Answer
$a)$ $\vec{u}\cdot\vec{v}=5\sqrt{3}$
$b)$ $\theta=30^{\circ}$
Work Step by Step
$\vec{u}=-5\vec{j}$ $,$ $\vec{v}=-\vec{i}-\sqrt{3}\vec{j}$
$a)$
To find $\vec{u}\cdot\vec{v}$, multiply corresponding components and add:
$\vec{u}\cdot\vec{v}=(0)(-1)+(-5)(-\sqrt{3})=5\sqrt{3}$
$b)$
The angle between two vectors is given by the following formula,
$$\theta=\cos^{-1}\dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}$$
where $\vec{u}\cdot\vec{v}$ is the dot product of $\vec{u}$ and $\vec{v}$, and $|\vec{u}|$ and $|\vec{v}|$ are their magnitudes.
Find the magnitude of both $\vec{u}$ and $\vec{v}$:
$|\vec{u}|=\sqrt{0^{2}+(-5)^{2}}=\sqrt{25}=5$
$|\vec{v}|=\sqrt{(-1)^{2}+(-\sqrt{3})^{2}}=\sqrt{1+3}=\sqrt{4}=2$
Substitute the known values into the formula and evaluate:
$\theta=\cos^{-1}\dfrac{5\sqrt{3}}{(5)(2)}=\cos^{-1}\dfrac{\sqrt{3}}{2}=30^{\circ}$
