The vector is of the form #a hat i + b hat j, a in RR, b in RR#. We need to find #a# and #b#.
The magnitude of a vector is defined as #sqrt(a^2+b^2)#.
The direction of a vector can be found using #arctan(b/a)#.
From the question, we know that #sqrt(a^2+b^2)=16# and #arctan(b/a)=120^@#. Since #90^@<120^@<180^@#, the vector lies in the second quadrant. In other words, #a<0# and #b>0#.
From #arctan(b/a)=120^@#, we know that #b/a=-sqrt(3)#, or #b=-asqrt(3)#.
Substitute this into #sqrt(a^2+b^2)=16#. We get #sqrt(a^2+(-asqrt(3))^2)=16#. Simplifying this, we get #sqrt(4a^2)=16#. Solving this gives #a=+-8#. However, we said before that #a<0#. Thus, #a=-8#.
Now, #b=-asqrt(3)=-(-8)sqrt(3)=8sqrt(3)#.
The vector can be written as #-8 hat i + 8sqrt(3) hat j#.
