƒ—โƒ— โˆ’ 2๐‘ โƒ—โƒ— =[ 2 4 ] โˆ’ 2. [ 1 โˆ’5 ] = [ 2 4 ] โˆ’ [ 2 โˆ’10 ] = [ 0 14 ]
  • 2. 2. Vektor Pada Bangun Ruang (๐‘น๐Ÿ‘) A. Vektor satuan ( ๐‘’ โƒ— ) ๐‘’ โƒ— = ๐‘Ž โƒ— |๐‘Ž โƒ—| Contoh : 1. Tentukan vektor satuan dari ๐‘ โƒ—โƒ— =[ 0 3 โˆ’4 ]! Jawab: ๐‘’ โƒ— = ๐‘ โƒ—โƒ— |๐‘ โƒ—โƒ—| == [ 0 3 โˆ’4 ] โˆš02+32(โˆ’4) 2 = [ 0 3 โˆ’4 ] โˆš0+9+16 = [ 0 3 โˆ’4 ] โˆš25 = [ 0 3 โˆ’4 ] 5 = [ 0 5 3 5 โˆ’ 4 5] =[ 0 3 5 โˆ’ 4 5 ] 2. Tentukan vektor satuan dari ๐‘ โƒ— =2๐‘– โƒ—+ ๐‘— โƒ— + ๐‘˜ โƒ—โƒ— Jawab: ๐‘ โƒ— =2๐‘– โƒ— + ๐‘— โƒ— + ๐‘˜ โƒ—โƒ—=[ ๐‘– ๐‘— ๐‘˜ ] = [ 2 1 1 ], maka ๐‘’ โƒ— = ๐‘ โƒ— |๐‘ โƒ—| = [ 2 1 1 ] โˆš22 +12 +12 = [ 2 1 1 ] โˆš4+1+1 = [ 2 1 1 ] โˆš6 = [ 2 โˆš6 1 โˆš6 1 โˆš6] B. Perkalian vektor dengan skalar Contoh: Diketahui ๐‘Ž โƒ— = [ 2 0 โˆ’4 ] ๐‘‘๐‘Ž๐‘› ๐‘ โƒ—โƒ— = [ 1 8 โˆ’5 ] Tentukan ; 1. 3๐‘Ž โƒ— 2. โˆ’2๐‘ โƒ—โƒ— Jawab: 1. 3๐‘Ž โƒ— =3. [ 2 0 โˆ’4 ] = [ 6 0 โˆ’12 ] 2. โˆ’2๐‘ โƒ—โƒ—=(-2). [ 1 8 โˆ’5 ] = [ โˆ’2 โˆ’16 10 ] C. Penjumalahan Vektor Contoh: Diketahui ๐‘Ž โƒ— = [ 2 4 3 ] ๐‘‘๐‘Ž๐‘› ๐‘ โƒ—โƒ— = [ 1 โˆ’5 โˆ’4 ] , ๐‘š๐‘Ž๐‘˜๐‘Ž: 1. ๐‘Ž โƒ— + ๐‘ โƒ—โƒ— =[ 2 4 3 ] + [ 1 โˆ’5 โˆ’4 ] = [ 3 โˆ’1 โˆ’1 ] 2. 3๐‘Ž โƒ—โƒ— + ๐‘ โƒ—โƒ— =3. [ 2 4 3 ] + [ 1 โˆ’5 โˆ’4 ] = [ 6 12 9 ] + [ 1 โˆ’5 โˆ’4 ] = [ 7 7 5 ] D. Selisih/pengurangan Vektor Contoh: Diketahui ๐‘Ž โƒ— = [ 2 4 1 ] ๐‘‘๐‘Ž๐‘› ๐‘ โƒ—โƒ— = [ 1 โˆ’5 3 ] , ๐‘š๐‘Ž๐‘˜๐‘Ž: 1. ๐‘Ž โƒ— โˆ’ ๐‘ โƒ—โƒ— =[ 2 4 1 ] โˆ’ [ 1 โˆ’5 3 ] = [ 1 10 โˆ’2 ] 2. ๐‘Ž โƒ—โƒ— โˆ’ 2๐‘ โƒ—โƒ— =[ 2 1 1 ] โˆ’ 2. [ 1 โˆ’5 3 ] = [ 2 4 1 ] โˆ’ [ 2 โˆ’10 6 ] = [ 0 14 โˆ’5 ]
