Oliy matematika kafedrasi
Berilgan ikki nuqtadan o’tuvchi to’g’ri chiziq tenglamasi
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fazoda togri chiziq va uning tenglamalari
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4. Berilgan ikki nuqtadan o’tuvchi to’g’ri chiziq tenglamasi. ) , , ( 1 1 1 1 z y x M va
) , , ( 2 2 2 2
y x M
o’tuvchi to’g’ri chiziq tenglamasi tekislikda berilgan ikki nuqtadan o’tuvchi to’g’ri chiziq tenglamasidagidek ushbu ko’rinishda
1 2 1 1 2 1 1 2 1 z z z z y y y y x x x x − − = − − = − −
(6) bo’ladi. 3-misol. Uchburchakning uchlari
) 1 , 2 , 3 ( −
, )
, 5 , 6 ( − B
va ) 3 , 4 , 5 ( −
berilgan.
mediananing kanonik tenglamasini yozing. Yechish.
nuqta
AC tomonni teng ikkiga bo’ladi. Kesmani berilgan nisbatda bo’lish formulasiga asosan:
16
2 2 3 1 , 3 2 4 2 , 4 2 5 3 = + = − = − − = + = D D D z y x . Demak, ) 2 , 3 , 4 ( −
bo’ladi. Mediana
va
D nuqtalardan o’tadi. (6) formulaga asosan: 7 2 7 5 3 5 6 4 6 + + = − − − = − − z y x
yoki 9 7 8 5 2 6 + = − − = − − z y x . Bu BD mediananing kanonik tenglamasidir. 1. Fazoda berilgan ikki to’g’ri chiziq orasidagi burchak va ularning parallellik hamda perpendikulyarlik shartlari. Ikki to’g’ri chiziq orasidagi burchak. Fazoda ikkita to’g’ri chiziq kanonik tenglamalari bilan berilgan bo’lsin:
; 1 1 1 1 1 1
z z n y y m x x − = − = − . 2 2 2 2 2 2
z z n y y m x x − = − = − Bu
to’g’ri chiziqlar orasidagi burchak, ularning yo’naltiruvchi vektorlari orasidagi burchakka teng bo’lib,
2 2 2 2 2 2 2 1 2 1 2 1 2 1 2 1 2 1 cos p n m p n m p p n n m m + + ⋅ + + + + = ϕ
(7) formula yordamida topiladi. Berilgan to’g’ri chiziqlar parallel bo’lsa,
2 1
1 2 1 p p n n m m = =
8) bo’lib, bu fazoda ikki to’g’ri chiziqning parallellik sharti deyiladi. To’g’ri chiziqlar perpendikulyar bo’lsa, yo’naltiruvchi vektorlar ham perpendikulyar bo’lib,
0 2
2 1 2 1 = ⋅ + ⋅ + ⋅ p p n n m m
(9) bo’ladi, bu ikki to’g’ri chiziqning perpendikulyarlik shartidir.
4-misol. 1 4 4 3 7 8 1 2 5 3 7 5 − − = − + = − − = − = − z y x va z y x
to’g’ri chiziqlar orasidagi burchakni toping. Yechish. Oldin to’g’ri chiziqlarning yo’naltiruvchi vektorlarini topamiz: → →
→ → → → → − − = + + =
j i s k j i s 4 7 , 5 7 2 1 . To’g’ri chiziqlar orasidagi burchak ularning yo’naltiruvchi 17
vektorlari orasidagi burchakka teng. (7) formulaga asosan: . 3127 . 0 cos , 3127
. 0 22 15 28 ) 1 ( ) 4 ( 7 1 5 7 ) 1 ( 1 ) 4 ( 5 7 7 cos
2 2 2 2 2 2 ≈ ≈ = − + − + ⋅ + + − ⋅ + − ⋅ + ⋅ = α ϕ
Jadvaldan 1 0 48 71 ≈ ϕ ekanligini topamiz. 5-misol. ) 3 , 1 , 2 ( 0 − M nuqtadan o’tib,
4 2 3 5 2 4 − = − = + z y x
to’g’ri chiziqqa parallel to’g’ri chiziqning kanonik tenglamasini yozing. Yechish. Izlanayotgan to’g’ri chiziq yo’naltiruvchi vektori uchun berilgan to’g’ri chiziq yo’naltiruvchi vektorini olish mumkin, chunki ular shartga ko’ra parallel, ya’ni → → → → + + =
j i s 4 3 2 yo’naltiruvchi vektor bo’ladi. Berilgan nuqtadan o’tib, →
yunaltiruvchi vektorga ega bo’lgan, izlanayotgan to’g’ri chiziq tenglamasi (3) ga asosan,
4 3 3 1 2 2 − = + = − z y x
bo’ladi. 6. Fazoda to’g’ri chiziq va tekislik orasidagi burchak hamda ularning parallellik va perpendikulyarlik shartlari. Fazoda to’g’ri chiziq va tekislik orasidagi burchak. Fazoda to’g’ri chiziq va tekislik orasidagi burchak
b, to’g’ri chiziqning tekislikdagi proyeksiyasi bilan to’g’ri chiziq orasidagi qo’shni burchaklardan biri olindi (14.2-chizma).
