How do you find the unit vector of v=4i+2j?

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Answer 1 · 2016-11-28T18:01:38

The answer is $=2/sqrt5i+1/sqrt5j$

The unit vector of $vecv$ is

$hatv=vecv/(∥vecv∥)$

$vecv=4i+2j=〈4,2〉$

$∥vecv∥=sqrt(16+2)=sqrt20=2sqrt5$

Therefore,

$hatv=1/(2sqrt5)〈4,2〉=〈2/sqrt5,1/sqrt5〉$

Answer 2 · 2016-11-28T18:02:59

$(4/sqrt20, 2/sqrt20)$

$vecv$ is (4,2) Its magnitude $||vec v|| =sqrt(4^2 +2^2)= sqrt20$

Hence unit vector would be $(4/sqrt20, 2/sqrt20)$