Question

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

Answer 1

`hatu = < -sqrt6/6, -sqrt6/6, sqrt6/3 >`. See below.

Explanation:

To normalize a vector is to find a unit vector in the same direction as the given vector, where a unit vector, `hat u`, is a vector with a magnitude of `1`. In other words, the normalized vector `hat u` of a non-zero vector u is a unit vector in the direction of u.

We can find a unit vector in the same direction as the given vector by dividing the given vector by its magnitude.

`hat u = u /| u|`

Where the magnitude of a vector is given by:

`|u|=sqrt((u_x)^2+(u_y)^2+(u_z)^2)`

Given the vector, u `=< -1,-1,2 >`,

`|u|=sqrt((-1)^2+(-1)^2+(2)^2)`

`|u|=sqrt(1+1+4)`

`|u|=sqrt6`

We then have

`hat u= (< -1,-1,2 >) /sqrt6`

`=>hat u = < -1/sqrt6,-1/sqrt6,2/sqrt6 >`

Or equivalently, by rationalizing the denominator:

`=> hatu = < -sqrt6/6, -sqrt6/6, sqrt6/3 >`