Answer
$\dfrac{25 \sqrt 3}{2} \ i - \dfrac{25}{2} \ j$
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: $p=||25|| \cos \ (330^\circ)= \dfrac{25 \sqrt 3}{2}$
and $v=||25|| \sin \ (330^\circ)=\dfrac{-25}{2}$
Therefore $v= \dfrac{25 \sqrt 3}{2} \ i - \dfrac{25}{2} \ j$
