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Excuse me. A (5,1). B (7,10). Vector AB (how to calculate? , Answer is (2,9)
Take for example vector A;(X1, Y1) vector B (X2, Y2), then vector AB (X2-X1, Y2-Y1)
So according to the title vector AB (7-5,10-1)=(2,9)
Given that a, b are two non-collinear vectors, and vector AB=λa+b, vector AC=a b,(λ, R), then A, B, C are collinear, what does the real number λ,μ satisfy Given that a, b are two non-collinear vectors, and vector A B=λa+b, vector AC=a b,(λ, R), then what is the condition that A, B, C are collinear,λ,μ satisfies? ] Given that a, b are two non-collinear vectors, and vector A B=λa+b, vector AC=a b,(λ, R), then A, B, C are collinear, what does the real number λ,μ satisfy Given that a, b are two non-collinear vectors, and vector A B =λ a + b, vector AC = a b,(λ, R), then if A, B, C are collinear, what is the condition that real numbers λ,μ satisfy? ] Given that a, b are two non-collinear vectors, and vector AB=λa+b, vector AC=a b,(λ, R), then A, B, C are collinear, what does the real number λ,μ satisfy Given that a, b are two non-collinear vectors, and vector A B =λ a + b, vector A C = a b,(λ, R), then what is the condition that the real numbers λ,μ satisfy when A, B, C are collinear? ]
Because A, B and C are collinear.
So AB=kAC
I.e.λa+b=k (a b)
So λ= k
1= Kμ
So λμ=1
Given that A, B and C are collinear, vector AC=-2/3 vector CB, and vector AB=h vector CA, then the real number h=?
Vector AC=-2/3 Vector CB
Vector CB=-3/2AC
Vector AB = Vector AC + Vector CB = Vector AC-3/2 Vector AC=-1/2 Vector AC=1/2 Vector CA
And vector A B = h vector CA, then real number h =1/2
Vector AC=-2/3 Vector CB
Vector CB=-3/2AC
Vector AB = Vector AC + Vector CB = Vector AC-3/2 Vector AC=-1/2 Vector AC=1/2 Vector CA
And vector AB=h vector CA, then real number h=1/2
If A, B, P are collinear, i.e. there is a real number t∈R, so that vector AP=t vector AB, if O is any point on the plane, then vector OP=_____________(represented by vector OA, vector OB)
Vector OP=vector OA+vector AP
= Vector OA + t vector AB
= Vector OA+t (vector OB-vector OA)
=(1-T) vector OA+t vector OB
If points A (-1,-1), B (1,3), C (x,5) are collinear, find the coordinates of point C and the value of the real number λ in the vector AB=λ vector BC If points A (-1,-1), B (1,3), C (x,5) are collinear, find the coordinates of point C and the value of the real number λ in the vector AB=λ vector BC
If points A (-1,-1), B (1,3), C (x,5) are collinear, find the coordinates of point C
Vector AB=(2,4), AC=(X+1,6),2/(X+1)=4/6
X =2, C (2,5)
The value of the real number λ in the vector AB =λ vector BC
Vector AB=(2,4), BC=(1,2), AB=2(1,2)=2BC, i.e. the value of λ is 2
In △ABC, D, E and F are the midpoint of AB, BC and CA respectively. BF and CD intersect at point O. Let vector AB = vector a and vector AC = vector b. (1) Prove that A, O and E are on the same straight line, and AO/OE=BO/OF=CO/OD=2 (2) The vector a, b is the vector AO Please use vector knowledge to solve, do not use junior high similar triangle In △ABC, D, E and F are the midpoint of AB, BC and CA respectively. BF intersects with CD at point O. Let vector AB = vector a and vector AC = vector b. (1) Prove that A, O and E are on the same straight line, and AO/OE=BO/OF=CO/OD=2 (2) The vector a, b is the vector AO Please use vector knowledge to solve, do not use junior high similar triangle
(1) Vector BO is collinear with vector BF, so BO=xBF,
According to the triangle addition rule: vector AO=AB+BO
=A + xBF =a + x (AF-AB)
= A + x (b/2-a)=(1-x) a +(x/2) b.
Vector CO is collinear with vector CD, so CO=yCD,
According to the triangle addition rule: vector AO=AC+CO
=B+yCD=b+y (AD-AC)
=B+y (a/2-b)=(y/2) a+(1-y) b.
So vector AO=(1-x) a+(x/2) b=(y/2) a+(1-y) b.
Then 1-x = y/2, x/2=1-y,
Solution x=2/3, y=2/3.
Vector BO=2/3BF, vector CO=2/3CD,
I.e. BO: OF=CO: OD=2.
Vector AO=(y/2) a+(1-y) b=1/3a+1/3b,
Another factor vector AE=AB+BE=a+1/2BC=a+1/2(AC-AB)
= A+1/2(b-a)=1/2a+1/2b,
So that vector AO=2/3 vector AE,
That is, vector AO is collinear with vector AE, so that three point A, O and E are collinear,
And AO: OE=2.(2) Vector AO=vector (a+b)/3
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