Answer
$a)$ $\vec{u}\cdot\vec{v}=-12$
$b)$ $\theta=180^{\circ}$
Work Step by Step
$\vec{u}=\langle-6,6\rangle$ $,$ $\vec{v}=\langle1,-1\rangle$
$a)$
To find $\vec{u}\cdot\vec{v}$, multiply corresponding components and add:
$\vec{u}\cdot\vec{v}=(-6)(1)+(6)(-1)=-6-6=-12$
$b)$
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 $\vec{v}$:
$|\vec{u}|=\sqrt{(-6)^{2}+(6)^{2}}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}$
$|\vec{v}|=\sqrt{(1)^{2}+(-1)^{2}}=\sqrt{1+1}=\sqrt{2}$
Substitute the known values into the formula and evaluate:
$\theta=\cos^{-1}\dfrac{-12}{(6\sqrt{2})(\sqrt{2})}=\cos^{-1}\dfrac{-12}{6(2)}=\cos^{-1}\dfrac{-12}{12}=...$
$...=\cos^{-1}(-1)=180^{\circ}$
