Question

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

Answer 1

`u=<-3/sqrt(178),12/sqrt(178),-5/sqrt(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|)`

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

`|v|=sqrt((v_x)^2+(v_y)^2+(v_z)^2)`

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

`|v|=sqrt(9+144+25)`

`|v|=sqrt(178)`

We now have:

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

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

Hope that helps!