Answer
$a)$ $\vec{u}\cdot\vec{v}=2$
$b)$ $\theta=45^{\circ}$
Work Step by Step
$\vec{u}=\langle2,0\rangle$ $,$ $\vec{v}=\langle1,1\rangle$
$a)$
To find $\vec{u}\cdot\vec{v}$, multiply corresponding components and add:
$\vec{u}\cdot\vec{v}=(2)(1)+(0)(1)=2+0=2$
$b)$
The angle between two vectors is given by the following formula:
$$\theta=\cos^{-1}\dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}$$
Therefore, obtain the magnitude of both $\vec{u}$ and $\vec{v}$:
$|\vec{u}|=\sqrt{2^{2}+0^{2}}=\sqrt{4}=2$
$|\vec{v}|=\sqrt{1^{2}+1^{2}}=\sqrt{1+1}=\sqrt{2}$
Substitute the known values into the formula and evaluate:
$\theta=\cos^{-1}\dfrac{2}{2\sqrt{2}}=\cos^{-1}\dfrac{1}{\sqrt{2}}=\cos^{-1}\dfrac{\sqrt{2}}{2}=45^{\circ}$
