Geometry with Trigonometry

(Marvins-Underground-K-12) #1

188 Vector and complex-number methods Ch. 11


Definition.ForO∈Π,letV(Π;O)be the set of all couples(O,Z)forZ∈Π.We
define thesum(O,Z 1 )+(O,Z 2 )of two couples to be(O,Z 3 )where mp(O,Z 3 )=
mp(Z 1 ,Z 2 ),sothat(O,Z 1 )↑(Z 2 ,Z 3 ). Thus + is a binary operation inV(Π;O).
We define theproduct by a number or scalart.(O,Z 1 ), of a numbert∈Rand
a couple, to be a couple(O,Z 4 )as follows.WhenZ 1 =Owe takeZ 4 =Ofor all
t∈R.WhenZ 1 =Owe takeZ 4 to be in the linel=OZ 1 and with|O,Z 4 |=|t||O,Z 1 |;
furthermore if≤lis the natural order for whichO≤lZ 1 ,wetakeO≤lZ 4 whent≥0,
andZ 4 ≤lOwhent<0. Thus product by a number is a function onR×V(Π;O)
intoV(Π;O).


COMMENT. To prove by synthetic means the basic properties of couples listed in
11.2.2 and 11.3.1, would be very laborious in covering all the cases. We establish in-
stead initial algebraic characterizations which allow an effective algebraic approach.


If O is the origin and Z 1 ≡(x 1 ,y 1 ),Z 2 ≡(x 2 ,y 2 ),then

(i)(O,Z 1 )+(O,Z 2 )=(O,Z 3 )where Z 3 ≡(x 1 +x 2 ,y 1 +y 2 ).

(ii)t.(O,Z 1 )=(O,Z 4 )where Z 4 ≡(tx 1 ,ty 1 ).

Proof.
(i)Forthiswehave0+x 3 =x 1 +x 2 , 0 +y 3 =y 1 +y 2.
(ii) We verify this as follows. Let(x 4 ,y 4 )=(tx 1 ,ty 1 ).When(x 1 ,y 1 )=( 0 , 0 )
clearly we have(x 4 ,y 4 )=(x 1 ,y 1 ).When(x 1 ,y 1 )=( 0 , 0 ), clearlyZ 4 ∈OZ 1 while


|O,Z 4 |^2 =(tx 1 )^2 +(ty 1 )^2 =t^2 |O,Z 1 |^2.

Now iflis not perpendicular toOIand the correspondence between the natural order
≤land the natural order≤OIonOI, under whichO≤OII, is direct thenx 1 <x 2. Thus
whent>0, we havetx 1 >0andsoO≤lZ 4 ;whent<0, we havetx 1 <0andso
Z 4 ≤lO. When the correspondence between the natural orders is inverse, we reach
the same conclusion. Whenlis perpendicular toOIwe project toOJinstead.


11.2.2 Vector space overR ..........................


Definition. A triple(V,+,.)is said to be avector spaceoverRif the following
hold:-


(i) First, + is a binary operation inV.

(ii) For alla,b,c∈V, (a+b)+c=a+(b+c).

(iii) There is ano∈V such that for alla∈V,

a+o=a,o+a=a.

(iv) Corresponding to eacha∈V, there is some−a∈V such that

(−a)+a=o,a+(−a)=o.
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