Question

How do you normalize `<3,-6,4>`?

Answer 1

`u=<3/sqrt(61),-6/sqrt(61),4/sqrt(61)>`

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,-6,4>`, we can calculate the magnitude of the vector:

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

`|v|=sqrt((3)^2+(-6)^2+(4)^2)`

`|v|=sqrt(9+36+16)`

`|v|=sqrt(61)`

We now have:

`u=(<3,-6,4>)/sqrt(61)`

`=>u=<3/sqrt(61),-6/sqrt(61),4/sqrt(61)>`

Hope that helps!