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 $= \frac{2}{\sqrt{5}} i + \frac{1}{\sqrt{5}} j$ $=2/sqrt5i+1/sqrt5j$

The unit vector of $\to v$ $vecv$ is

$ˆ v = \frac{\to v}{∥ ∥ \to v ∥ ∥}$ $hatv=vecv/(∥vecv∥)$

$\to v = 4 i + 2 j = ⟨ 4, 2 ⟩$ $vecv=4i+2j=〈4,2〉$

$∥ ∥ \to v ∥ = \sqrt{16 + 2} = \sqrt{20} = 2 \sqrt{5}$ $∥vecv∥=sqrt(16+2)=sqrt20=2sqrt5$

Therefore,

$ˆ v = \frac{1}{2 \sqrt{5}} ⟨ 4, 2 ⟩ = ⟨ \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} ⟩$ $hatv=1/(2sqrt5)〈4,2〉=〈2/sqrt5,1/sqrt5〉$

Answer 2 · 2016-11-28T18:02:59

$(\frac{4}{\sqrt{20}}, \frac{2}{\sqrt{20}})$ $(4/sqrt20, 2/sqrt20)$

$\to v$ $vecv$ is (4,2) Its magnitude $∣ ∣ ∣ ∣ \to v ∣ ∣ ∣ ∣ = \sqrt{4^{2} + 2^{2}} = \sqrt{20}$ $||vec v|| =sqrt(4^2 +2^2)= sqrt20$

Hence unit vector would be $(\frac{4}{\sqrt{20}}, \frac{2}{\sqrt{20}})$ $(4/sqrt20, 2/sqrt20)$