Geometry with Trigonometry

190 Vector and complex-number methods Ch. 11

11.3 Scalar or dot products .......................

11.3.1 .....................................

Z 2

O

Z 1

πOZ 1 (Z 2 )

Figure 11.3.

Z 2

O

Z 1

πOZ 1 (Z 2 )

Definitions.Wedefineascalar product,ordot product,(O,Z 1 ).(O,Z 2 )as follows.
IfZ 1 =Othen(O,Z 1 ).(O,Z 2 )=0; otherwiseZ 1 =Oand we set

(O,Z 1 ).(O,Z 2 )=

{

|O,Z 1 ||O,πOZ 1 (Z 2 )|,ifπOZ 1 (Z 2 )∈[O,Z 1 ,

−|O,Z 1 ||O,πOZ 1 (Z 2 )|,ifπOZ 1 (Z 2 )∈OZ 1 \[O,Z 1.

Clearly the scalar product is a function onV(Π;O)×V(Π;O)intoR.
Thenorm‖a‖of a vectora=(O,Z)is defined to be the distance|O,Z|.
The scalar product has the following properties:-
(i)If Zj≡(xj,yj)for j= 1 , 2 then a.b=(O,Z 1 ).(O,Z 2 )=x 1 x 2 +y 1 y 2.

(ii)For all a,b∈V(Π;O),a.b=b.a.

(iii)For all a,b,c∈V(Π;O),a.(b+c)=a.b+a.c.

(iv)For all a,b∈V(Π;O)and all t∈R,t.(a.b)=(t.a).b.

(v)For all a=o,a.a> 0 , while o.o= 0.

(vi)For all a,‖a‖=√a.a.

Proof. (i) IfZ 1 =O,thenx 1 =y 1 =0sothatx 1 x 2 +y 1 y 2 =0 as required.

Suppose then thatZ 1 =O. Write
l=OZ 1 and letmbe the line
through the pointOwhich is per-
pendicular tol. Define the closed
half-planeH 5 ={X:πl(X)∈
[O,Z 1 }and letH 6 be the other
closed half-plane with edgem.
Nowl≡−y 1 x+x 1 y=0andm≡
x 1 x+y 1 y=0.

O

I

J Z 1

l

m

H 6 H 5

Figure 11.4.

Then asZ 1 ∈H 5 ,

H 5 ={Z≡(x,y):x 1 x+y 1 y≥ 0 }, H 6 ={Z≡(x,y):x 1 x+y 1 y≤ 0 }.