Question

How do you normalize <2,0,-1>?

Answer 1

`hatu=<(2sqrt(5))/5,0,(-sqrt(5))/5>`

Explanation:

To normalize a vector is to find unit vector (vector with magnitude/length of one) in the same direction as the given vector. This can be accomplished by dividing the given vector by its magnitude.

`hatu=vecv/(|vecv|)`

Given `vecv=<2,0,-1>`, we can calculate the magnitude of the vector:

`abs(vecv)=sqrt((v_x)^2+(v_y)^2+(v_z)^2)`

`=>=sqrt((2)^2+(0)^2+(-1)^2)`

`=>=sqrt(4+0+1)`

`=>=sqrt(5)`

We now have:

`hatu=(<2,0,-1>)/sqrt(5)`

`=>hatu=<2/sqrt(5),0,-1/sqrt(5)>`

We can also rationalize the denominator on the `hatx` `(hati)` and `hatz` `(hatk)` component:

`=>hatu=<(2sqrt(5))/5,0,(-sqrt(5))/5>`

Hope that helps!