Answer
Horizontal component $=\dfrac{\sqrt{3}}{2}$
Vertical component $=-\dfrac{3}{2}$
$\vec{v}=\dfrac{\sqrt{3}}{2}\vec{i}-\dfrac{3}{2}\vec{j}$
Work Step by Step
$|\vec{v}|=\sqrt{3}$ $,$ $\theta=300^{\circ}$
Multiply the magnitude of the vector $\vec{v}$ by the cosine of $\theta$ to obtain its horizontal component:
Horizontal component
$|\vec{v}|\cos\theta=\sqrt{3}\cos300^{\circ}=\sqrt{3}\Big(\dfrac{1}{2}\Big)=\dfrac{\sqrt{3}}{2}$
Multiply the magnitude of the vector $\vec{v}$ by the sine of $\theta$ to obtain its vertical component:
Vertical component
$|\vec{v}|\sin\theta=\sqrt{3}\sin300^{\circ}=\sqrt{3}\Big(-\dfrac{\sqrt{3}}{2}\Big)=-\dfrac{(\sqrt{3})^{2}}{2}=-\dfrac{3}{2}$
Write the vector in terms of $\vec{i}$ and $\vec{j}$. Do so by adding the product between its horizontal component and $\vec{i}$ and the product between its vertical component and $\vec{j}$:
$\vec{v}=\dfrac{\sqrt{3}}{2}\vec{i}-\dfrac{3}{2}\vec{j}$
