ProblemShow that if u and v are vectors thenbisects the angle between u and v. First notice that we only need to show that (u+v) bisects the angle between u and v when |u|=|v|=1. The general case will follow from here, for if the result is true for vectors of length 1 then, will bisect the angle between u/|u| and v/|v| which is the same as the angle between u and v (we are only changing their lengths). Also, if a vector b bisects an angle then so does any scalar multiple of it. Thus, multiplying by (|u|*|v|) we obtain the original result. OK but we still need to show that u+v does the trick. That is not difficult to see. We loose no generality if we choose a coordinate system where: |
> u := vector([1,0]); v := vector([cos(t),sin(t)]);
u := [ 1, 0 ]
v := [ cos(t), sin(t) ]
| The vector v forms an angle of "t" with u, |
> angle(u,u+v);
1 + cos(t)
arccos(----------------------------)
2 2 1/2
((1 + cos(t)) + sin(t) )
| so this is the angle between u and u+v... is this t/2 ? First, simplify with trig, |
> ang := simplify(",trig);
1 + cos(t)
ang := arccos(-----------------)
1/2
(2 + 2 cos(t))
| we are getting there... but not yet. Now let's use the formula for cos(t) in terms of cos(t/2). Let's make s=t/2 and simplify. |
> ang := subs(t=2*s,ang);
1 + cos(2 s)
ang := arccos(-------------------)
1/2
(2 + 2 cos(2 s))
1/2 2
4 cos(s)
ang := arccos(1/2 ------------)
2 1/2
(cos(s) )
arccos(csgn(cos(s)) cos(s))
|
The inside of arccos is nothing but |cos(s)| and since
we are assuming that 0 < s < Pi/2, the above is just s.
Q.E.D. |