Let vecu=(x,y) be the reqd. unit vector.
:. ||vecu||=1 rArr x^2+y^2=1.................(1).
Given that, Angle btwn. vecu and vecv is pi/3, we take these vectors' Dot Product, to get,
vecu*vecv=||u||||v||cos(hat(vecu, vecv))
:. (x,y)*(3,4)=1(sqrt(3^2+4^2))cos(pi/3)
:. 3x+4y=1*5*1/2=5/2 rArr 3x=5/2-4y
rArr x=1/3(5/2-4y).......................(2).
Using (2) in (1), we get,
1/9(5/2-4y)^2+y^2=1rArr25/4-20y+16y^2+9y^2=9
rArr 25y^2-20y=9-25/4.
To make the L.H.S. complete square, we add 4 on both sides.
:. 25y^2-20y+4=9-25/4+4.
:. (5y-2)^2=27/4
:. 5y-2=+-3sqrt3/2, i.e., 5y=2+-3sqrt3/2, so, y=2/5+-3sqrt3/10
By (2), then, x=1/3{5/2-4(2/5+-3sqrt3/10)}.
Thus, the reqd. unit vectors are, (3/10-2/5sqrt3,2/5+3sqrt3/10), or,
(3/10+2sqrt3/5, 2/5-3sqrt3/10).
An Alternative Method to solve this problem, is, instead of starting
with vecu=(x,y), we may suppose that,
vecu=(costheta,sintheta), where, we may, preferably restrict
theta in [0,pi/2].