# t01sol - 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 1 1. (Revision) ( i ) Given that A = (1 , 1), B = (2 , 3) and C = (4 , 5), find −→ AB , −→ BC and −→ CA . Check your answers by calculating the sum of these three vectors. ( ii ) Find an equation for the line through the point (1 , 1) in the direction given by the vector bracketleftbigg 2 3 bracketrightbigg . Solution. ( i ) −→ AB = bracketleftbigg 2 − 1 3 − 1 bracketrightbigg = bracketleftbigg 1 2 bracketrightbigg , −→ BC = bracketleftbigg 4 − 2 5 − 3 bracketrightbigg = bracketleftbigg 2 2 bracketrightbigg , −→ CA = bracketleftbigg 1 − 4 1 − 5 bracketrightbigg = bracketleftbigg − 3 − 4 bracketrightbigg . This can be checked by finding the sum of −→ AB + −→ BC + −→ CA. We have bracketleftbigg 1 2 bracketrightbigg + bracketleftbigg 2 2 bracketrightbigg + bracketleftbigg − 3 − 4 bracketrightbigg = bracketleftbigg bracketrightbigg . ( ii ) The line ax + by + c = 0 has direction vector bracketleftbigg − b a bracketrightbigg . So the line through (1 , 1) in the direction bracketleftbigg 2 3 bracketrightbigg has an equation of the form 3 x − 2 y + c = 0 , where 3 − 2 + c = 0, that is, the equation of the line is 3 x − 2 y − 1 = 0. 2. Determine (in each case) if the given formula defines α as a transformation of the plane, and if it does, find a formula for the inverse transformation α − 1 . ( i ) α ( x, y ) = ( − y, x + 2), ( ii ) α ( x, y ) = (ln x, y ), ( iii ) α ( x, y ) = ( x − 2 y, − y ), ( iv ) α ( x, y ) = ( x, y + x 2 ). Solution. ( i ) Let ( u, v ) be any point in E . Then α ( x, y ) = ( u, v ) ⇐⇒ ( x, y ) = ( v − 2 , − u ) . That is, given any point ( u, v ) ∈ E , there is a unique point ( x, y ), namely ( x, y ) = ( v − 2 , − u ), such that α ( x, y ) = ( u, v ).This shows that α is bijective. So α is a transformation and α − 1 ( u, v ) = ( v − 2 , − u ), or in more usual notation, α − 1 ( x, y ) = ( y − 2 , − x ). ( ii ) α ( x, y ) is not defined for all ( x, y ) ∈ E . Hence α is not a transformation of the plane. 2 ( iii ) Let ( u, v ) be any point in E . Then α ( x, y ) = ( u, v ) ⇐⇒ ( x − 2 y, − y ) = ( u, v ) ⇐⇒ ( x, y ) = ( u − 2 v, − v ). This shows that α is bijective. So α is a transformation and α − 1 ( x, y ) = ( x − 2 y, − y ). (In this case, α = α − 1 .) ( iv ) Let ( u, v ) be any point in E . Then α ( x, y ) = ( u, v ) ⇐⇒ ( x, y + x 2 ) = ( u, v ) ⇐⇒ ( x, y ) = ( u, v − u 2 ). This shows that α is bijective. So α is a transformation and α − 1 ( x, y ) = ( x, y − x 2 )....
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t01sol - The University of Sydney Math3061 Geometry and...

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