Explanation:
In this exercise we have the following vector:
[tex]\vec{v}=(8,-4)[/tex]
So we need to find a unit vector [tex]\vec{u}[/tex] which is defined as follows:
[tex]\vec{u}=\frac{\vec{v}}{\mid \vec{v} \mid}[/tex]
Finding the magnitude of the vector:
[tex]\mid \vec{v} \mid=\sqrt{(8)^2+(-4)^2} \\ \\ \mid \vec{v} \mid=\sqrt{64+16} \\ \\ \mid \vec{v} \mid=\sqrt{80} \\ \\ \mid \vec{v} \mid=4\sqrt{5}[/tex]
Then, by substituting the unit vector is:
[tex]\vec{u}=\frac{(8,-4)}{4\sqrt{5}} \\ \\ \vec{u}=\frac{(8,-4)}{4\sqrt{5}} \\ \\ \vec{u}=(\frac{8}{4\sqrt{5}},-\frac{4}{4\sqrt{5}}) \\ \\[/tex]
[tex]\vec{u}=(\frac{8}{4\sqrt{5}}\frac{\sqrt{5}}{\sqrt{5}},-\frac{4}{4\sqrt{5}}\frac{\sqrt{5}}{\sqrt{5}}) \\ \\ \\ \boxed{\vec{u}=\left(\frac{2\sqrt{5}}{5},-\frac{5}{\sqrt{5}}\right)}[/tex]
