# t03sol - The University of Sydney Math3061 Geometry and...

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Unformatted text preview: The University of Sydney Math3061 Geometry and Topology Web page: www.maths.usyd.edu.au/u/UG/SM/MATH3061/ 2009 Tutorial 3 1. Given that ℓ is the line through (1 , 0) with direction u = bracketleftBig 2 1 bracketrightBig , find the image of the point ( x, y ) under the glide reflection γ ℓ, u . Solution. We note, to begin with, that Q = (1 , 0) is a point on ℓ and that u ⊥ = 1 √ 5 bracketleftbigg − 1 2 bracketrightbigg is a unit vector perpendicular to ℓ . Let P = ( x, y ) and σ ℓ ( P ) = ( x ′ , y ′ ). Then −→ PQ = bracketleftbigg 1 − x − y bracketrightbigg and −−−−→ Pσ ℓ ( P ) = bracketleftbigg x ′ − x y ′ − y bracketrightbigg . We have −−−−→ Pσ ℓ ( P ) = bracketleftbigg x ′ − x y ′ − y bracketrightbigg = 2( −→ PQ. u ⊥ ) u ⊥ = 2 parenleftbigg − (1 − x ) − 2 y √ 5 parenrightbigg × 1 √ 5 bracketleftbigg − 1 2 bracketrightbigg = 2 5 ( x − 2 y − 1) bracketleftbigg − 1 2 bracketrightbigg = bracketleftbigg ( − 2 x + 4 y + 2) / 5 (4 x − 8 y − 4) / 5 bracketrightbigg Therefore x ′ = x + ( − 2 x + 4 y + 2) / 5 = (3 x + 4 y + 2) / 5 and y ′ = y + (4 x − 8 y − 4) / 5 = (4 x − 3 y − 4) / 5 . Thus σ ℓ ( P ) = ((3 x + 4 y + 2) / 5 , (4 x − 3 y − 4) / 5) . This completes the reflection part of the glide reflection. Now u = bracketleftbigg 2 1 bracketrightbigg . Thus γ ℓ, u ( P ) = τ u σ ℓ ( P ) = τ u ( 3 x +4 y +2 5 , 4 x − 3 y − 4 5 ) = (2 + 3 x +4 y +2 5 , 1 + 4 x − 3 y − 4 5 ) = ( 3 x +4 y +12 5 , 4 x − 3 y +1 5 ) . 2 2. If the three lines ℓ , m and n are neither all parallel nor concurrent, then σ ℓ σ m σ n is a glide reflection. Find the line b and the vector u such that σ ℓ σ m σ n = γ b, u in the following circumstances. ( i ) ℓ , m and n are the lines x = 1, y = 0 and x + y = 2. ( ii ) ℓ , m and n are the lines y = x , y = 0 and x + y = 1....
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t03sol - The University of Sydney Math3061 Geometry and...

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