Answer
$\vec v=\frac{3}{2}i+\frac{3\sqrt 3}{2}j$
Work Step by Step
1. As the vector is in the XY-plane, we can assume the vector has a form $\vec v=ai+bj$
2. The magnitude is $\sqrt {a^2+b^2}=3$ or $a^2+b^2=9$
3. The direction angle is $60^\circ$ (quadrant I), we have $\frac{b}{a}=tan60^\circ=\sqrt 3$, thus $b=\sqrt 3a$ (both positive)
4. Combine the above, we have $4a^2=9$, thus $a=\frac{3}{2}$ and $b=\frac{3\sqrt 3}{2}$
5. The vector is $\vec v==\frac{3}{2}i+\frac{3\sqrt 3}{2}j$