14.2-chizma. To’g’ri chiziq p z z n y y m x x 0 0 0 − = − = − kanonik tenglamasi bilan tekislik 0 =
+ +
Cz By Az umumiy tenglamasi bilan n s 1 ϕ 2 ϕ 18
berilgan bo’lsin. 1 ϕ burchakni topish uchun to’g’ri chiziqning yo’naltiruvchi vektori → →
→ + + = k p j n i m s vektor bilan tekislikning normal vektori orasidagi 2 ϕ burchakni hisoblaymiz:
2
2 2 2 2 2 cos n p m C B A Cp Bn Am + + ⋅ + + + + = ϕ . 1 ϕ burchak
2
burchakni 2 / π gacha to’ldiradi. Demak, 1 1
sin ) 2 / ( cos cos ϕ ϕ π ϕ = − =
Shunday qilib, 2 2 2 2 2 2 1 sin n p m C B A Cp Bn Am + + ⋅ + + + + = ϕ
(10) bo’ladi. (10) fazoda to’g’ri chiziq va tekislik orasidagi burchakni topish formulasi bo’ladi. To’g’ri chiziq tekislikka parallel bo’lsa ( )
n m s , , → va
( )
B A n , , →
vektorlar perpendikulyar bo’lib,
0 =
+ Cp Bn Am
(11) tenglik o’rinli bo’ladi. (11) tenglikka to’g’ri chiziq va tekislikning parallellik sharti deyiladi. To’g’ri chiziq tekislikka perpendikulyar bo’lsa, ( )
n m s , , → va
( )
B A n , , → vektorlar parallel bo’ladi va
p C n B m A = =
(12) munosabat kelib chiqadi. (12) tenglik to’g’ri chiziq va tekislikning perpendikulyarlik sharti bo’ladi. (11) shart bajarilmasa to’g’ri chiziq va tekislik kesishadi. Kesishish nuqtasini topish uchun, ushbu
= + + + − = − = − 0 , 0 0 0
Cz By Ax p z z n y y m x x
uch noma’lumli tenglamalar sistemasini yechish kerak bo’ladi. 6-misol. ) 4 , 1 , 5 ( − A va
) 3 , 1 , 6 ( −
nuqtalardan o’tuvchi to’g’ri chiziq bilan 0 3
2 = − + −
y x tekislik orasidagi burchakni toping. Yechish. AB nuqtalardan o’tuvchi to’g’ri chiziqning yo’naltiruvchi vektori sifatida ( )
1 , 0 , 1 → → =
s ni olamiz. Tekislikning normal vektori ( )
, 2 , 2 − → n bo’lganligi uchun (10) formulaga asosan: 19
2 2 2 3 3 1 4 4 1 1 1 1 ) 2 ( 0 1 2 sin
2 2 = = + + ⋅ + ⋅ + − ⋅ + ⋅ = ϕ
, . 45 , 2 / 2 sin
0 = = ϕ ϕ
7 –misol. = − + + = − + + 0 4 2 2 , 0 7 3 3 2 z y x z y x
to’g’ri chiziqni yasang. Yechish. Ma’lumki to’g’ri chiziqni yasash uchun u o’tadigan ikkita nuqtani aniqlash yetarli. Buning uchun to’g’ri chiziqning koordinat tekisliklari bilan kesishish nuqtalarini topamiz. Bu nuqtalarga
deyiladi. To’g’ri chiziqning
tekislikdagi izini topish uchun berilgan sistemada 0 = z deb olamiz, ya’ni
= − + − + . 0 4 2 , 7 3 2
x y x
Bu sistemani y x , noma’lumlarga nisbatan yechsak, 1 ,
= =
x
bo’ladi. Demak, berilgan to’g’ri chiziqning XOY koordinata tekisligidagi izi )
, 1 , 2 ( 1 M nuqta bo’ladi. Endi to’g’ri chiziqning
tekislikdagi izini topamiz. Buning uchun berilgan tenglamalar sistemasida 0 = y deb, hosil bo’lgan sistemani yechib, 1 , 2 = = z x topamiz. Demak, to’g’ri chiziqning XOZ tekislikdagi izi ) 1
