143
Masalan,
KKKH
OA
vektorning
koordinatalari
vektor oxiri
A
ning koordinatalari bilan tola
aniqlanadi, yani vektor oxirining koordina-
talariga teng boladi.
Agar
A
(
x
;
y
) bolsa,
OA
KKK
H
(
x
;
y
)
boladi.
1 - x u l o s a .
Agar vektor oxirining koor-
dinatalari vektorning koordinatalari bilan teng
bolsa, u holda
berilgan vektorning boshi
koordinatalar boshida boladi (244-b rasm).
2 - x u l o s a .
Agar
H
a
(
a
1
;
a
2
) vektor bilan
uning oxiri bolgan
B
(
x
2
;
y
2
) nuqtasi koor-
dinatalari berilgan bolsa, u holda vektor boshi
A
(
x
1
;
y
1
)
nuqtaning koordi-
natalarini topish uchun
B
nuqtaning koordinatalaridan
H
a
(
a
1
;
a
2
) vektorning mos
koordinatalarini ayirish kifoya:
x
1
=
x
2
−
a
1
;
y
1
=
y
2
−
a
2
.
3- x u l o s a .
Agar
H
a
(
a
1
;
a
2
) vektor bilan uning boshi bolgan
A
(
x
1
;
y
1
) nuq-
tasi koordinatalari berilgan bolsa, u holda vektor oxiri
B
(
x
2
;
y
2
) nuqtaning koor-
dinatalarini topish uchun
A
nuqtaning koordinatalariga
H
a
(
a
1
;
a
2
) vektorning mos
koordinatalarini qoshish kifoya:
x
2
=
x
1
+
a
1
;
y
2
=
y
1
+
a
2
.
M a s a l a .
A
(
−
1; 5) nuqta
H
a
(2;
−
3) vektorning boshi bolsa, bu vektor oxiri
B
ning koordinatalarini toping.
Y e c h i l i s h i
.
Berilgan malumotlarni songgi munosabatlarga qoyib, izla-
nayotgan koordinatalarni topamiz:
x
2
=
−
1
+
2
=
1,
y
2
=
5
+
(
−
3)
=
2.
J a v o b :
B
(1; 2).
549.
1) Koordinatalar oqidagi birlik vektorlar qanday belgilanadi?
2) Boshi koordinatalar boshida bolgan
vektorning koordinatalari
nimaga teng?
550.
Vektorlarning koordinatalarini yozing:
1)
=
H
H
H
4 5
a
i
j
;
2)
=
+
H
H
H
4
5
a
i
j
; 3)
= −
H
H
7
b
j
;
4)
= −
H
H
3
c
i
.
551.
1)
A
(2; 5) va
B
(4; 2); 2)
A
(3;
−
4) va
B
(1;
−
6); 3)
A
(
−
5;
−
3) va
B
(
−
1; 3)
nuqtalar berilgan.
KKKH
AB
vektorning koordinatalarini toping.
552.
1)
A
(
−
3; 0) va
B
(5;
−
4); 2)
A
(0;
−
4) va
B
(7;
−
2) nuqtalar berilgan.
KKKH
BA
va
KKKH
AB
vektorlarning koordinatalarini toping.
553.
B e r i l g a n :
A
(1;
−
1),
B
(2; 0),
C
(
−
1; 3). Agar: 1)
=
KKKH KKKKH
B
D
A
C
;
2)
=
KKKKH KKKKH
A
D
B
C
bolsa,
D
nuqtaning koordinatalarini toping.
y
A
2
y
2
y
1
A
1
y
2
–
y
1
x
x
1
x
2
x
2
–
x
1
O
245
Savol, masala va topshiriqlar
144
554.
A
(5;
−
3) nuqta
a
H
(
−
7;
−
8) vektorning boshi bolsa, bu vektor oxiri
(
B
) ning koordinatalarini toping.
555.
