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 }.