BAB IV VEKTOR

1. BAB IV VEKTOR
1. Hitunglah panjang dari vektor-vektor berikut
a. 𝑘 = [
6
−8
]
b. 𝑙 = [
9
11
]
c. 𝑚 = [
4
6
−2
]
d. 𝑛 = [
6
9
−7
]
Jawaban :
𝑘 = [
6
−8
]
|𝑘| = √62 + 82
= √36 + 64 = √100 = 10
𝑙 = [
9
11
]
|𝑙| = √92 + 112
√81 + 121 = √202
𝑚 = [
4
6
−2
]
|𝑚| = √42 + 62 + −22
√16 + 36 + 4 = √56 = 2√14
𝑛 = [
6
9
−7
]
|𝑛| = √62 + 92 + −72
√36 + 81 + 49 = √166
2. Hitunglah vektor satuan dari vektor-vektor berikut:
a. 𝑎 = [
4
8
]
b. 𝑏 = [
−5
7
]
c. 𝑐 = [
8
3
−1
]

2. BAB IV VEKTOR
d. 𝑑 = [
5
3
−8
]
Jawaban:
a) |𝑎| = √42 + 82 = 4√5
𝑒 =
𝑎
|𝑎|
=
1
4√5
[
4
8
]
=
[
1
√5
2
√5]
b) |𝑏| = √−52 + 72 = √74
𝑒 =
𝑏
|𝑏|
=
1
√74
[
−5
7
]
=
[
−5
√74
7
√74]
c) | 𝑐| = √82 + 32 + −12 = √74
𝑒 =
𝑐
| 𝑐|
=
1
√74
[
8
3
−1
]
=
[
8
√74
3
√74
−1
√74]
d) | 𝑑| = √52 + 32 + −82 = √25 + 9 + 64 = 7√2
𝑒 =
𝑑
| 𝑑|
=
1
7√2
[
5
3
−8
]

3. BAB IV VEKTOR
=
[
5
7√2
3
7√2
−8
7√2]
3. Diketahui 𝑎 = 6𝑖 − 𝑥𝑗 + 10𝑘 dan 𝑏 = 8𝑖 + 7𝑗 − 9𝑘. Tentukanlah nilai x agar
vektor 𝑎 tegak lurus dengan vektor 𝑏 .
Jawaban:
𝑎 = 6𝑖 − 𝑥𝑗 + 10𝑘
𝑏 = 8𝑖 + 7𝑗 − 9𝑘
vektor 𝑎tegak lurus dengan vektor 𝑏, maka:
𝑎 . 𝑏 = 0
(6𝑖 − 𝑥𝑗 + 10𝑘)(8𝑖 + 7𝑗 − 9𝑘) = 0
48 − 7𝑥 − 90 = 0
−7𝑥 = 42
𝑥 = −6
4. Diketahui |𝑎| = 8 , |𝑏| = 9, sudut yang dibentuk kedua vektor 30°.
Tentukanlah 𝑎 . 𝑏
Jawaban:
𝑎 . 𝑏 = |𝑎|.|𝑏|cos30°
= 8 . 9 cos 30°
= 72
1
2
√3 = 36√3

4. BAB IV VEKTOR
5. Diketahui 𝑎 = [
−3
4
5
] 𝑑𝑎𝑛 𝑏 = [
7
−2
6
] . Tentukanlah 𝑎 . 𝑏
Jawaban:
𝑎 . 𝑏 = [
−3
4
5
] [
7
−2
6
]
= (−21) + (−8) + 30 = 1
6. Jika 𝑎 = 6𝑖 + 7𝑗 − 2𝑘 𝑑𝑎𝑛 𝑏 = 4𝑖 + 7𝑗 − 9𝑘. Tentukanlah 𝑎 x 𝑏
Jawaban:
𝑎 x 𝑏 = |
𝑖 𝑗 𝑘
6 7 −2
4 7 −9
|
= (−63𝑖 + (−8)𝑗 + 42𝑘) − (−141𝑖 + (−54)𝑗 + 28𝑘)
= −49𝑖 + 46𝑗 + 14𝑘
7. Hitunglah besar sudut antara 𝑎 = 3𝑖 − 4𝑗 + 6𝑘 𝑑𝑎𝑛 𝑏 = 5𝑖 + 2𝑗 − 3𝑘
Jawaban:
𝑐𝑜𝑠𝜃 =
𝑎 . 𝑏
|𝑎|. |𝑏|
=
3(5) + (−4)2 + 6(−3)
√32 + −42 + 62. √52 + 22 + −32
= −
11
48,15
= −0,23
𝑐𝑜𝑠𝜃 = −0,23
𝜃 = 𝑎𝑟𝑐 cos(−0,23)
𝜃 = 103,3°
8. Diketahui 𝑎 = 5𝑖 − 𝑗 + 2𝑘 𝑑𝑎𝑛 𝑏 = 3𝑖 + 4𝑗 − 5𝑘.

5. BAB IV VEKTOR
Tentukanlah:
a) Proyeksi vektor 𝑎 pada vektor 𝑏
b) Panjang proyeksi vektor 𝑎 pada vektor 𝑏
Jawaban:
a) Proyeksi vektor 𝑎pada vektor 𝑏
(𝑎 . 𝑏)
|𝑏|
𝟐 𝑏 =
5(3) + (−1)4 + 2(−5)
(√32 + 42 + −52)
2 (3𝑖 + 4𝑗 − 5𝑘)
=
1(3𝑖 + 4𝑗 − 5𝑘)
(√50)
2
=
3
50
𝑖 +
2
25
𝑗 −
1
10
𝑘
b) Panjang proyeksi vektor 𝑎pada vektor𝑏
𝑎 . 𝑏
|𝑏|
=
1
√50
=
1
5√2
1
5√2
.
5√2
5√2
=
5√2
50
=
√2
10