A
(
−
1;
−
3),
B
(2;
−
4),
C
(
−
3;
−
1) va
D
(5; 2) nuqtalar berilgan.
KKKKH
A
C
va
KKKKH
D
B
vektorlar tengmi?
556.
Agar: 1)
A
(
−
2;
−
3),
B
(
−
3;
−
1); 2)
A
(
m
;
n
),
B
(
−
m
;
−
n
) bolsa,
KKKH
BA
vektorning koordinatalari nimaga teng boladi?
Bizga
a
H
(
x
1
,
y
1
) va
b
H
(
x
2
,
y
2
), yani vektorlar koordinatalari
bilan berilgan
bolsin. Koordinatalari bilan berilgan vektorlarni qoshish, ayirish va songa
kopaytirish amallari bilan tanishamiz.
1. Koordinatalari bilan berilgan vektorlarni qoshish.
Avval sodda holni qaraylik.
a
H
va
b
H
vek-
torlar
Ox
oqiga kollinear bolsin. Bunda
y
1
=
y
2
=
0,
a
H
(
x
1
)
=
x
1
·
i
H
v a
b
H
(
x
2
)
=
x
2
·
i
H
(246- rasm).
Bu yerda
a
H
+
b
H
vektorning
moduli
a
H
va
b
H
vektorlarning modullari yigindisiga
teng boladi va
a
H
+
b
H
vektor ham
Ox
oqi-
ga kollinear. Shuning uchun
a
H
+
b
H
=
(
x
1
+
x
2
)
i
H
.
Demak, yigindi vektor
a
H
+
b
H
vektorning koordinatasi qoshiluvchi
a
H
va
H
b
vektorlarning mos koordinatalari yigindisiga teng ekan. Kollinear vektorlarni
qoshish uchun ularning mos koordinatalarini qoshish kifoya.
Endi
ixtiyoriy
a
H
(
x
1
,
y
1
) va
b
H
(
x
2
,
y
2
) vektorlar yigindisini koraylik:
+ =
⋅ +
⋅
+
⋅ +
⋅
= ⋅ +
⋅ +
⋅ +
⋅ =
H
H
H
H
H
H
H
H
H
H
1
1
2
2
1
1
2
2
(
) (
)
a
b
x i
y j
x i
y
j
x i
y j
x i
y
j
=
+
+
+
H
H
2
1
2
(
)
(
)
x
x i
y
y j
.
Demak,
+
H
H
a
b
vektorning koordinatalari
+
+
1
2
1
2
(
;
)
x
x
y
y
ga teng.
Shunday qilib,
vektorlarni qoshish uchun ularning mos koordinatalarini qo-
shish kifoya ekan.
1- m a s a l a .
a
H
(3; 5) va
b
H
(2, 7) vektorlar yigindisini toping.
Y e c h i l i s h i
.
(3
;
5) 3
5
a
i
j
=
+
H
H
H
;
(
2
;
7)
2
7
b
i
j
=
+
H
H
H
;
(
)
(
)
3
2
5 7
5
1
2
a
b
i
j
i
j
+ =
+
+
+
=
+
H
H
H
H
H
H
.
4 5- m a v z u . KOORDINATALARI BERILGAN VEKTORLAR
USTIDA AMALLAR
y
x
O
i
H
j
H
a
H
(
x
1
)
b
H
(
x
2
)
246
145
Demak,
+
H
H
a
b
vektorning koordinatalari (5; 12) ga teng.
Bu masala yechimini koordinatalar tekisligida tekshirib koring.
2. Koordinatalari bilan berilgan vektorlarni ayirish.
Koordinatalari bilan berilgan vektorlarni ayirish
uchun ularning mos koordi-
natalarini ayirish kifoya, yani:
(
) (
) (
)
−
=
−
−
H
H
H
1
1
2
2
1
2
1
2
;
;
;
a
x
y
b
x
y
c
x
x
y
y
.
