L
vektor aylanma harakatning tezligini,
a
H
vektor esa tezlanishni ifodalashi mumkin. Biroq bu vektorlarni fizika nuqtayi
nazaridan qoshish manoga ega emas.
Shunday bolsa-da, fizikada tezlik yoki tezlanishlarni vektorlar deb tog-
ridan-togri aytiladi. Gap nima togrisida ketayotganligi aniq tasavvur qilinsa,
bunday soz erkinligi umumiylikka hech bir ziyon keltirmaydi. Xuddi shunga
oxshash biz oz vaqtida uchburchak tomonining uzunligini, qisqalik uchun, oddiy-
gina qilib uning tomoni deb aytishga kelishib olgan edik va hokazo.
47- m a v z u .
VEKTORLARNING FIZIK VA GEOMETRIK
TALQINLARI
149
573.
Quyidagi davo togrimi: ixtiyoriy ikkita
a
H
va
b
H
vektor
a
k
b
H
H
⋅
=
teng-
likni qanoatlantiradigan
k
son mavjud bolganda va faqat shundagina
kollinear boladi?
574.
Trapetsiyaning orta chizigi asoslariga parallel va ular uzunligining
yarmiga teng ekanini isbot qiling.
575.
AB
C
uchburchak berilgan.
A
1
,
B
1
,
C
1
uchburchak
B
C
,
A
C
va
AB
to-
monlarining ortalari,
O
tekislikning ixtiyoriy nuqtasi. Quyidagi
tenglikni isbotlang:
1
1
1
O
A
O
B
O
C
O
A
O
B
O
C
+
+
=
+
+
KKKH KKKH KKKH
KKKH KKKKH KKKKH
.
576.
D
v a
E
nuqtalar
AB
C
uchburchak
AB
v a
B
C
tomonlarining ortalari.
(
)
=
−
1
2
D
E
BC BA
KKKH
KKKH KKKH
ekanini isbotlang.
577.
K
nuqta
AB
CD
parallelogramm
A
D
tomonining ortasi.
KKKKH
K
C
vektorni
KKKH
AB
va
KKKKH
A
D
vektorlar orqali ifodalang.
578.
B
(4; 2) nuqta
a
H
(
−
2; 3) vektorning oxiri bolsa, bu vektor boshi
(
A
) ning koordinatalarini toping.
579.
A
(
−
2; 3) nuqta
a
H
(
−
3; 8) vektorning boshi bolsa, bu vektor oxiri
(
B
) ning koordinatalarini toping.
580.
Agar: 1)
A
(0; 1),
B
(1; 0); 2)
A
(
−
2; 1),
B
(
−
4; 3) bolsa,
KKKH
AB
vektor-
ning koordinatalari nimaga teng boladi?
581.
a
H
(
−
4; 4) va
b
H
(
−
4; 5) vektorlar berilgan.
= −
H
H
H
c
a
b
vektorning koor-
dinatalarini toping.
582.
A
(2; 4),
B
(3; 6) va
C
(6; 14) nuqtalar berilgan.
+
KKKH KKKKH
AB A
C
vektorning
koordinatalarini toping.
583.
= − −
H H
H
5
a
i
j
v a
= −
H
H
1,
5
b
j
vektorlar berilgan. 1)
= +
H
H
H
4
c
a
b
;
2)
= −
+
H
H
H
2
3
c
a
b
vektorning koordinatalarini toping.
584.
Agar: 1)
a
H
(2; 1) va
b
H
(
−
3; 4);
2)
a
H
(2;
−
0,5) va
b
H
(3; 2) bolsa,
a
H
va
b
H
vektorlar skalar kopaytmasini toping.
a
H
b
H
c
H
A
247
a
H
H
L
248
5- § ga doir qoshimcha mashqlar
150
585.
Tekislikda tortta
A
,
B
,
C
va
D
nuqtalarni belgilang. Isbotlang:
+
=
+
KKKH KKKKH KKKKH KKKKH
AB B
C
A
D DC
. Xuddi shunga oxshash tenglik tuzing.
586.
