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]#.