2- m a s a l a .
a
H
(
−
3; 5) va
b
H
(3; 3) vektorlar ayirmasini toping.
Y e c h i l i s h i
.
(
)
(
)
(
)
(
)
−
−
−
=
− −
− −
=
−
H
H
H
H
3;5
3; 3
3 3; 5 ( 3)
6; 8
a
b
c
c
.
3. Koordinatalari bilan berilgan vektorni songa kopaytirish.
Koordinatalari bilan berilgan vektorni songa kopaytirish amali bilan tani-
shamiz.
a
H
(
x
1
,
y
1
) vektorning
k
songa kopaytmasi
b
H
=
k
a
H
ni topamiz:
b
H
=
(
)
⋅ = ⋅
+
=
+
=
H
H
H
H
H
H
1
1
1
1
1
1
(
)
;
.
k
a
k
x
i
y j
k
x
i
k
y j
b
k
x
k
y
Demak,
vektorni songa kopaytirish uchun uning
koordinatalarini shu songa
kopaytirish yetarli ekan.
3- m a s a l a .
a
H
(3; 5) vektorga qarama-qarshi
b
H
vektorni toping.
Y e c h i l i s h i
.
a
H
vektorga qarama-qarshi
b
H
vektor quyidagiga teng:
b
H
= −
a
H
=
(
−
1)
a
H
= −
1 ·
a
H
(3; 5)
=
b
H
(
−
1 · 3;
−
1 · 5)
=
b
H
(
−
3;
−
5).
Demak,
a
H
(3; 5) va
b
H
(
−
3;
−
5) vektorlar qarama-qarshi vektorlardir.
Umuman:
= − = −
⋅ +
⋅
= − ⋅ −
⋅ = −
−
H
H
H
H
H
H
H
1
1
1
1
1
1
(
)
(
;
)
b
a
x i
y j
x i
y j
b
x
y
.
4- m a s a l a .
Agar
a
H
(
−
3; 4) bolsa,
b
H
=
4
a
H
vektorning
koordinatalarini
toping.
Y e c h i l i s h i
.
b
H
=
4
a
H
=
4 ·
a
H
(
−
3; 4)
=
b
H
(4 · (
−
3); 4 · 4)
=
b
H
(
−
12; 16).
557.
1) Vektorning koordinatalari deganda nimani tushunasiz?
2) Koordinatalari berilgan vektorlar ustida chiziqli amallar qanday baja-
riladi?
558.
Agar
a
H
(
−
4; 8) va
b
H
(1;
−
4) bolsa, shu vektorlar: 1) yigindisining;
2) ayirmasinining koordinatalarini toping.
559.
a
H
(
−
2; 6) va
b
H
(
−
2; 4) vektorlar berilgan. 1)
a
H
+
b
H
; 2)
a
H
−
b
H
;
3)
b
H
−
a
H
; 4)
−
a
H
−
b
H
vektorning koordinatalarini toping.
Savol, masala va topshiriqlar
146
560.
a
H
(2; 3) va
b
H
(
−
1; 0) vektorlar berilgan. 1) 2
a
H
+
b
H
; 2)
a
H
−
3
b
H
;
3) 2
b
H
−
a
H
; 4)
−
2
b
H
−
4
a
H
vektorning koordinatalarini toping.
561.
a
H
(2;
−
3) va
b
H
(
−
2;
−
3) vektorlar berilgan. 1)
2
c
a
b
= −
H
H
H
; 2)
= −
+
2
c
a
b
H
H
H
;
3)
= −
−
H
H
H
3
2
c
a
b
vektorning koordinatalarini toping.
562.
= −
−
2
3
a
i
j
H
H
H
va
= −
2
b
j
H
H
vektorlar berilgan.
1)
2
c
a
b
=
−
H
H
H
; 2)
= −
+
H
H
H
4
3
c
a
b
vektorning koordinatalarini toping.
563.
j
i
a
H
H
H
2
2
+
−
=
va
3
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