Answer
$v=-\frac{3}{2}i-\frac{3\sqrt3}{2}j.$
Work Step by Step
If a vector $v=ai+bj$ makes an angle of $\alpha$ with the positive x-axis and has magnitude $||v||=\sqrt{a^2+b^2}$, then $a=||v||cos\alpha$ and $v=||v||sin\alpha$.
Hence here $a=||3||cos(240^\circ)=-\frac{3}{2}$ and $v=||3||sin(240^\circ)=-\frac{3\sqrt3}{2}$. Therefore $v=-\frac{3}{2}i-\frac{3\sqrt3}{2}j.$
