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How do you find a unit vector in the direction of v: v = - 5i + 2j?

1 Answer

Sep 22, 2016

$- \frac{5}{\sqrt{29}} i + \frac{2}{\sqrt{29}} j$ $-5/sqrt29i+2/sqrt29j$ .

Explanation:

A unit vector in the direction of $\to v$ $vecv$ , is denoted by $ˆ \to v$ $hat(vecv)$ ,

and, is defined by,

$ˆ \to v = \frac{\to v}{∣ ∣ ∣ ∣ \to v ∣ ∣ ∣ ∣}, provided, \to v \neq \to 0$ $hat(vecv)=vecv/||vecv||", provided, "vecvnevec0$ .

Here, $\to v = - 5 i + 2 j = (- 5, 2) \neq \to 0$ $vecv=-5i+2j=(-5,2) nevec0$

$\Rightarrow ∣ ∣ ∣ ∣ \to v ∣ ∣ ∣ ∣ = \sqrt{{(- 5)}^{2} + {(2)}^{2}} = \sqrt{29}$ $rArr ||vecv||=sqrt{(-5)^2+(2)^2}=sqrt29$ .

$:. hat(vecv)=-5/sqrt29i+2/sqrt29j$ .

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