Question 166634
Since *[Tex \LARGE \vec{v}=-5i+12j], this means that *[Tex \LARGE \vec{v}=<-5,12>]



So for any vector *[Tex \LARGE \vec{z}=<a,b>] the magnitude or norm of that vector is:


*[Tex \LARGE ||\vec{z}||=\sqrt{a^2+b^2}]




So in this case



*[Tex \LARGE ||\vec{v}||=\sqrt{(-5)^2+(12)^2}=\sqrt{25+144}=\sqrt{169}=13]



So *[Tex \LARGE ||\vec{v}||=13]



Geometrically, this means that the length of *[Tex \LARGE \vec{v}] is 13 units



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To find the unit vector *[Tex \LARGE \vec{u}] that has the same direction as *[Tex \LARGE \vec{v}], simply divide EACH component of the vector *[Tex \LARGE \vec{v}] by the length of the vector *[Tex \LARGE \vec{v}].



So...

 
*[Tex \LARGE \vec{u}=\frac{\vec{v}}{||\vec{v}||}=\frac{<-5,12>}{13}=<-\frac{5}{13},\frac{12}{13}>]



So the unit vector *[Tex \LARGE \vec{u}] that has the same direction as *[Tex \LARGE \vec{v}] is



*[Tex \LARGE \vec{u}=<-\frac{5}{13},\frac{12}{13}>]