How do you normalize $<3,1,5>$ ?

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Answer 1 · 2015-12-15T15:16:24

$<3/sqrt35,1/sqrt35,5/sqrt35>$

First, you have to calculate for the magnitude of the vector .

$||v||=sqrt(3^2+1^2+5^2)=sqrt35$

Second, divide the magnitude by each of the component of the vector.

$1/sqrt35 <3,1,5> = <3/sqrt35,1/sqrt35,5/sqrt35>$

How do you normalize <3,1,5> <3,1,5>?