Answer
$a)$ $\vec{u}\cdot\vec{v}=4$
$b)$ $\theta\approx60^{\circ}$
Work Step by Step
$\vec{u}=2\vec{i}+\vec{j}$ $,$ $\vec{v}=3\vec{i}-2\vec{j}$
$a)$
To find $\vec{u}\cdot\vec{v}$, multiply corresponding components and add:
$\vec{u}\cdot\vec{v}=(2)(3)+(1)(-2)=6-2=4$
$b)$
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 $\vec{v}$:
$|\vec{u}|=\sqrt{2^{2}+1^{2}}=\sqrt{4+1}=\sqrt{5}$
$|\vec{v}|=\sqrt{3^{2}+(-2)^{2}}=\sqrt{9+4}=\sqrt{13}$
Substitute the known values into the formula and evaluate:
$\theta=\cos^{-1}\dfrac{4}{(\sqrt{5})(\sqrt{13})}=\cos^{-1}\dfrac{4}{\sqrt{65}}\approx60^{\circ}$