Agar: 1)
A
(0; 1) va
B
(1; 0); 2)
A
(
−
2; 1) va
B
(
−
4; 2); 3)
A
(
−
3;
−
1) va
B
(
−
3;
−
12); 4)
A
(
p
;
q
) va
B
(
−
p
;
−
q
) bolsa,
KKKH
AB
vektorning koor-
dinatalarini va uzunligini toping.
587.
a
H
(1; 3),
H
b
(
−
2; 4),
H
c
(
−
1;
−
3),
KH
d
(
−
4; 4),
KH
p
(3; 9),
H
q
(
−
1; 2) vektorlar
berilgan. Shular ichidan: 1) yonalishdosh vektorlarni; 2) bir juft
qarama-qarshi yonalgan vektorlarni toping.
1.
AB
CD
parallelogramm.
O
A
C
va
B
D
diagonallarning kesishish nuqtasi.
+
KKKKH KKKH
B
C
O
A
ni toping.
A)
KKKKH
O
C
;
B)
KKKKH
B
O
;
D)
KKKH
O
B
;
E)
KKKH
C
O
.
2.
MKP
C
parallelogramm.
E
MP
va
K
C
diagonallarning kesishish nuqtasi.
−
KKKKKH KKKH
MK
E
P
ni toping.
A)
KKKKKH
MK
;
B)
KKKKH
K
C
;
D)
KKKH
C
E
;
E)
KKKKH
E
K
.
3.
PE
kesma
MPK
uchburchakning medianasi.
−
KKKKH KKKKH
E
K MP
ni toping.
A)
KKKKH
PK
;
B)
KKKH
P
E
;
D)
KKKH
E
P
;
E)
KKKH
KP
.
4.
A
D
AB
C
uchburchakning medianasi.
−
KKKH KKKKH
C
A
D
B
ni toping.
A)
KKKH
BA
;
B)
KKKH
AB
;
D)
KKKH
D
A
;
E)
A
D
KKKH
.
5.
a
H
(7; 3) va
H
b
(5; 2) vektorlar berilgan. |
H
a
+
H
b
| ni hisoblang.
A) 9;
B) 5;
D) 8;
E) 13.
6.
A
(2; 4),
B
(3; 6) va
C
(6; 14) nuqtalar berilgan.
+
KKKH KKKKH
|
|
AB A
C
ni hisoblang.
A) 14;
B) 12;
D) 10;
E) 13.
7.
a
H
(
−
3; 1) va
H
b
(5;
−
6) vektorlar berilgan.
= −
H
H
H
3
c
b
a
vektorning koordina-
talarini toping.
A) (14; 9);
B) (4; 3);
D) (14; 3);
E) (9; 3).
8.
A
(
−
3; 0) va
B
(
−
5; 4) nuqtalar berilgan.
KKKH
BA
vektorning koordinatalarini
toping.
A) (
−
8;
−
4);
B) (
−
8; 4);
D) (2;
−
4);
E) (8;
−
4).
9.
−
H
(
2
;
3
)
a
v a
− −
H
(
2
;
3
)
b
vektorlar berilgan.
= −
H
H
H
2
m
a
b
vektorning koordi-
natalarini toping.
A) (
−
3; 6);
B) (6; 3);
D) (2;
−
3);
E) (
−
2;
−
9).
7- TEST
151
10.
H
(
3
;
2
)
a
va
−
H
(
0
;
1
)
b
vektorlar berilgan.
−
+
H
H
2
4
a
b
vektorning modulini toping.
A) 10;
B) 6;
D) 8;
E) 3.
11.
Ifodani soddalashtiring:
−
−
KKKKH KKKH KKKKH
A
D CD
A
C
.
A)
KH
O
;
B)
KKKH
D
A
;
D)
KKKKH
2
A
C
;
E)
KKKH
C
A
.
12.
Ifodani soddalashtiring:
−
+
KKKKH KKKKH KKKKH
AK B
C
K
C
.
A)
KH
O
;
B)
KKKH
AB
;
D)
KKKKH
2
K
C
;
E)
KKKKH
A
C
.
13.