0 , 2 ( 2
bo’ladi. Topilgan 1
va
2 M nuqtalardan to’g’ri chiziq o’tkazamiz.
1.Fazoda to’g’ri chiziqning vektorli tenglamasini toping. A) →
→ + = s t r r 0
В ) + = + = + = tp z z tn y y tm x x 1 1 1 , , 20
D) p z z n y y m x x 1 1 1 − = − = − E)
+ = + = nz y y mz x x 1 1 ,
2. Fazoda to’g’ri chiziqning parametrik tenglamasini toping. A)
+ = + = + =
z z tn y y tm x x 1 1 1 , ,
В )
z z n y y m x x 1 1 1 − = − = − D)
+ = + = nz y y mz x x 1 1 ,
E) → → → + =
t r r 0
3. Fazoda to’g’ri chiziqning kanonik tenglamasini toping. A)
1 1 1 − = − = − В ) + = + = + =
z z tn y y tm x x 1 1 1 , , D)
+ = + = nz y y mz x x 1 1 ,
E) = + + + = + + + 0 , 0 2 2 2 2 1 1 1 1 D z C y B x A D z C y B x A
4. Fazoda to’g’ri chiziqning umumiy tenglamasini toping. A) = + + + = + + + 0 , 0 2 2 2 2 1 1 1 1
z C y B x A D z C y B x A
В )
z z n y y m x x 1 1 1 − = − = − D)
+ = + = nz y y mz x x 1 1 ,
E) + = + = + =
z z tn y y tm x x 1 1 1 , , 5. Fazoda to’g’ri chiziqning proyeksiyalarga nisbatan tenglamasini toping. A)
+ = + =
y y mz x x 1 1 ,
В ) p z z n y y m x x 1 1 1 − = − = − D)
+ = + = + =
z z tn y y tm x x 1 1 1 , ,
E) 1 2 1 1 2 1 1 2 1 z z z z y y y y x x x x − − = − − = − − 6. Fazoda berilgan ikki nuqtadan o’tuvchi to’g’ri chiziqning tenglamasini toping. 21
A) 1 2 1 1 2 1 1 2 1 z z z z y y y y x x x x − − = − − = − − В ) + = + = + =
z z tn y y tm x x 1 1 1 , , D)
+ = + = nz y y mz x x 1 1 ,
E) p z z n y y m x x 1 1 1 − = − = − 7. Fazoda p z z n y y m x x 1 1 1 − = − = − to’g’ri chiziq va 0 = + + + D Cz By Ax
tekislikning parallellik shartini toping. A) 0 = + +
Bn Am
В )
C n B m A = = D)
0 = + + p C n B m A
E) 0 = − −
Bn Am
8. Fazoda ; 1 1 1 1 1 1 p z z n y y m x x − = − = −
. 2
2 2 2 2 p z z n y y m x x − = − = − Ikkita to’g’ri chiziqlar orasidagi burchak qanday topiladi? A) 2
2 1 2 2 2 1 2 1 2 1 2 1 2 1 2 1 cos
p n m p n m p p n n m m + + ⋅ + + + + = ϕ
B) 2 1 2 1 2 2 2 1 2 1 2 1 2 1 2 1 2 1 sin p n m p n m p p n n m m + + ⋅ + + + + = ϕ
D) 2 1 2 1 2 2 2 1 2 1 2 1 2 1 2 1 2 1 cos p n m p n m p p n n m m + + ⋅ + + − − = ϕ
E) 2 1 2 1 2 2 2 1 2 1 2 1 2 1 2 1 2 1 cos p n m p n m p p n n m m + + ⋅ + + + + + = ϕ
9. Fazoda p z z n y y m x x 0 0 0 − = − = − to’g’ri chiziq va 0 = + + + D Cz By Az
tekislik orasidagi burchak qanday topiladi? A) 2 2 2 2 2 2 sin
n p m C B A Cp Bn Am + + ⋅ + + + + = ϕ
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B) 2 2 2 2 2 2 cos
n p m C B A Cp Bn Am + + ⋅ + + + + = ϕ
D) 2 2 2 2 2 2 sin n p m C B A Cp Bn Am + + ⋅ + + + + + = ϕ
E) 2 2 2 2 2 2 sin
n p m C B A Cp Bn Am + + ⋅ + + − − = ϕ
10 Fazoda p z z n y y m x x 1 1 1 − = − = − to’g’ri chiziq va 0 = + + + D Cz By Ax
tekislikning perpendikulyarlik shartini toping. A) p C n B m A = =
В ) 0 = + + Cp Bn Am
D) 0 = + + p C n B m A
E)
0 = − − Cp Bn Am
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