Geometry with Trigonometry

206 Vector and complex-number methods Ch. 11

This circle is called theincirclefor the triangle; its centreZ 15 is called theincentre
of the triangle. The three side-lines are tangents to the circle with the points of contact
being the feet of the perpendiculars. For the incentre, by (11.5.2) we have the formula

Z 15 =

a

a+b+c

Z 1 +

b

a+b+c

Z 2 +

c

a+b+c

Z 3. (11.5.5)

11.6 Mobilecoordinates ..........................

NOTE. In standard vector notation, the vector product or cross product takes us out
of the planeΠand into solid geometry. Sensed-area gives us half of the magnitude
of the vector product and we use that instead. Without the vector product, however,
we have not got orientation of the planeΠby vector means. We go on to supply this
lack.
However the standard vector operations can be awkward in dealing with perpen-
dicularity and distance, and can involve quite a bit of trigonometry, so we also set out
a method of reducing unwieldy calculations.

11.6.1 Grassmann’s supplement of a vector .................

Given any Z=O, we show that there is a unique W such that

|O,W|=|O,Z|,OW⊥OZ,δF(O,Z,W)> 0.

Proof. WithZ≡(x,y),W≡(u,v)these require

x^2 +y^2 =u^2 +v^2 ,ux+vy= 0 ,xv−yu> 0.

By the middle one of these ∣
∣∣
∣

xy

−vu

∣∣

∣

∣=^0 ,

so the rows of this are linearly dependent. Thus we haver(x,y)+s(−v,u)=( 0 , 0 ),
for some(r,s)=( 0 , 0 ). We cannot haver=0 as that would implyW=Oand so
Z=O.Then
x=

s

r

v,y=−

s

r

u,

so by the first property above

x^2 +y^2 =

s^2

r^2

(x^2 +y^2 ).

Thus we have eithers/r=1, so thatu=−y,v=x,forwhich2δF(O,Z,W)=x^2 +
y^2 >0, or we haves/r=−1, so thatu=y,v=−x,forwhich2δF(O,Z,W)=
−(x^2 +y^2 )<0. Thus the unique solution isu=−y,v=x.