Calculus (3rd Edition)

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

Chapter 13 - Vector Geometry - 13.1 Vectors in the Plane - Exercises - Page 650: 46


$$\left\langle\frac{1}{\sqrt 5},-\frac{2}{\sqrt 5}\right\rangle.$$

Work Step by Step

The unit vector in the direction opposite to $ v $ is given by $$-e_v= -\frac{v}{\|v\|}=-\frac{\left\langle-2,4\right\rangle}{\sqrt{4+16}}=\left\langle\frac{2}{2\sqrt 5},-\frac{4}{2\sqrt 5}\right\rangle.$$ Simplifying, we get: $$\left\langle\frac{1}{\sqrt 5},-\frac{2}{\sqrt 5}\right\rangle.$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.