Observe that, #vec u=(cos60^@i+sin60^@j)# is a Unit Vector,
making an angle of #60^@# with the #+ve# direction of X-Axis; &,
#vecv=3vecu.#
From this, we can immediately conclude that the magnitude of
#vecv=||vecv||=3#, and, the Direction Angle #theta=60^@#.
However, we may formally solve the Problem as follows :-
#vec v=3(cos60^@i+sin60^@j)=3{(1/2)i+(sqrt3/2)j}#
#=(3/2)i+((3sqrt3)/2)j#
#:. ||vecv||=sqrt{(3/2)^2+((3sqrt3)/2)^2}#
#=3/2sqrt(1+3)=3/2*2=3 #.
Suppose that the Direction Angle of #vecv# is #theta.#
Then, #/_(vecv, veci)=theta.#
#rArr vecv.veci=||vecv||||veci||cos theta#
Here, #vecv=(3/2,(3sqrt3)/2), veci=(1,0)#
#(3/2)(1)+((3sqrt3)/2)(0)=(3)(1)costheta#
#rArr 3/2=3costheta rArr costheta=1/2#
#rArr theta=60^@#.
