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: 43



Work Step by Step

Since the vector is in the direction $\langle-1,-1\rangle$, then it will be in the form $ au=\langle-a,-a\rangle $. Now, since the length is $4$, then we have $$\sqrt{a^2+a^2}=\sqrt{2a^2}=4\Longrightarrow 2a^2=16\Longrightarrow a=\sqrt{8}=2\sqrt{2}.$$ Hence the required vector is given by the components $$\langle-2\sqrt{2},-2\sqrt{2}\rangle.$$
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