Ifodani soddalashtiring:
−
−
KKKH KKKKH KKKH
C
B A
C
BA
.
A)
KH
O
;
B)
KKKKH
B
C
;
D)
KKKH
2
C
B
;
E)
KKKH
C
A
.
14.
Ifodani soddalashtiring:
+
+
KKKH KKKKH KKKH
C
B A
C
BA
.
A)
KH
O
;
B)
KKKH
C
A
;
D)
KKKKH
A
C
;
E)
BA
KKKH
.
Vektor tushunchasi XIX asrning ortalarida bir vaqtda bir nechta mate-
matikning ishlarida uchraydi. Tekislikda vektorlar bilan ish korishni ilk bor
1835-yili italiyalik olim
Bellivitis
(18031880) boshlab berdi. Bundan tashqari,
K. Gauss
(17771855) 1831- yili «Bikvadratik solishtirmalar nazariyasi» nomli
asarida hamda
Y. Argan
(17681822) va
K. Vessel
(17451818)ning kompleks
sonlarni geometrik tasvirlashga doir ishlarida vektor tushunchasi aytib otil-
gan. Nihoyat,
V. Gamilton
(18051865) va
R. Grassman
(18541901)larning
vektorlar ustida amallar bajarishga doir ishlari vujudga keldi. Birinchi bolib
Gamilton vektor va skalar kattaliklarni farq qilishni tushuntirdi. Gamilton-
ning osha ishida «skalar», «vektor» atamalari yuzaga keldi. «Vektor» ata-
masini Gamilton lotincha
vehere
«tashimoq», «sudramoq» sozidan hosil
qilgan (1845),
vektor
«tashuvchi», «eltuvchi» demakdir.
1806- yili Argan yonalgan kesmalarni harf ustiga chiziq qoyish bilan
belgilagan. Vektorlarning boshi va oxirini korsatish uchun
A. Myobius
(17901868) uni
AB
korinishda belgilagan. Grassman vektorlarni «kesmalar»
deb atagan, u koordinata oqlari boyicha yonalgan
e
1
,
e
2
birlik vektorlarni
va vektorlarni
x
1
e
1
+
x
2
e
2
korinishida tasvirlashni tavsiya qilgan. Gamilton va
J. Gibbs
(18391903) vektorlarni yunoncha harflar bilan belgilagan. Vektorlar-
ni qora harflar bilan belgilashni 1891- yili
A. Xevisayd
(18501925) taklif etgan.
Vektorning uzunligini |
AB
| korinishda belgilashni 1905- yili
R. Gans
(1880)
kiritgan. «Modul» sozini 1814- yili lotincha
modulus
«olchov» sozidan
Argan hosil qilgan. Keyinchalik uni
A. Koshi
(17891857) ishlatgan. Bu atama
XX asrda keng qollanila boshlangan.
T a r i x i y m a l u m o t l a r
152
8- SINFDA OTILGAN MAVZULARNI TAKRORLASH UCHUN
MASHQLAR
588.
ABCD
parallelogrammda: 1) agar
BC
tomon
AB
dan 8 sm uzun, pe-
rimetri esa 64 sm ga teng bolsa, tomonlarni; 2) agar
.
A
=
55° bolsa,
burchaklarni toping.
589.
Agar parallelogrammning perimetri 2 m ga teng va: 1) qoshni tomonlari
ayirmasi 1 sm ga teng; 2) qoshni tomonlarining nisbati 2 ga teng;
3) ikkita teng yonli uchburchaklardan tashkil topgan bolsa, parallelo-
gramm tomonlari nimaga teng?
590.
ABCD
parallelogramm
A
burchagining bissektrisasi
BC
tomonni
P
nuq-
tada kesadi va shu bilan birga
BP
=
PC
. Agar parallelogrammning
perimetri 42 sm ga teng bolsa, uning tomonlarini toping.
591.
Ikkitta
ABCD
va
ANCP
parallelogrammni yasang.
1)
AC
,
BD
va
NP
kesmalar bir nuqtada kesishishini isbotlang.
2)
BNDP
tortburchak parallelogramm ekanini isbotlang.
592.
