How do you normalize (3i - j - 2k) (3i−j−2k)?
1 Answer
Feb 26, 2017
Divide by its length to get:
(3sqrt(14))/14i-sqrt(14)/14j-sqrt(14)/7k3√1414i−√1414j−√147k
Explanation:
Divide it by its length:
abs(abs(3i-j-2k)) = sqrt(3^2+(-1)^2+(-2)^2) = sqrt(9+1+4) = sqrt(14)||3i−j−2k||=√32+(−1)2+(−2)2=√9+1+4=√14
So the unit length vector in the same direction as
1/sqrt(14) (3i-j-2k) = sqrt(14)/14(3i-j-2k)1√14(3i−j−2k)=√1414(3i−j−2k)
color(white)(1/sqrt(14) (3i-j-2k)) = (3sqrt(14))/14i-sqrt(14)/14j-sqrt(14)/7k1√14(3i−j−2k)=3√1414i−√1414j−√147k
