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A point O in triangle ABC proves that vector OA+vector OB+vector OC is equal to vector 0.
0
Given OB vector =λOA vector +μOC vector, if A, B and C are collinear, prove:=1
A, B, C are collinear,
Vector AB // Vector AC
Has a unique real number μ such that vector AB =μ vector AC
I.e. vector OB-vector OA=μ(vector OC-vector OA)
Vector OB=(1-μ) vector OA +μ vector OC
OB vector =λOA vector +μOC vector
∴λ=1-μ, I.e.=1.
The three points A, B and C are collinear, the vector OA is β times the vector OB-Ω times the vector OC, and β-Ω=? I can not thank you enough!
Vector OA=β-fold vector OB-Ω-fold vector OC vector OA-vector OB=(β-1) vector OB-Ω-fold vector OC, vector BA=(β-1) vector OB-Ω-fold vector OC=(β-1) vector OB-Ω-fold vector OC-[Ω-(β-1)] vector OC=(β-1) vector OC-[Ω-(β-1)] times vector BC-[Ω-(β-1)] vector OC by...
A, B and C are collinear, vector OA=a times vector OB+1/3 vector OC (a belongs to R), then a=
0
P is a point on the plane of △ABC. If the vector PA·PB=PB·PC=PC·PA, what is the heart of P? Why?
PA·PB=PB·PC
PA·PB-PB·PC=0
PB·(PA-PC)=0
PB·CA=0
PB⊥CA
The same can be derived: PC⊥ABPA⊥BC
(Take two equations from any of the three equations, move to, do the operation, and get the answer)
PA·PB=PB·PC
PA·PB-PB·PC=0
PB·(PA-PC)=0
PB·CA=0
PB⊥CA
The same can be exported: PC⊥ABPA⊥BC
(Take two equations from any of the three equations, move to, do the operation, and get the answer)
In triangle ABC, AB=2, AC=3, D is the midpoint of edge BC, then vector AD* vector BC=?
You can extend the line AD, do DE = AD, and then connect BE, CE. So ABCE is a parallelogram.
Vector AD =1/2 vector AE
Vector AD·vector BC=1/2 vector AE·vector BC
=1/2(Vector AC + vector AB)(vector AC - vector AB)
=1/2(Modulus of AC - modulus of AB)
=1/2(9-4)
=2.5
AD can be extended, do DE = AD, and then connect BE, CE. So ABCE is a parallelogram.
Vector AD =1/2 vector AE
Vector AD·vector BC=1/2 vector AE·vector BC
=1/2(Vector AC + vector AB)(vector AC - vector AB)
=1/2(Modulus of AC - modulus of AB)
=1/2(9-4)
=2.5
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