Respuesta :
[tex]\bf \textit{unit vector}
\\\\
\cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}}\ ,\ \cfrac{b}{\sqrt{a^2+b^2}}\\\\
-------------------------------\\\\
\cfrac{\ \textless \ 2,-6\ \textgreater \ }{\sqrt{(2)^2+(-6)^2}}\implies \cfrac{2,-6}{\sqrt{40}}\implies \cfrac{2,6}{2\sqrt{10}}\implies \cfrac{2}{2\sqrt{10}}\ ,\ \cfrac{-6}{2\sqrt{10}}
\\\\\\
\boxed{\cfrac{1}{\sqrt{10}}\ ,\ \cfrac{-3}{\sqrt{10}}}[/tex]
[tex]\bf -------------------------------\\\\ \cfrac{\ \textless \ -4,7\ \textgreater \ }{\sqrt{(-4)^2+(7)^2}}\implies \cfrac{-4,7}{\sqrt{65}}\implies \boxed{\cfrac{-4}{\sqrt{65}}\ ,\ \cfrac{7}{\sqrt{65}}}[/tex]
[tex]\bf -------------------------------\\\\ \cfrac{\ \textless \ -4,7\ \textgreater \ }{\sqrt{(-4)^2+(7)^2}}\implies \cfrac{-4,7}{\sqrt{65}}\implies \boxed{\cfrac{-4}{\sqrt{65}}\ ,\ \cfrac{7}{\sqrt{65}}}[/tex]
Answer:
Here, given vectors,
[tex]v_1=(2, -6)[/tex]
[tex]\implies \overrightarrow{v_1}=2i-6j[/tex]
Thus, the unit vector would be,
[tex]\hat{v_1}=\frac{\overrightarrow{v_1}}{|\overrightarrow{v_1}|}[/tex]
[tex]=\frac{2i-6j}{\sqrt{2^2+6^2}}[/tex]
[tex]=\frac{2i-6j}{\sqrt{4+36}}[/tex]
[tex]=\frac{2i-6j}{\sqrt{40}}[/tex]
[tex]=(\frac{2}{\sqrt{40}}, -\frac{6}{\sqrt{40}})[/tex]
Now, [tex]v_2=(-4, 7)[/tex]
[tex]\implies \overrightarrow{v_2}=-4i+7j[/tex]
Thus, the unit vector would be,
[tex]\hat{v_2}=\frac{\overrightarrow{v_2}}{|\overrightarrow{v_2}|}[/tex]
[tex]=\frac{-4i+7j}{\sqrt{4^2+7^2}}[/tex]
[tex]=\frac{-4i+7j}{\sqrt{16+49}}[/tex]
[tex]=\frac{-4i+7j}{\sqrt{65}}[/tex]
[tex]=(-\frac{4}{\sqrt{65}}, \frac{7}{\sqrt{65}})[/tex]
