Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 9 - Polar Coordinates; Vectors - 9.6 Vectors in Space - 9.6 Assess Your Understanding - Page 625: 42



Work Step by Step

We know that if $v=ai+bj+ck$ and $w=di+ej+fk$, then $v+w=(a+d)i+(b+e)j+(c+f)k$ and that if $z$ is a constant, then $zv=(za)i+(zb)j+(zc)k$. Also, the magnitude of a vector $v=ai+bj+ck$ is: $||v||=\sqrt{a^2+b^2+c^2}$. Hence here: $||v+w||=||(3i-5j+2k)+(-2i+3j-2k)||=||i-2j||=\sqrt{1^2+(-2)^2}=\sqrt{1+4}=\sqrt{5}$.
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