Agar tortburchakning ikki juft teng tomonlari bolsa, bu tortburchak
har doim ham parallelogramm boladimi?
593.
Parallelogramm burchaklaridan birining bissektrisasi ozi kesib otadigan
tomonni 7 sm va 9 sm li kesmalarga boladi. Shu parallelogrammning
perimetrini toping.
594.
Togri tortburchak diagonallarining kesishish nuqtasidan uning tomon-
lariga otkazilgan perpendikularlar, mos ravishda, 5 sm va 7 sm ga teng.
Bu togri tortburchakning perimetri va yuzini toping.
595.
Togri tortburchak diagonallarining kesishish nuqtasidan uning tomon-
lariga otkazilgan perpendikularlar, mos ravishda, 4 sm va 6 sm ga teng.
Bu togri tortburchakning perimetrini va yuzini toping.
596.
1)
ABCD
parallelogrammda
∠
A
=
75°. Parallelogrammning qolgan bur-
chaklari nimaga teng?
2) Parallelogrammning ikkita qarama-qarshi burchaklarining yigindisi
220° ga teng. Shu parallelogrammning burchaklari nimaga teng?
597.
Agar
ABCD
rombda
∠
B
=
100° va
AB
=
15 sm bolsa, uning perimetri va
burchaklarini toping.
598.
Togri tortburchak diagonallarining kesishish nuqtasidan uning tomon-
lariga otkazilgan perpendikularlar, mos ravishda, 4 sm va 11 sm ga teng.
Bu togri tortburchakning yuzini toping.
599.
ABCD
romb berilgan.
AC
va
BD
diagonallar, mos ravishda, 30 sm va
12 sm ga teng. Rombning yuzini toping.
600.
1)
ABCD
teng yonli trapetsiyada
BC
=
20 sm,
AB
=
24 sm va
.
D
=
60°
bolsa, uning
AD
asosini toping.
2) Teng yonli trapetsiyaning burchaklaridan biri 105° ga teng. Tra-
petsiyaning qolgan burchaklarini toping.
153
601.
Trapetsiyaning ketma-ket olingan burchaklarining nisbati quyidagicha
bolishi mumkinmi: 1) 7 : 4 : 3 : 5; 2) 8 : 7 : 13 : 14?
602.
Togri burchakli trapetsiyaning asoslari
a
va
b
ga, burchaklaridan biri
esa
α
ga teng. Agar: 1)
a
=
7 sm,
b
=
4 sm,
α =
60° bolsa, katta yon to-
monni toping; 2)
a
=
15 sm,
b
=
10 sm,
α =
45° bolsa, kichik yon to-
monni toping.
603.
Parallelogrammning yuzi 40 sm
2
ga, tomonlari esa 10 sm va 8 sm ga
teng. Shu parallelogrammning ikkala balandligini toping.
604.
ABCD
rombning diagonallari 15 sm va 36 sm ga teng.
AC
diagonalida
P
nuqta shunday olinganki, unda
AP
:
PC
=
4 : 1 nisbatda.
APD
uch-
burchakning yuzini toping.
605.
Teng yonli togri burchakli uchburchakning gipotenuzasi 20 sm ga
teng. Shu uchburchakning yuzini toping.
606.
Teng yonli trapetsiyaning perimetri 32 sm ga, yon tomoni 5 sm ga, yuzi
esa 44 sm
2
ga teng. Trapetsiyaning balandligini toping.
607.
Togri burchakli trapetsiyaning yuzi 120 sm
2
ga, perimetri 56 sm ga,
kichik yon tomoni esa 6 sm ga teng. Katta yon tomonini toping.
608.
ABCD
togri tortburchak
C
uchining bissektrisasi
AD
tomonni
P
nuq-
tada kesadi. Agar
AP
=
10 sm,
PD
=
14 sm ga teng bolsa, shu togri
tortburchakning yuzini toping.
609.
Togri tortburchak bilan parallelogramm bir asosga va bir xil perimetr-
ga ega. Shu parallelogramm bilan togri tortburchakning yuzlarini taq-
qoslang.
610.
