160
Yechish: parallelogram qoidasiga koβra : |πβ + πββ| =d
1
, |
πβ β πββ| =d
2
Vektor xossalaridan foydalansak: |
πβ + πββ|=β(πβ + πββ)
2
=β
πβ
2
+ 2π
ββββπββ + π
2
βββββ
|πβ β πββ|=β(πβ β πββ)
2
=β
πβ
2
β 2π
ββββπββ + π
2
βββββ
d
1
=
β16 + 2Ξ4Ξ3Ξπππ 120
0
+ 9 = β13
d
2
=
β16 β 2Ξ4Ξ3Ξπππ 120
0
+ 9 = β37
2-Masala. Uchlari A(
1; 3; 5), π΅(β3; 4; 7) va C(4; 6; 3) nuqtalarda boβlgan
uchburchak yuzasini toping.
Parallelogrammning yuzasi.
Fazoda berilgan
πβ va πββ
vektorlarning vektor
koβpaytmasining moduli |πβ Γ πββ| son jihatdan shu πβ va πββ
vektorlarga qurilgan
parallelogrammning yuzasiga teng.
Πemak, πβ va πββ
vektorlarga qurilgan uchburchakning yuzi shu vektorlar vektor
koβpaytmasi modulining yarmiga teng.
Yechish. ABC uchburchak yuzasi
π΄π΅
ββββββ va π΄πΆ
ββββββ vektorlarga qurilgan
parallelogramm yuzining yarmisiga teng.
π΄π΅
ββββββ = (β3 β 1; 4 β 3; 7 β 5) = (β4; 1; 2)
π΄πΆ
ββββββ = (4 β 1; 6 β 3; 8 β 5) = (3; 3; 3)
π΄π΅
ββββββ Γ π΄πΆ
ββββββ = (β3; 18; β15)
|π΄π΅
ββββββ Γ π΄πΆ
ββββββ| = β9 + 324 + 225 = 3β62
πβ + π
πβ
πββ
πβ β πββ
πβ
πββ
161
π =
3
2
β62
3-masala. ABC ucburchakning
A,
B, C burchaklari berilgan boβlib,
M nuqta
BC tomonning oβrtasi boβlsa
BAM
burchakni toping.
Yechish: faraz qilaylik
< BAM=Ξ±,
PB=c, AC=b, BC=a
AM
ββββββββ β (AB
ββββββ+AC
ββββββ) ekanligidan
cosΞ±=
AB
βββββββ(AB
βββββββ+AC
ββββββ)
|AB
βββββββ||AB
βββββββ+AC
ββββββ|
=
AB
βββββββ
2
+(AB
βββββββ Ξ AC
ββββββ)
c(c
2
+b
2
+2bc cosA)
va
b
c
=
sinB
sinC
dan
cosΞ±=
sinC+sinB+sinA
βsin
2
B+sin
2
C+sin
2
A
bundan Ξ±=arccos
sinC+sinB+sinA
βsin
2
B+sin
2
C+sin
2
A