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: 41

Answer

$\sqrt{105}$.

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: $||v-w||=||(3i-5j+2k)-(-2i+3j-2k)||=||5i-8j+4k||=\sqrt{5^2+(-8)^2+4^2}=\sqrt{25+64+16}=\sqrt{105}$.
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