Geometry with Trigonometry

Sec. 11.3 Scalar or dot products 191

But by 6.6.1(ii),

πl(Z 2 )≡

(

x 2 +

y 1

y^21 +x^21

(−y 1 x 2 +x 1 y 2 ),y 2 −

x 1

y^21 +x^21

(−y 1 x 2 +x 1 y 2 )

)

=

(

x 1

x 1 x 2 +y 1 y 2

x^21 +y^21

,y 1

x 1 x 2 +y 1 y 2

x^21 +y^21

)

.

Thus

|O,Z 1 |^2 |O,πl(Z 2 )|^2 =(x^21 +y^21 )

[

x^21

(x 1 x 2 +y 1 y 2 )^2

(x^21 +y^21 )^2

+y^21

(x 1 x 2 +y 1 y 2 )^2

(x^21 +y^21 )^2

]

=(x 1 x 2 +y 1 y 2 )^2 ,

so that|O,Z 1 ||O,πl(Z 2 )|=|x 1 x 2 +y 1 y 2 |.

IfZ 2 ∈H 5 so that

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

andx 1 x 2 +y 1 y 2 ≥0sothat|x 1 x 2 +y 1 y 2 |=x 1 x 2 +y 1 y 2 , clearly

(O,Z 1 ).(O,Z 2 )=x 1 x 2 +y 1 y 2.

IfZ 2 ∈H 6 \mwe haveπl(Z 2 )∈l\[O,Z 1 .Then

(O,Z 1 ).(O,Z 2 )=−|O,Z 1 ||O,πl(Z 2 )|

and x 1 x 2 +y 1 y 2 ≤0, so that |x 1 x 2 +y 1 y 2 |=−(x 1 x 2 +y 1 y 2 ). Clearly again
(O,Z 1 ).(O,Z 2 )=x 1 x 2 +y 1 y 2.
(ii) Leta=(O,Z 1 ),b=(O,Z 2 ). Then by (i) of the present theorem,a.b=x 1 x 2 +
y 1 y 2 , b.a=x 2 x 1 +y 2 y 1 , and clearly these are equal.
(iii) Leta=(O,Z 1 ),b=(O,Z 2 ),c=(O,Z 3 ). Then by 11.2.1(i)b+c=(O,Z 4 )
whereZ 4 ≡(x 2 +x 3 ,y 2 +y 3 ). Then by (i) abovea.(b+c)=x 1 (x 2 +x 3 )+y 1 (y 2 +y 3 ),
whilea.b+a.c=(x 1 x 2 +y 1 y 2 )+(x 1 x 3 +y 1 y 3 ), and these are equal.
(iv) Leta=(O,Z 1 ),b=(O,Z 2 ).Thent.(a.b)=t(x 1 x 2 +y 1 y 2 ). But by 11.2.1(ii),
t.a=(O,Z 4 )whereZ 4 ≡(tx 1 ,ty 1 )and so(t.a).b=(tx 1 )x 2 +(ty 1 )y 2 , which is equal
to the earlier expression.
(v) Ifa=(O,Z)thena.a=x^2 +y^2 .This is positive when(x,y)=( 0 , 0 ), and equal
to 0 forx=y=0.
(vi) This follows immediately.
NOTE. Note that 11.2.2(i) to (v) make(V,+)acommutative group. In textbooks
on algebra it is proved that there is not a second element which has the property (iii);
we shall refer tooas thenull vector. It is also a standard result that for eacha∈V
there is not a second element with the property (iv); we call−athe inverseofa.
Subtraction−is defined by specifying thedifferenceb−a=b+(−a);then−
is a binary operation onV.Ifa=(O,Z 1 ),b=(O,Z 2 ),then−a=(O,Z 3 )where
Z 3 ≡(−x 1 ,−y 1 ), and consequentlyb−a=(O,Z 4 )whereZ 4 ≡(x 2 −x 1 ,y 2 −y 1 ).
Thus(O,Z 2 )−(O,Z 1 )=(O,Z 4 )if and only if(Z 1 ,Z 2 )↑(O,Z 4 ).