Question

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

Answer 1

Divide by its length to get:

`(3sqrt(14))/14i-sqrt(14)/14j-sqrt(14)/7k`

Explanation:

Divide it by its length:

`abs(abs(3i-j-2k)) = sqrt(3^2+(-1)^2+(-2)^2) = sqrt(9+1+4) = sqrt(14)`

So the unit length vector in the same direction as `3i-j-2k` is:

`1/sqrt(14) (3i-j-2k) = sqrt(14)/14(3i-j-2k)`

`color(white)(1/sqrt(14) (3i-j-2k)) = (3sqrt(14))/14i-sqrt(14)/14j-sqrt(14)/7k`