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Problem

Show that if u and v are vectors then
|v| u + |u| v

bisects 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,

(u/|u|) + (v/|v|)

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:

i=u, and j on the plane containing u and v

> 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)) > ang := arccos(expand(cos(ang))); 1/2 2 4 cos(s) ang := arccos(1/2 ------------) 2 1/2 (cos(s) ) > simplify("); 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. >