Answer
$v=\frac{25\sqrt3}{2}i-\frac{25}{2}j.$
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 $a=||25||cos(330^\circ)=\frac{25\sqrt3}{2}$ and $v=||25||sin(330^\circ)=-\frac{25}{2}$. Therefore $v=\frac{25\sqrt3}{2}i-\frac{25}{2}j.$
