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: 44


The required vector is $ \frac{12}{5}i+ \frac{9}{5}j $.

Work Step by Step

Since the vector is in the direction of $ v $ then it will be in the form $ av= 4ai+3aj $. Now, since the length is $3$, then we have $$\sqrt{16a^2+9a^2}=\sqrt{25a^2}=3\Longrightarrow 5a=3\Longrightarrow a= \frac{3}{5}.$$ The required vector is $ \frac{12}{5}i+ \frac{9}{5}j $.
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