How do you normalize (-2i- 3j + 2k) (−2i−3j+2k)?

1 Answer
Morgan
Mar 5, 2017

u=<-2/sqrt(17),-3/sqrt(17),2/sqrt(17)>u=<−2√17,−3√17,2√17>

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=<-2,-3,2>v=<−2,−3,2>, 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((-2)^2+(-3)^2+(2)^2)|v|=√(−2)2+(−3)2+(2)2

|v|=sqrt(4+9+4)|v|=√4+9+4

|v|=sqrt(17)|v|=√17

We now have:

u=(<-2,-3,2>)/sqrt(17)u=<−2,−3,2>√17

=>u=<-2/sqrt(17),-3/sqrt(17),2/sqrt(17)>⇒u=<−2√17,−3√17,2√17>

Hope that helps!

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