Presentasi Matematika Kelas Xii Vektor
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![Page 1: Presentasi Matematika Kelas Xii Vektor](https://reader037.fdocument.pub/reader037/viewer/2022102523/55cf936e550346f57b9d80b6/html5/thumbnails/1.jpg)
VEKTOR
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Setelah menyaksikan tayangan ini anda dapat
Menentukan penyelesaianoperasi aljabar vektor
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Vektor adalahbesaran
yang mempunyaibesar dan arah
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Besar vektor artinya panjang vektor
Arah vektor artinya sudut yang dibentuk
dengan sumbu X positifVektor disajikan dalam bentuk
ruas garis berarah
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A
B
ditulis vektor AB atau u A disebut titik pangkalB disebut titik ujung
u45 X
Gambar Vektor
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Notasi Penulisan Vektor Bentuk vektor kolom:
43
u
02
1PQatau
Bentuk vektor baris: 4 ,3 AB atau 0 ,3 ,2 v
Vektor ditulis dengan notasi: i, j dan k misal : a = 3i – 2j + 7k
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VEKTOR DI R2
Vektor di R2 adalah
vektor yang terletak di satu bidangatau
Vektor yang hanya mempunyaidua komponen yaitu x dan y
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VEKTOR DI R2
OA PA OP
O Pij
X
A(x,y)Y
OP = xi; OQ= yjJadi
OA =xi + yjatau
a = xi + yj
ax
y
i vektor satuan searahsumbu Xj vektor satuan searahsumbu Y
Q OA OQ OP
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Vektor di R3
Vektor di R3
adalah Vektor yang terletak di ruang dimensi tiga
atau Vektor yang mempunyai
tiga komponen yaitu x, y dan z
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Misalkan koordinat titik T di R3
adalah (x, y, z) maka OP = xi;OQ = yj dan OS = zk
X
Y
Z
T(x,y,z)
Oxi
yj
zk
PQ
S
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X
Y
Z
T(x,y,z)
Ot
P
QR(x,y)
S
xi
yj
zk
OP + PR = OR atauOP + OQ = OR
OR + RT = OT atauOP + OQ + OS = OT
Jadi OT = xi + yj + zk atau t = xi + yj + zk
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Vektor Posisi
Vektor posisi adalah
Vektor yang titik pangkalnya O(0,0)
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X
Y
O
Contoh:
A(4,1)
B(2,4)
Vektor posisi
titik A(4,1) adalah
14
a OA
Vektor posisi titik B(2,4) adalahji 42 b OB
a
b
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Panjang vektor
Dilambangkan dengan tanda ‘harga mutlak’
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Di R2, panjang vektor:
2
1
aa
a
atau a = a1i + a2j Dapat ditentukan dengan
teorema Pythagoras
22
21 a aa
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Di R2, panjang vektor:
2
1
aa
a
atau a = a1i + a2j Dapat ditentukan dengan
teorema Pythagoras
22
21 a aa
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Di R3 , panjang vektor:
222 y x zv
zyx
v
atau v = xi + yj + zk Dapat ditentukan dengan
teorema Pythagoras
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Contoh:1. Panjang vektor:
43
a
adalah 22 4 3a = 25 = 5
2. Panjang vektor: 2k -j i2 v
adalah 222 )2(1 2 v
= 9 = 3
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Vektor Satuanadalah suatu vektor yang
panjangnya satu
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Vektor satuan searah sumbu X,
sumbu Y , dan sumbu Z berturut-turut
adalah vektor i , j dan k
100
dan 010
,001
kji
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Vektor Satuan
dari vektor a = a1i + a2j+ a3k
adalah
23
22
21
321 aaa
kajaiaaa ee aa
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Contoh: Vektor Satuan dari vektor a = i - 2j+ 2k adalah….Jawab:
aaea
222 2)2(1
22
kjiea
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222 2)2(1
22
kjiea
322
kji
ea
kjiea 32
32
31
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ALJABAR VEKTORKesamaan vektorPenjumlahan vektorPengurangan vektorPerkalian vektor dengan
bilangan real
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Kesamaan VektorMisalkan: a = a1i + a2j + a3k danb = b1i + b2j + b3k
Jika: a = b , maka a1 = b1
a2 = b2 dana3 = b3
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Contoh
Diketahui: a = i + xj - 3k danb = (x – y)i - 2j - 3kJika a = b, maka x + y = ....
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Jawab:a = i + xj - 3k danb = (x – y)i - 2j - 3ka = b1 = x - yx = -2; disubstitusikan1 = -2 – y; y = -3Jadi x + y = -2 + (-3) = -5
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Penjumlahan Vektor
aaa
a
3
2
1
bbb
b
3
2
1
Misalkan: dan
Jika: a + b = c , maka vektor
33
22
11
cbababa
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Contoh
1-2p-3
a
36p
b
Diketahui:
Jika a + b = c , maka p – q =....
dan 2
4q5-
c
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2 45
3)1(6 2
3qp
p
jawab: a + b = c
24
5
36p
1-2p-3
q
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2 45
3)1(6 2
3qp
p
3 + p = -5 p = -8 -2p + 6 = 4q16 + 6 = 4q 22 = 4q q = 5½;Jadi p – q = -8 – 5½ = -13½
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Pengurangan Vektor
Jika: a - b = c , maka c =(a1 – b1)i + (a2 – b2)j + (a3 - b3)k
Misalkan: a = a1i + a2j + a3k danb = b1i + b2j + b3k
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X
Y
O
A(4,1)
B(2,4)
a
b
Perhatikan gambar:
32-
vektor posisi:titik A(4,1) adalah:
14
a
titik B(2,4) adalah:
42
b
vektor AB =
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Jadi secara umum: ab AB
14
42
ab
32-
14
a
42
b
32-
AB
vektor AB =
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Contoh 1
Jawab:
Diketahui titik-titik A(3,5,2) danB(1,2,4). Tentukan komponen-komponen vektor AB
232
253
- 421
ab AB
232
AB Jadi
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Contoh 2Diketahui titik-titik P(-1,3,0)dan Q(1,2,-2). Tentukan panjang vektor PQ(atau jarak P ke Q)
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Jawab: P(1,2,-2)
Q(-1,3,0)
PQ = q – p =
21
2
2-21
- 031-
221
p
031
q
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21
2 PQ
222 21)2(PQ
39PQ Jadi
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Perkalian Vektor dengan Bilangan Real
aaa
a
3
2
1
Misalkan:
Jika: c = m.a, maka
3
2
1
3
2
1
.
.
.c
amamam
aaa
m
dan m = bilangan real
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Contoh
Diketahui:
Vektor x yang memenuhi a – 2x = 3b adalah....Jawab:misal
41
232
61
2
3
2
1
xxx
61-2
a
41-2
b
dan
x
3
2
1
xxx
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41
232
61
2
3
2
1
xxx
123
6
222
61
2
3
2
1
xxx
2 – 2x1 = 6 -2x1 = 4 x1= -2-1 – 2x2 = -3 -2x2 = -2 x2 = 16 – 2x3 = 12 -2x3 = 6 x3 = -3Jadi
3
12
xvektor
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SELAMAT BELAJAR
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