How do you normalize (3i−j−2k)?

1 Answer
George C.
Feb 26, 2017

Divide by its length to get:

3√1414i−√1414j−√147k

Explanation:

Divide it by its length:

||3i−j−2k||=√32+(−1)2+(−2)2=√9+1+4=√14

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

1√14(3i−j−2k)=√1414(3i−j−2k)

1√14(3i−j−2k)=3√1414i−√1414j−√147k

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