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How do you find the unit vector in the direction of the vector a = 2i + 3j?

1 Answer

Oct 26, 2016

$\to ˆ a = \frac{\sqrt{13}}{13} (2 \to i + 3 \to j)$ $vec(hata)=sqrt13/13(2veci+3vecj)$

Explanation:

unit vector in direction of $\to a$ $veca$ is given by

$\to ˆ a = \frac{\to a}{| a |}$ $vec(hata)=veca/|a|$

so for $\to a = 2 \to i + 3 \to j$ $veca=2veci+3vecj$

$\to ˆ a = \frac{1}{\sqrt{2^{2} + 3^{2}}} (2 \to i + 3 \to j)$ $vec(hata)=1/sqrt(2^2+3^2)(2veci+3vecj)$

$\to ˆ a = \frac{1}{\sqrt{13}} (2 \to i + 3 \to t j)$ $vec(hata)=1/sqrt13(2veci+3vectj)$

$\to ˆ a = \frac{\sqrt{13}}{13} (2 \to i + 3 \to j)$ $vec(hata)=sqrt13/13(2veci+3vecj)$

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