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Normalize the vector <3,4>.
First we have to find out how long the vector is.
Since the vector has magnitude 5 and we want it to have magnitude 1, the vector is 5 times too long.
We need to ``divide" the vector <3,4 > by 5, which is the same as multiplying by .
We claim v is a unit vector. Check by finding its magnitude:
Yes, v is a unit vector.
Hmm. The normalization of <3,4 > is
This happens to be the same thing as
Normalize the vector
Find out how long the vector is.
This vector is too short. We want it to have length 1, but it only has length .
If we multiply the vector by 2, we'll get a new vector with magnitude 1.
The magnitude of v is
Therefore v is a unit vector, which is what we wanted.
Remember your fraction division: multiplying by 2 is the same thing as dividing by .
We found the normalization of .
Since multiplying by 2 is the same thing as dividing by , we could also write the vector as
which happens to be the same thing as
If we're given a vector v and asked to normalize it, find the vector
This is guaranteed to be a unit vector pointing in the same direction as v (unless v is 0, but that's not a vector). What direction does <0,0> point?. It doesn't.