Answer
$a)$ $\vec{u}\cdot\vec{v}=0$
$b)$ $\theta=90^{\circ}$
Work Step by Step
$\vec{u}=\vec{i}+\sqrt{3}\vec{j}$ $,$ $\vec{v}=-\sqrt{3}\vec{i}+\vec{j}$
$a)$
To find $\vec{u}\cdot\vec{v}$, multiply corresponding components and add:
$\vec{u}\cdot\vec{v}=(1)(-\sqrt{3})+(\sqrt{3})(1)=-\sqrt{3}+\sqrt{3}=0$
$b)$
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 the magnitudes of $\vec{u}$ and $\vec{v}$, respectively.
Find the magnitude of both $\vec{u}$ and $\vec{v}$:
$|\vec{u}|=\sqrt{1^{2}+(\sqrt{3})^{2}}=\sqrt{1+3}=\sqrt{4}=2$
$|\vec{v}|=\sqrt{(-\sqrt{3})^{2}+1^{2}}=\sqrt{3+1}=\sqrt{4}=2$
Substitute the known values into the formula and evaluate:
$\theta=\cos^{-1}\dfrac{0}{(2)(2)}=\cos^{-1}0=90^{\circ}$