  • 3. E. Materi Tambahan Contoh; 1. Tentukan vektor satuannya a. ๐‘‘ โƒ— = 4๐‘– โƒ—โˆ’ ๐‘— โƒ— + ๐‘˜ โƒ—โƒ— b. ๐‘ โƒ— = โˆ’4๐‘— โƒ—โˆ’ 2๐‘˜ โƒ—โƒ— c. ๐‘ž โƒ— = ๐‘– โƒ— โˆ’ 5๐‘˜ โƒ—โƒ— Jawab. a. ๐‘‘ โƒ— = 4๐‘– โƒ—โˆ’ ๐‘— โƒ— + ๐‘˜ โƒ—โƒ— = [ ๐‘– ๐‘— ๐‘˜ ] = [ 4 โˆ’1 1 ], maka ๐‘’ โƒ— = ๐‘ โƒ— |๐‘‘ โƒ—| = [ 4 โˆ’1 1 ] โˆš42+(โˆ’1)2 +12 = [ 4 โˆ’1 1 ] โˆš16+1+1 = [ 4 โˆ’1 1 ] โˆš18 = [ 4 โˆš18 โˆ’ 1 โˆš18 1 โˆš18 ] b. ๐‘ โƒ— = โˆ’4๐‘— โƒ—โˆ’ 2๐‘˜ โƒ—โƒ—=[ ๐‘– ๐‘— ๐‘˜ ] = [ 0 โˆ’4 โˆ’2 ], maka ๐‘’ โƒ— = ๐‘ โƒ— |๐‘ โƒ—| = [ 0 โˆ’4 โˆ’2 ] โˆš02 +(โˆ’4)2 +(โˆ’2)2 = [ 0 โˆ’4 โˆ’2 ] โˆš0+16+4 = [ 0 โˆ’4 โˆ’2 ] โˆš20 = [ 0 โˆš20 โˆ’ 4 โˆš20 โˆ’ 2 โˆš20] = [ 0 โˆ’ 4 โˆš20 โˆ’ 2 โˆš20 ] c. ๐‘ž โƒ— = ๐‘– โƒ— โˆ’ 5๐‘˜ โƒ—โƒ— = [ ๐‘– ๐‘— ๐‘˜ ] = [ 1 0 โˆ’5 ], maka ๐‘’ โƒ— = ๐‘ โƒ— |๐‘ž โƒ—โƒ—| = [ 1 0 โˆ’5 ] โˆš12 +02 +(โˆ’5)2 = [ 1 0 โˆ’5 ] โˆš1+0+25 = [ 1 0 โˆ’5 ] โˆš26 = [ 1 โˆš26 0 โˆš26 โˆ’ 5 โˆš26] = [ 1 โˆš26 0 โˆ’ 5 โˆš26 ] ๏ถ Latihan Soal 1. Tentukan vektor satuannya a. ๐‘Ÿ โƒ— =[ โˆ’6 8 ] b. ๐‘ โƒ—โƒ— = [ 0 4 ] c. ๐‘ โƒ— = 3๐‘– โƒ— + ๐‘— โƒ— + ๐‘˜ โƒ—โƒ— d. ๐‘‘ โƒ— = ๐‘– โƒ— โˆ’ ๐‘— โƒ— + ๐‘˜ โƒ—โƒ— e. ๐‘ โƒ— = โˆ’4๐‘— โƒ— + 2๐‘˜ โƒ—โƒ— ---Selamat mengerjakan---