Vektorlar. Ular ustida chiziqli amallar 1-misol
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- VEKTORLARNING ARALASH KO’PAYTMASI c b a , , vektorlarning aralash ko‘paytmasi
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1-misol. a va b lar orasidagi burchak 3 2
bo’lib, 10 =
va 2 = b bo’lsa, ( ) (
) b a b a − + 3 2 ni hisoblang. Skalyar ko’paytmaning xossalariga ko’ra ( ) (
) ( )
= − + = − + = − + 2 2 2 2 2 , cos
5 3 2 5 3 3 2 b b a b a a b b a a b a b a
242 8 50 300 4 2 3 2 cos
2 10 5 100 3 = − − = − + = . ⚫
1. ( )
3 , , 1 , 2 = = = = b a b a ma’lum, b a c 3 2 − = ning modulini toping. 2-misol. Ushbu k j i a 6 6 7 − + = va
k j i b 9 2 6 + + = vektorlar kubning kirralari bo’la oladimi va kubning uchinchi qirrasini toping. a va b kubning qirralari bo’lishi uchun ular o’zaro orthogonal bo’lib, uzunliklari teng bo’lishi kerak. 0
b a -ortoganallik shartiga ko’ra 0 54
42 9 6 2 6 6 7 = − + = − + = b a , bundan b a ⊥
ekan. 11 36 36 49 = + + =
, 11
4 36 = + + = b , ya’ni
b a = . Endi kubning uchunchi ( )
y x c ; ; = qirrasini topamiz. c a ⊥ bo’lgani uchun 0 =
a , ya’ni 0 6
7 = − + z y x ; shu bilan birga c b ⊥ bo’lgani uchun 0 =
b , ya’ni 0 9
6 = + + z y x ; 11 = = = c b a ligidan 121 2
2 = + + z y x . c vektorning koordinatalarni quyidagicha topamiz:
= − = = − = − = = = + + = − = = + + = + + = − + 2 , 2 9 , 9 6 , 6 121
4 81 9 2 9 3 121 0 3 2 6 0 6 6 7 1 2 1 2 1 2 2 2 2 2 2
z y y x x z z z z y z x z y x z y x z y x
( ) ( ) 2 ; 9 ; 6 2 9 6 = − − = k j i c .
( ) (
) 2 ; 1 ; 4 , 1 ; 3 ; 2 − −
A va
( ) 2 ; 0 ; 1 C berilgan. Quyidagilarni toping: a) C uchning ichki burchagi; b)
пр -? a) C uchning ichki burchagi bu ( )
=
CA; . Bu vektorlarning koordinatalarini topamiz: ( ) (
) ( ) ( ) . 3 ; 3 ; 1 2 1 ; 0 3 ; 1 2 , 4 ; 1 ; 3 2 2 ; 0 1 ; 1 4 − = − − − − = − = − − − − = CA CB
Ularning modullarini topamiz: 19 9 9 1 , 26 16 1 9 = + + = = + + =
CB . Bulardan ( ) ( ) 494
18 arccos
, 494
18 19 26 3 4 3 1 1 3 cos = = − − + + = =
CB CA CB . b) Quyidagi o’rinli 19 18 пр = = CA CA CB CB CA .
⚫
( )
) ( ) 12 ; 4 ; 3 , 5 ; 4 ; 1 , 1 ; 6 ; 3 − = − = − − = c b a berilgan bo’lsa, ( )
a + c пр ni toping. 3. ( )
( ) 0 60 , , , , 1 , 1 = = ⊥ = = = b c a c b a c b a shartlarni qanoatlantiruvchi komplanar bo’lmagan b a,
va c vektorlar berilgan bo’lsa. Quyidagilarni toping: a) ( ) ( ) a c b a − − 2 ; b)
( ) 2 c b a + + . VEKTORLARNI VEKTORIAL KO’PAYTIRISH Fazodagi a va b vektorlarning vеktorial ko‘paytmasi:
sin b a c formula bilan aniqlanadi. Agar a va
vektorlar koordinatalari ( )
) z y x z y x b b b b a a a a ; ; , ; ; = =
bilan berilgan bo’lsa, u holda − = = y x y x z x z x z y z y z y x z y x b b a a b b a a b b a a b b b a a a k j i b a ; ; . a va
b vektorlardan tuzilgan parallelogrammning yusasini hisoblash b a S = . 1-misol. ( )
= = = = 6 5 , , 6 , 2
a b a shartlarni qanoatlantiruvchi a va b lar berilgan, quyidagilarni toping: a)
; b)
( ) (
) b a b a 4 3 2 − + . a) 6 2 1 6 2 sin = = = = b a b a S . Shu bilan birga S e b a = bundan
e b a = 6 . b) ( ) ( ) ( ) ( ) (
) ( ) = − + − = − +
b a b b a a a b a b a 12 3 8 2 4 3 2
( ) (
) ( ) ( ) ( ) b a b a b a a b b a − = − − = + − = 11 3 8 3 8 . Bundan
(
) ( ) ( ) 66 6 11 11 11 4 3 2 = = = − = − + b a b a b a b a .
