Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - Chapter Review Exercises - Page 701: 3


$$ e_v= \left\langle -\frac{2}{\sqrt{29}},\frac{5}{\sqrt{29}}\right \rangle .$$

Work Step by Step

Since $ v=\langle -2,5\rangle $, then the unit vector in the direction of $ v $ is $ e_v $, which is given by $$ e_v=\frac{v}{\|v\|}=\frac{\langle -2,5\rangle}{\sqrt{4+25}}=\left\langle -\frac{2}{\sqrt{29}},\frac{5}{\sqrt{29}}\right \rangle .$$
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