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

1 Answer
Feb 26, 2017

Answer:

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#