Answer
$a)$ $\vec{u}\cdot\vec{v}=1$
$b)$ $\theta\approx86^{\circ}$
Work Step by Step
$\vec{u}=\vec{i}+3\vec{j}$ $,$ $\vec{v}=4\vec{i}-\vec{j}$
$a)$
To find $\vec{u}\cdot\vec{v}$, multiply corresponding components and add:
$\vec{u}\cdot\vec{v}=(1)(4)+(3)(-1)=4-3=1$
$b)$
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 $\vec{v}$:
$|\vec{u}|=\sqrt{1^{2}+3^{2}}=\sqrt{1+9}=\sqrt{10}$
$|\vec{v}|=\sqrt{4^{2}+(-1)^{2}}=\sqrt{16+1}=\sqrt{17}$
Substitute the known values into the formula and evaluate:
$\theta=\cos^{-1}\dfrac{1}{(\sqrt{10})(\sqrt{17})}=\cos^{-1}\dfrac{1}{\sqrt{170}}\approx86^{\circ}$
