194 Vector and complex-number methods Ch. 11
and so for any pointZ 0 ≡F(x 0 ,y 0 ),
x−x 0 =p(x 1 −x 0 )+q(x 2 −x 0 )+r(x 3 −x 0 ),
y−y 0 =p(y 1 −y 0 )+q(y 2 −y 0 )+r(y 3 −y 0 ).ButZ≡F′(x−x 0 ,y−y 0 ),whereF′=tO,Z 0 (F). Hence
−−→
Z 0 Z=p−−→
Z 0 Z 1 +q−−→
Z 0 Z 2 +
r−Z− 0 →Z 3 , withp+q+r=1.
NOTATION. Where a vector equation is independent of the origin, as in−→
OZ=
p−−→
OZ 1 +q−−→
OZ 2 +r−−→
OZ 3 , withp+q+r=1, it is convenient to write this asZ=pZ 1 +
qZ 2 +rZ 3 withp+q+r=1. In particular, in (11.3.1) we writeZ=( 1 −t)Z 1 +tZ 2.
Z 1
Z 2
Z 3
O Z
Figure 11.5.Figure 11.5 caters for whenOandZ 1 are taken as origins, a similar diagram would
cater for whenZ 0 andZ 1 are origins, and then a combination of the two would give
the stated result.
11.4.5 .....................................
We also use the notationδF(Z 1 ,Z 2 ,pZ 4 +qZ 5 +rZ 6 )forδF(Z 1 ,Z 2 ,Z 3 )where
−−→
OZ 3 =
p−−→
OZ 4 +q−−→
OZ 5 +r−−→
OZ 6 andp+q+r=1. We can then write the conclusion of 10.5.3
as
δF(Z 1 ,Z 2 ,( 1 −s)Z 4 +sZ 5 )=( 1 −s)δF(Z 1 ,Z 2 ,Z 4 )+sδF(Z 1 ,Z 2 ,Z 5 ).The more general resultδF(Z 1 ,Z 2 ,pZ 4 +qZ 5 +rZ 6 )
=pδF(Z 1 ,Z 2 ,Z 4 )+qδF(Z 1 ,Z 2 ,Z 5 )+rδF(Z 1 ,Z 2 ,Z 6 ),