⚫
( )
) 1 ; 2 ; 1 , 2 ; 1 ; 3 − = − − =
a bo’lsa, ( )
a a + 2 ning koordinatasini toping. 2. ( )
3 2 , , 2 , 1 = = = b a b a ma’lum. b a va
( ) (
) a b b a − + 3 2 hisoblang. 2-misol. Uchlari ( ) ( ) ( ) 2 ; 1 ; 2 , 1 ; 2 ; 3 , 0 ; 2 ; 1 −
B A bo’lgan uchburchakning yuzasini toping. ABC uchburchakning yuzasi AB va
AC vektorlardan tuzilgan parallelogram yuzasining yarmiga teng, ya’ni 2
AB S = . Vektorlarni topamiz: ( ) ( ) 2 ; 1 ; 3 , 1 ; 0 ; 2 − − = =
AB . U holda k j i i j k j k j i AC AB 2 7 4 2 3 2 1 3 1 0 2 − − = + − − − = − − = , 54 4 49 1 = + + = AC AB
2 6 3 2 54 =
= S .
⚫
( )
; 2 ; 0 O nuqtaga ( )
; 4 ; 2 − =
kuch qo’yilgan bo’lsin. ( ) 3 ; 2 ; 1 − A nuqtaga nisbatan kuch momentini aniqlang.
( ) 3 ; 2 ; 1 −
nuqtaga nisbatan kuch moment bu
= . OA va izlanayotgan M vektorlarning koordinatalarini topamiz. ( ) 2 ; 0 ; 1 − = OA
( ) 4 ; 9 ; 8 , 4 9 8 8 5 4 4 5 4 2 2 0 1 = + + = + + + = − − =
k j i i j k j k j i M . ⚫ 3. k j i a 5 2 + − = va k j b 7 5 − = vektorlardan yasalgan uchburchak yuzini hisoblang. 4. 30 , 20 , 3 = = = b a b a berilgan, b a ni toping. 5. ( ) 1 ; 2 ; 2 −
= a va
( ) 6 ; 3 ; 2 =
vektorlar orasidagi burchak sinusini toping.
VEKTORLARNING ARALASH KO’PAYTMASI c b a , , vektorlarning aralash ko‘paytmasi: ( ) с b a kabi belgilanadi. c b a , , larda hosil qilingan parallelopiped hajmi: ( ) с b a H S V = = asos
Agar b a , va c vektorlar ( )
) , ; ; , ; ; z y x z y x b b b b a a a a = = ( )
y x c c c c ; ; = bilan berilagan bo’lsin, u holda ( )
y x z y x z y x z y x z y x z y x z y x c c c b b b a a a b b b a a a c c c Zc Yc Xc c b a = = + + =
Uchta vektorning komplanarlik sharti: 0 = z y x z y x z y x c c c b b b a a a
( )
y x a a a a , , = , ( ) z y x b b b b , , = va
( )
y x c c c c , , = vektorlardan yasalgan parallelepipedning V hajmi: z y x z y x z y x c c c b b b a a a V = . ( ) z y x a a a a , , = , ( ) z y x b b b b , , = va
( )
y x c c c c , , = vektorlardan tuzilgan piramidaning V hajmi: z y x z y x z y x c c c b b b a a a V 6 1 =
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