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