Finding Dual Space Coordinates for Projective Geometry Points

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SUMMARY

The discussion focuses on finding dual space coordinates for points A, B, and C in projective geometry, specifically within a 3-dimensional vector space V with a given basis. The points are expressed in homogeneous coordinates as A=[2,1,0], B=[0,1,1], and C=[-1,1,2]. The dual basis f1, f2, f3 is defined such that f_i(v_j)=δ_{ij}. The lines AB, BC, and CA are derived with equations indicating relationships between the basis vectors, leading to the coordinates AB=[1,a,-2], BC=[1,b,1], and CA=[2,c,3]. The validity of the answer is questioned, highlighting a lack of responses from the community.

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I have this question from Nigel Hitchin's notes in PG:
"Let V be 3-dimensional vector space with basis v1,v2,v3 and let A,B,C be points in the projective space P(V) be expressed in homogeneous coordinates relative to this basis:
A=[2,1,0], B=[0,1,1], C=[-1,1,2].
Find the coordinates wrt to the dual basis of the three points in the dual space P(V') (where V' is dual space of V) which represent the lines AB, BC,CA."

The dual basis would be f1,f2,f3, where f:V->F for some F field.
such that f_i(v_j)=\delta_{ij} (the delta of kroncker).
So for A: v=2v1+v2, f=\sum_{i=1}^{3}c_i f_i=\sum_{i=1}^{3}f(v_i) f_i
B: v=v2+v3, the line AB satisfies: 2v1=-v3
C: v=-v1+v2+2v3 the line CA satisfies 3v1=2v3 and BC satisfies v1=v3.

So for AB=[1,a,-2] for some a scalar, BC=[1,b,1] and for CA=[2,c,3].

Is my answer valid or not?

Thanks in advance.
 
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