A treatise on the analytical geometry of the point, line, by John Casey

By John Casey

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Consider, in the affine space R3 , the affine frames R and R given by R = {(0, 0, 0); ((1, 0, 0), (0, 1, 0), (0, 0, 1))}, R = {(−1, 0, 0); ((1, 1, 0), (0, −1, 0), (0, 0 − 1))}. (a) Given the point P with coordinates (1, 2, −1) in R, determine the coordinates of P in R . (b) Find, with respect to R , the equation of the plane Π, given, with respect to R, by the equation 2x − y + z + 2 = 0. (c) Find, with respect to R , the equations of the straight line r given, with respect to R, by the equations 2x + y = 0, x − 2y + z = 1.

Affine Spaces It is now clear that the role played by P1 in the definition of barycenter can be played by any of the points Pi , i = 1, . . , r. That is, we also have −−→ 1 −−→ G = Pi + (Pi P1 + · · · + Pi Pr ). r The barycenter of two points is called the midpoint between them. That is, the midpoint between P1 and P2 is the point 1 −−−→ G = P1 + P1 P2 . 1 Computations in Coordinates Let R be an affine frame of A, and let us denote by Pi = (xi1 , . . , xin ), i = 1, . . , r, G = (g1 , . . , gn ) the coordinates of the points Pi and G in R.

N. Equivalently, every hyperplane parallel to a1 x1 + · · · + an xn = b has equation a1 x1 + · · · + an xn = b , with b ∈ k. 24 The above result on parallel hyperplanes can also be obtained as a consequence of the following lemma. 25 Let f, g : k n −→ k be surjective linear maps such that ker f = ker g. Then there exists a λ ∈ k such that f = λg. Proof Surjectivity implies dim ker f = dim ker g = n − 1. Let (e1 , . . , en−1 ) be a basis of ker f = ker g, and let (e1 , . . , en−1 , en ) be a basis of k n .

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