192 Vector and complex-number methods Ch. 11
COMMENT. Now that we have set up our couples we call(O,Z)aposition vector
with respect to the pointO, and we adopt the standard notation
−→
OZfor(O,Z).
Position vectors can be used for many geometrical purposes instead of Cartesian
coordinates, or complex coordinates and complex-valued distances. We would note
that by 6.1.1(iv) and 11.2Z 0 =mp(Z 1 ,Z 2 ) if and only if
−−→
OZ 0 =^12 (
−−→
OZ 1 +
−−→
OZ 2 );by10.1.1(v) thatZ 1 Z 2 ‖Z 3 Z 4 if and only if
−−→
OZ 4 −
−−→
OZ 3 =t(−−→
OZ 2 −
−−→
OZ 1 )for somet= 0
inR, and by 9.7.1(ii) and 6.5.1 Corollary (ii) thatZ 1 Z 2 ⊥Z 3 Z 4 if and only if(−OZ−→ 1 −
−−→
OZ 2 ).(
−−→
OZ 3 −
−−→
OZ 4 )=0. Most importantly, from parametric equations of a linex=
x 1 +t(x 2 −x 1 ),y=y 1 +t(y 2 −y 1 )(t∈R),wehavethatZ∈Z 1 Z 2 if and only if
−OZ→=−OZ−→
1 +t(−OZ−→
2 −
−OZ−→
1 )=(^1 −t)−OZ−→
1 +t−OZ−→
2 (11.3.1)
for somet∈R.
COMMENT. It is usual, in modern treatments, to define vectors to be the equiv-
alence classes for equipollence. This defines free vectors. Position vectors are then
defined by taking a specific pointOinΠso that we have a pointed plane, and then
concentrating on the representatives of the form(O,Z)for the vectors. But if our
objective is to introduce position vectors, it is wasteful of effort to set up the free
vectors, and in fact the use of free vectors and subsequent specialisation to position
vectors can be a confusing route to position vectors.
11.4 Components of a vector .......................
11.4.1 Components ............................
Given non-collinear pointsZ 1 ,Z 2 ,Z 3 , we wish to obtain an expression
−−→
Z 1 Z=p−−→
Z 1 Z 2 +q−−→
Z 1 Z 3.
For this we need
(x 2 −x 1 )p+(x 3 −x 1 )q=x−x 1 ,
(y 2 −y 1 )p+(y 3 −y 1 )q=y−y 1.We obtain the solutions
p=δF(Z 1 ,Z,Z 3 )
δF(Z 1 ,Z 2 ,Z 3 ),q=δF(Z 1 ,Z 2 ,Z)
δF(Z 1 ,Z 2 ,Z 3 ),
andsohave
−−→
Z 1 Z=
δF(Z 1 ,Z,Z 3 )
δF(Z 1 ,Z 2 ,Z 3 )−−→
Z 1 Z 2 +
δF(Z 1 ,Z 2 ,Z)
δF(Z 1 ,Z 2 ,Z 3 )