Answer
$\alpha=\beta=45^\circ,\gamma=90^\circ$, $\vec v=\sqrt 2(cos45^\circ i+cos45^\circ j+cos90^\circ k)$
Work Step by Step
1. Given $\vec v=i+j$, we have $||\vec v||=\sqrt {1+1+0}=\sqrt 2$
2. We have $\alpha=cos^{-1}(\frac{\sqrt 2}{2})=45^\circ$
3. We have $\beta=cos^{-1}(\frac{\sqrt 2}{2})=45^\circ$
4. We have $\gamma=cos^{-1}(0)=90^\circ$
5. We have $\vec v=\sqrt 2(cos45^\circ i+cos45^\circ j+cos90^\circ k)$