Answer
$v=\frac{15\sqrt2}{2}i-\frac{15\sqrt2}{2}j.$
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 $a=||15||cos(315^\circ)=\frac{15\sqrt2}{2}$ and $v=||15||sin(315^\circ)=-\frac{15\sqrt2}{2}$. Therefore $v=\frac{15\sqrt2}{2}i-\frac{15\sqrt2}{2}j.$
