Answer
$\sqrt{2}$
Work Step by Step
The magnitude of a vector $v=ai+bj$ is: $||v||=\sqrt{a^2+b^2}$.
Hence, here we have
$||v||
\\=\sqrt{(-1)^2+(-1)^2}
\\=\sqrt{1+1}
\\=\sqrt{2}$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 9 - Polar Coordinates; Vectors - 9.4 Vectors - 9.4 Assess Your Understanding - Page 606: 40
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