160
Yechish: parallelogram qoidasiga ko’ra : |𝑎⃗ + 𝑏⃗⃗| =d
1
, |
𝑎⃗ − 𝑏⃗⃗| =d
2
Vektor xossalaridan foydalansak: |
𝑎⃗ + 𝑏⃗⃗|=√(𝑎⃗ + 𝑏⃗⃗)
2
=√
𝑎⃗
2
+ 2𝑎
⃗⃗⃗⃗𝑏⃗⃗ + 𝑏
2
⃗⃗⃗⃗⃗
|𝑎⃗ − 𝑏⃗⃗|=√(𝑎⃗ − 𝑏⃗⃗)
2
=√
𝑎⃗
2
− 2𝑎
⃗⃗⃗⃗𝑏⃗⃗ + 𝑏
2
⃗⃗⃗⃗⃗
d
1
=
√16 + 2·4·3·𝑐𝑜𝑠120
0
+ 9 = √13
d
2
=
√16 − 2·4·3·𝑐𝑜𝑠120
0
+ 9 = √37
2-Masala. Uchlari A(
1; 3; 5), 𝐵(−3; 4; 7) va C(4; 6; 3) nuqtalarda bo’lgan
uchburchak yuzasini toping.
Parallelogrammning yuzasi.
Fazoda berilgan
𝑎⃗ va 𝑏⃗⃗
vektorlarning vektor
ko‘paytmasining moduli |𝑎⃗ × 𝑏⃗⃗| son jihatdan shu 𝑎⃗ va 𝑏⃗⃗
vektorlarga qurilgan
parallelogrammning yuzasiga teng.
Дemak, 𝑎⃗ va 𝑏⃗⃗
vektorlarga qurilgan uchburchakning yuzi shu vektorlar vektor
ko’paytmasi modulining yarmiga teng.
Yechish. ABC uchburchak yuzasi
𝐴𝐵
⃗⃗⃗⃗⃗⃗ va 𝐴𝐶
⃗⃗⃗⃗⃗⃗ vektorlarga qurilgan
parallelogramm yuzining yarmisiga teng.
𝐴𝐵
⃗⃗⃗⃗⃗⃗ = (−3 − 1; 4 − 3; 7 − 5) = (−4; 1; 2)
𝐴𝐶
⃗⃗⃗⃗⃗⃗ = (4 − 1; 6 − 3; 8 − 5) = (3; 3; 3)
𝐴𝐵
⃗⃗⃗⃗⃗⃗ × 𝐴𝐶
⃗⃗⃗⃗⃗⃗ = (−3; 18; −15)
|𝐴𝐵
⃗⃗⃗⃗⃗⃗ × 𝐴𝐶
⃗⃗⃗⃗⃗⃗| = √9 + 324 + 225 = 3√62
𝑎⃗ + 𝑏
𝑎⃗
𝑏⃗⃗
𝑎⃗ − 𝑏⃗⃗
𝑎⃗
𝑏⃗⃗
161
𝑆 =
3
2
√62
3-masala. ABC ucburchakning
A,
B, C burchaklari berilgan bo‘lib,
M nuqta
BC tomonning o‘rtasi bo‘lsa
BAM
burchakni toping.
Yechish: faraz qilaylik
< BAM=α,
PB=c, AC=b, BC=a
AM
⃗⃗⃗⃗⃗⃗⃗⃗ ⇈ (AB
⃗⃗⃗⃗⃗⃗+AC
⃗⃗⃗⃗⃗⃗) ekanligidan
cosα=
AB
⃗⃗⃗⃗⃗⃗⃗(AB
⃗⃗⃗⃗⃗⃗⃗+AC
⃗⃗⃗⃗⃗⃗)
|AB
⃗⃗⃗⃗⃗⃗⃗||AB
⃗⃗⃗⃗⃗⃗⃗+AC
⃗⃗⃗⃗⃗⃗|
=
AB
⃗⃗⃗⃗⃗⃗⃗
2
+(AB
⃗⃗⃗⃗⃗⃗⃗ · AC
⃗⃗⃗⃗⃗⃗)
c(c
2
+b
2
+2bc cosA)
va
b
c
=
sinB
sinC
dan
cosα=
sinC+sinB+sinA
√sin
2
B+sin
2
C+sin
2
A
bundan α=arccos
sinC+sinB+sinA
√sin
2
B+sin
2
C+sin
2
A
