Answer
$a)$ $\vec{u}\cdot\vec{v}=13$
$b)$ $\theta\approx56^{\circ}$
Work Step by Step
$\vec{u}=\langle2,7\rangle$ $,$ $\vec{v}=\langle3,1\rangle$
$a)$
To find $\vec{u}\cdot\vec{v}$, multiply corresponding components and add:
$\vec{u}\cdot\vec{v}=(2)(3)+(7)(1)=6+7=13$
$b)$
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 $\vec{v}$:
$|\vec{u}|=\sqrt{2^{2}+7^{2}}=\sqrt{4+49}=\sqrt{53}$
$|\vec{v}|=\sqrt{3^{2}+1^{2}}=\sqrt{9+1}=\sqrt{10}$
Substitute the known values into the formula and evaluate:
$\theta=\cos^{-1}\dfrac{13}{(\sqrt{53})(\sqrt{10})}=\cos^{-1}\dfrac{13}{\sqrt{530}}\approx56^{\circ}$
