Solution.
By the rule of adding vectors we get
O A ⃗ + A B ⃗ = O B ⃗ , \vec{OA}+\vec{AB}=\vec{OB}, O A + A B = OB ,
from here A B ⃗ = O B ⃗ − O A ⃗ = b − a . \vec{AB}=\vec{OB}-\vec{OA}=b-a. A B = OB − O A = b − a .
By the rule of adding vectors we get
O B ⃗ + B C ⃗ = O C ⃗ , \vec{OB}+\vec{BC}=\vec{OC}, OB + BC = OC ,
from here B C ⃗ = O C ⃗ − O B ⃗ = c − b . \vec{BC}=\vec{OC}-\vec{OB}=c-b. BC = OC − OB = c − b .
D D D is the midpoint of B C BC BC , then B D ⃗ = c − b 2 . \vec{BD}=\frac{c-b}{2}. B D = 2 c − b .
By the rule of adding vectors we get
O D ⃗ = O B ⃗ + B D ⃗ = b + c − b 2 = b + c 2 . \vec{OD}=\vec{OB}+\vec{BD}=b+\frac{c-b}{2}=\frac{b+c}{2}. O D = OB + B D = b + 2 c − b = 2 b + c .
By the rule of adding vectors we get
A D ⃗ = A B ⃗ + B D ⃗ = b − a + c − b 2 = b + c − 2 a 2 . \vec{AD}=\vec{AB}+\vec{BD}=b-a+\frac{c-b}{2}=\frac{b+c-2a}{2}. A D = A B + B D = b − a + 2 c − b = 2 b + c − 2 a .
E E E is the point on A D AD A D such that A E : E D = 2 : 1 , AE:ED=2:1, A E : E D = 2 : 1 , then A E ⃗ = 2 3 A D ⃗ = 2 3 b + c − 2 a 2 = b + c − 2 a 3 . \vec{AE}=\frac{2}{3}\vec{AD}=\frac{2}{3}\frac{b+c-2a}{2}=\frac{b+c-2a}{3}. A E = 3 2 A D = 3 2 2 b + c − 2 a = 3 b + c − 2 a .
By the rule of adding vectors we get
O E ⃗ = O A ⃗ + A E ⃗ = a + b + c − 2 a 3 = a + b + c 3 . \vec{OE}=\vec{OA}+\vec{AE}=a+\frac{b+c-2a}{3}=\frac{a+b+c}{3}. OE = O A + A E = a + 3 b + c − 2 a = 3 a + b + c .
Answer.
O D ⃗ = b + c 2 \vec{OD}=\frac{b+c}{2} O D = 2 b + c
O E ⃗ = a + b + c 3 \vec{OE}=\frac{a+b+c}{3} OE = 3 a + b + c
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