Uchburchakning tomonlari 21, 72 va 75 ga teng. Shu uchburchakning
yuzini toping.
611.
ABC
da
AE
va
BD
balandliklar.
AC
=
20 sm,
BD
=
16 sm va
BC
=
32 sm.
AE
ni toping.
612.
Teng yonli trapetsiyaning diagonali 50 sm ga, balandligi esa 30 sm ga
teng. Shu trapetsiyaning yuzini toping.
613.
Aylanaga ichki chizilgan
BAC
burchak 45° ga teng, u
BC
yoyga tiraladi.
BOC
burchakni toping, bunda
O
aylana markazi.
614.
Togri burchakli
ABC
uchburchakda (
∠
C
=
90°)
∠
A
=
30°,
!
AC
=
.
Markazi
A
nuqtada va radiusi 2,2 ga teng bolgan aylana otkazilgan.
Shu aylana
BC
tomon bilan nechta umumiy nuqtaga ega?
615.
Tashqi chizilgan tortburchakning ikkita qarama-qarshi tomonlarining
yigindisi 35 sm ga teng. Shu tortburchakning perimetrini toping.
616.
Biror
ABCD
parallelogrammni chizing. Vektorlarni yasang:
1)
AB BC
+
KKKH KKKH
;
2)
AD DC
+
KKKH KKKKH
;
3)
−
KKKH KKKKH
AB AD
;
4)
−
KKKKH KKKH
DB DA
.
617.
Quyidagi vektorlar kollinearmi: 1)
a
H
(2; 1) va
b
H
(4; 2); 2)
a
H
(1; 3) va
b
H
(1; 3); 3)
a
H
(3; 2) va
b
H
(3; 2); 4)
a
H
(0; 1) va
b
H
(1; 0)?
618.
Vektorlar yigindisini toping:
+
+
+
+
+
KKKKH KKKKH KKKH KKKKH KKKKKH KKKKH
B
H HK T
P
MT KM
P
Q
.
619.
KKKH
F
K
vektorni
KKKH
E
F
va
KKKKH
EK
vektorlar orqali ifodalang.
620.
A
(
−
1; 2),
B
(
−
4;
−
2),
C
(
−
1; 3),
D
(
−
4;
−
2) bolsin. Hisoblang:
1)
⋅
KKKH KKKH
AB CD
;
2)
⋅
KKKKH KKKH
AC BD
;
3)
⋅
KKKKH KKKKH
AD BC
;
4)
⋅
KKKH KKKKH
CA DB
.
1.
Togri burchakli uchburchakning katetlari 6 va 8 ga teng. Uning gipote-
nuzasiga tushirilgan balandligini toping.
A) 4,8;
B) 5;
D) 4,5;
E) 4,7.
2.
Tortburchakning burchaklari ozaro 3 : 5 : 4 : 6 nisbatda. Tortburchakning
kichik burchagini toping.
A) 80°;
B) 30°;
D) 60°;
E) 40°.
3.
Qavariq tortburchakning diagonallari uni nechta uchburchakka ajratadi?
A) 4;
B) 5;
D) 6;
E) 8.
4.
Togri tortburchakning eni 5 ga teng, boyi undan 7 ga ortiq. Togri
tortburchakning perimetrini hisoblang.
A) 32;
B) 34;
D) 24;
E) 26.
5.
Har bir ichki burchagi 156° bolgan qavariq kopburchakning nechta to-
moni bor?
A) 10;
B) 15;
D) 6;
E) 12.
6.
Togri tortburchakning boyi 20% va eni 10% ga orttirilsa, uning yuzi
necha foiz ortadi?
A) 20%;
B) 35%;
C) 27%;
D) 32%.
7.
Rombning yuzi 24 ga, diagonallaridan biri esa 6 ga teng. Uning tomonini
toping.
A) 10;
B) 5;
C) 8;
D) 4,8.
8.
Rombning balandligi 5 ga, diagonallarining kopaytmasi 80 ga teng. Uning
perimetrini toping.
A) 32;
B) 16;
C) 24;
D) 28.
9.
Do'stlaringiz bilan baham: |