How do you normalize (- 3 i + 12j -5k)(−3i+12j−5k)?

1 Answer
Morgan
Mar 5, 2017

u=<-3/sqrt(178),12/sqrt(178),-5/sqrt(178)>u=<−3√178,12√178,−5√178>

Explanation:

In normalizing the vector we are finding a unit vector (magnitude/length of one) in the same direction as the given vector. This can be accomplished by dividing the given vector by its magnitude.

u=v/(|v|)u=v|v|

Given v=<-3,12,-5>v=<−3,12,−5>, we can calculate the magnitude of the vector:

|v|=sqrt((v_x)^2+(v_y)^2+(v_z)^2)|v|=√(vx)2+(vy)2+(vz)2

|v|=sqrt((-3)^2+(12)^2+(-5)^2)|v|=√(−3)2+(12)2+(−5)2

|v|=sqrt(9+144+25)|v|=√9+144+25

|v|=sqrt(178)|v|=√178

We now have:

u=(<-3,12,-5>)/sqrt(178)u=<−3,12,−5>√178

=>u=<-3/sqrt(178),12/sqrt(178),-5/sqrt(178)>⇒u=<−3√178,12√178,−5√178>

Hope that helps!

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