Question

URGENTEEE 911
Ejercicio de Vectores

Answer 1

$\displaystyle \textbf{a)} \\[4pt] \vec{a}=(a_1,a_2) \ \Rightarrow \ \vec{u}_a=\frac{\vec{a}}{\|\vec{a}\|}=\left( \frac{a_1}{\|\vec{a}\|} \ , \ \frac{a_2}{\|\vec{a}\|}\right) \\[4pt]$

$\displaystyle \vec{b}=(4,2) \ \rightarrow \ \|\vec{b}\|=\sqrt{4^2+2^2}=2\sqrt{5} \\[4pt] \Rightarrow \ \vec{u}_b=\left( \frac{4}{2\sqrt{5}} \ , \, \frac{2}{2\sqrt{5}}\right)=\left( \frac{2}{\sqrt{5}} \ , \, \frac{1}{\sqrt{5}}\right)$

$\displaystyle \vect{u}\cdot \vect{v}=\|\vect{u}\|\|\vect{v}\|\cos\theta \\[6pt] \vec{a}=\left(\frac{1}{4},-3 \right) \ \rightarrow \ \|\vec{a}\|=\sqrt{\left(\frac{1}{4}\right)^2+(-3)^2}=\frac{\sqrt{145}}{4} \\[4pt] \vec{b}=(4,2) \rightarrow \ \|\vec{b}\|=2\sqrt{5} \\[6pt] \text{reemplazando en la f\'{o}rmula} : \\[4pt] \left(\frac{1}{4},-3 \right) \cdot (4,2)=\frac{\sqrt{145}}{4} (2\sqrt{5} )\cos\theta  \\[6pt]$ 

$\displaystyle 1-6=\frac{\sqrt{145}}{4} (2\sqrt{5} )\cos\theta \\[4pt] \Rightarrow \cos\theta =\frac{-2}{\sqrt{29}} \ \Rightarrow \ \theta = \arccos\left( \frac{-2}{\sqrt{29}} \right) \approx 111.8\°$