\(v = \langle -2, 2 \rangle, | v | = 2\sqrt{2}\)
\(v\) is in quadrant II and\(\ \tan \theta = - 1\), so \(\theta = 135^\circ\). The direction vector is \(\dfrac{v}{|v|}= \left\langle -\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right\rangle\).
\(3v = \langle -6, 6 \rangle\) and \(-2v = \langle 4, -4 \rangle\).