FractionalGeometry-Chap2

47 point sets for prxs 271 left and prxs 272 right

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Unformatted text preview: f . That 1a u1 1 b = u2 1 e u3 Both have the same of eq (2.13) are a x1 y1 b = x2 y2 e x3 y3 and x1 x2 x3 y1 y2 y3 1 c v1 1 d = v2 1f v3 (2.13) coefficient matrix. If this matrix is invertible, the solutions −1 1 u1 1 u2 1 u3 and c x1 d = x2 f x3 y1 y2 y3 −1 1 v1 1 v2 1 v3 (2.14) 79 2.7. TRANSFORMATIONS FROM POINT IMAGES To see that the coefficient matrix is invertible, view the points P1 , P2 , and P3 as lying in the z = 0 plane in 3-dimensional space. Form the vectors S = P2 − P1 and R = P3 − P1 , so S = x2 − x1 , y2 − y1 , 0 and R = x3 − x1 , y3 − y1 , 0 Then |S × R| is the determinant of the coefficient matrix. But also |S × R| = |S ||R| sin(θ), where θ is the angle between S and T . The vectors S and R are parallel if and only if the points P1 , P2 , and P3 are collinear. Consequently, the requirement that the initial points are not collinear has as a consequence that |S × R| = 0 and so the coefficient matrix has non-zero determinant. That is, the coefficient matrix is invertible. Finally, to conv...
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