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If there are three different points on the line L, so that the equation x ^ 2 vector OA + X vector ob + vector BC = vector 0 has a solution, (o is not on L), find the real number solution set x^2*OA+x*OB+BC=0 BC=-(x^2*OA+x*OB) BC=OC-OB OC-OB=-(x^2*OA+x*OB) OC= - x^2*OA - x*OB + OB Because three points are collinear - x^2 - x* +1=1 - x^2 - x*=0 x(x+1)=0 X = 0 or 1 Because when x = 0, the three points coincide, which is not in line with the meaning of the topic So x = - 1 Why? OC= - x^2*OA - x*OB + OB Because three points are collinear So - x ^ 2 - x * + 1 = 1
If ob = xoa + yoc, there must be x + y = 1
A. If B and C are collinear, then AB = λ AC, AC is a nonzero vector and λ is a real number
OB-OA=λ(OC-OA) OB=(1-λ)OA+λOC X=1-λ Y=λ
So x + y = 1
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