OK=1 - катет против угла в 30 градусов равен половине гипотенузы
ОК=MA=1
KM=BC=2
OA=4
и
vector{OA}=4*vector{m}
vector{OB}=2*vector{n}
( рис. 2)
1)
Проводим BF || CA ( рис. 3)
vector{BF}=vector{CA}
По правилу треугольника
vector{OB}+vector{BF}=vector{OF} ⇒ vector{BF}=vector{OF} - vector{OB}
vector{CA}=2*vector{m} -2*vector{n}
[b]vector{AC}=-2*vector{m} +2*vector{n}[/b]
2)
vector{OM}=vector{OB}+vector{BM}=[b]2*vector{n}+vector{m}
[/b]
3)
vector{ON}=vector{OA}+vector{AN}=4*vector{m}+(1/2)vector{AC}=4*vector{m}-vector{m} +vector{n}=[b]3vector{m}+vector{n}[/b] [/b]
4)
vector{MN}=(1/2)*vector{BA}=(1/2)*(vector{BO}+vector{OA})=[b]-vector{n}+2*vector{m}[/b]