Answer
$a)$ $\vec{u}\cdot\vec{v}=-10$
$b)$ $\theta\approx153^{\circ}$
Work Step by Step
$\vec{u}=3\vec{i}+4\vec{j}$ $,$ $\vec{v}=-2\vec{i}-\vec{j}$
$a)$
To find $\vec{u}\cdot\vec{v}$, multiply corresponding components and add:
$\vec{u}\cdot\vec{v}=(3)(-2)+(4)(-1)=-6-4=-10$
$b)$
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 $\vec{v}$:
$|\vec{u}|=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5$
$|\vec{v}|=\sqrt{(-2)^{2}+(-1)^{2}}=\sqrt{4+1}=\sqrt{5}$
Substitute the known values into the formula and evaluate:
$\theta=\cos^{-1}\dfrac{-10}{5\sqrt{5}}=\cos^{-1}\dfrac{-2}{\sqrt{5}}\approx153^{\circ}$
