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Collinear problem of plane vector Let vector AB=√2/2×(a+5b), vector BC=-2a+8b, vector CD=3*(a-b), then the three points of collinear are
A:
A, B, D collinear,
BD=BC+CD=(-2a+8b)+3(a-b)=a+5b,
So √2 A B = BD, A, B, D are collinear,
It is also easy to prove that any other three points are not collinear when a and b are not collinear.
Condition of collinearity of plane vectors a, b? 1. There is real number n, b=na 2. There is a real number m, n, ma+nb=0, which is not all zero. (A, b are all vectors, the sign can not be printed) Which one is right? Why?
2, Because 1 is present at n =0, the b vector is the 0 vector.
2, Because 1 exists when n =0, b vector is 0 vector.
2, Because 1 exists when n =0, the b vector is 0.
Urgent: O is a point in triangle ABC, vector OA+vector OB+vector OC=0. Try to prove that O is the center of gravity of triangle ABC.
BD‖OC, CD‖OB, link OD, OD and BC intersect at G, then BG=CG (parallelogram diagonal bisects each other)
Vector OB + vector OC = vector OD,
Vector OB+Vector OC=-Vector OA, Vector OD=-Vector OA
A, O, G on a straight line ===> AG is the centerline on the BC edge
In the same way, the extension line of BO and CO is also the center line of △ABC.
O is the center of gravity of the triangular ABC
If point C satisfies vector OC=a phasor OA+b vector OB, and a+b=1, then point ABC is collinear, how to prove
Vector OC=a phasor OA+b vector OB
Vector OC=a Vector OA+(1-a) Vector OB
Vector OC = a vector OA + vector OB - a vector OB
Vector OC-vector OB=a vector OA-a vector OB
Vector OC-vector OB=a (vector OA-vector OB)
Vector BC=a vector BA
Vector BC parallel to vector BA
A, B and C are collinear
Given that O is the center of gravity of ΔABC, prove that vector OA+vector OB+vector OC=0
A (x1, y1), B (x2, y2), C (x3, y3)
The center of gravity coordinates are
O=(( x1+x2+x3)/3,(y1+y2+y3)/3)
OA=(x1-(x1+x2+x3)/3, y1-(y1+y2+y3)/3)
OB=(x2-(x1+x2+x3)/3, y2-(y1+y2+y3)/3)
OC=(x3-(x1+x2+x3)/3, y3-(y1+y2+y3)/3)
OA+OB+OC=0
The inner point O of triangle ABC satisfies, a vector OA + b vector OB + c vector OC =0 vector, proving that O is inner
Let △BOC=(1/2)*a*r=(1/2)*|OB*||OC|*sin∠BOCa=(|OB*|OC|/r)*sin∠BOC, the same as b=(|OC*|OA|/r)*sin∠COA, c=(|OA*|OB|/r)*sin∠AOBa*OA+b*OB+c*OC=(|OB*|OC|/r)*sin∠BOC*OA+(|OC*|O...
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