How do you normalize $(4 i + 4 j + 2 k)$ ?

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Answer 1 · 2018-04-02T17:57:58

Dividing each component by his module

Module of vector is $sqrt(4^2+4^2+2^2)=sqrt36=6$

So the normalized is $4/6veci+4/6vecj+2/6veck=2/3veci+2/3vecj+1/3veck$

Check the answer $sqrt((2/3)^2+(2/3)^2+(1/3)^2)=sqrt1=1$

How do you normalize (4 i + 4 j + 2 k) (4 i + 4 j + 2 k)?