Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - 13.3 Dot Product and the Angle Between Two Vectors - Exercises - Page 667: 68

Answer

$\textbf{a}=⟨4,2,2⟩+⟨0,-3,3⟩$

Work Step by Step

We are given the vectors $a=<4,-1,5>$ and $b=<2,1,1>$. We have: $\textbf{a}_{\parallel \textbf{b}}=(\frac{\textbf{a}\cdot\textbf{b}}{\textbf{b}\cdot\textbf{b}})\textbf{b}=\frac{4\times2+(-1)\times1+5\times1}{2^{2}+1^{2}+1^{2}}⟨2,1,1⟩=⟨4,2,2⟩$ $\textbf{a}_{\perp \textbf{b}}=\textbf{a}-\textbf{a}_{\parallel \textbf{b}}=⟨4,-1,5⟩-⟨4,2,2⟩$ $=⟨0,-3,3⟩$ Therefore, the decomposition of $\textbf{a}$ with respect to $\textbf{b}$ is $\textbf{a}=\textbf{a}_{\parallel \textbf{b}}+\textbf{a}_{\perp \textbf{b}}$ $=⟨4,2,2⟩+⟨0,-3,3⟩$
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