Answer
$ -\dfrac{3}{2} \ i - \dfrac{3 \sqrt 3}{2} \ j$
Work Step by Step
Let us consider that a vector $v$ is given by $v=pi+qj$ and makes an angle of $\alpha$ with the positive $x$-axis. The magnitude of the vector can be determined using the formula:
$||v||=\sqrt{p^2+q^2} $
Then we find $p=||v||cos\alpha$ and $v=||v||sin\alpha$.
Here, we have: $p=||3|| \cos \ (240^\circ)= -\dfrac{3}{2}$
and $v=||3|| \sin \ (240^\circ)=\dfrac{-3 \sqrt 3}{2}$
Therefore $v= -\dfrac{3}{2} \ i - \dfrac{3 \sqrt 3}{2} \ j$
