Respuesta :

Answer:

Step-by-step explanation:

v=<3,-1>

[tex]||v||=\sqrt{x^{2} +y^{2} } \\\sqrt{(3)^{2}+(-1)^{2} } \\\sqrt{9+1} =\sqrt{10} \\u=\frac{1}{||v||} *v\\u=\frac{1}{\sqrt{10} } *<3,-1>\\u=<\frac{3}{\sqrt{10} } ,-\frac{1}{\sqrt{10} } >\\[/tex]