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 3dimensional 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 nonzero determinant. That is,
the coeﬃcient matrix is invertible.
Finally, to conv...
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This document was uploaded on 02/14/2014 for the course MATH 290B at Yale.
 Fall '14
 AmandaFolsom
 Geometry, Fractal Geometry, Limits, The Land

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