Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.2 - Vectors - Exercises 12.2 - Page 704: 1

Answer

$\lt9, -6\gt$ and $3 \sqrt {13}$

Work Step by Step

Since, $3u=3\lt 3, -2 \gt =\lt9, -6\gt$ The magnitude of a vector is: $|n|=\sqrt{n_1^2+n_2^2}$ Now, $|\lt9, -6\gt|=\sqrt{9^2+(-6)^2}=\sqrt {117}=3 \sqrt {13}$ Thus, answers are: $\lt9, -6\gt$ and $3 \sqrt {13}$
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.