Answer
$a)$ $\vec{u}\cdot\vec{v}=-1$
$b)$ $\theta\approx97^{\circ}$
Work Step by Step
$\vec{u}=\langle3,-2\rangle$ $,$ $\vec{v}=\langle1,2\rangle$
$a)$
To find $\vec{u}\cdot\vec{v}$, multiply corresponding components and add:
$\vec{u}\cdot\vec{v}=(3)(1)+(-2)(2)=3-4=-1$
$b)$
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 $\vec{v}$:
$|\vec{u}|=\sqrt{3^{2}+(-2)^{2}}=\sqrt{9+4}=\sqrt{13}$
$|\vec{v}|=\sqrt{1^{2}+2^{2}}=\sqrt{1+4}=\sqrt{5}$
Substitute the known values into the formula and evaluate:
$\theta=\cos^{-1}\dfrac{-1}{(\sqrt{13})(\sqrt{5})}=\cos^{-1}\dfrac{-1}{\sqrt{65}}\approx97^{\circ}$
