Answer
$$\mathbf{v}=\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$$
Work Step by Step
$(\mathbf{u}+\mathbf{v})_x=u_x+v_x$, and $(\mathbf{u}+\mathbf{v})_y=u_y+v_y$
$u_x=\|\mathbf{u}\|\cos{\theta_u}=1\cos{45^\circ}=\frac{\sqrt{2}}{2}$
$u_y=\|\mathbf{u}\|\sin{\theta_u}=1\sin{45^\circ}=\frac{\sqrt{2}}{2}$
$(\mathbf{u}+\mathbf{v})_x=\|(\mathbf{u}+\mathbf{v})\|\cos{\theta_{u+v}}=\sqrt{2}\cos{90^\circ}=0$
$(\mathbf{u}+\mathbf{v})_y=\|(\mathbf{u}+\mathbf{v})\|\sin{\theta_{u+v}}=\sqrt{2}\sin{90^\circ}=\sqrt{2}$
$v_x=(\mathbf{u}+\mathbf{v})_x-u_x=0-\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{2}$
$v_y=(\mathbf{u}+\mathbf{v})_y-u_y=\sqrt{2}-\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$
