Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Section 12.2 - Vectors - 12.2 Exercises - Page 805: 22

Answer

$\vec{a}$ $+$ $\vec{b}$ $=$ $\lt$ $13$, $-1$, $-3$ $\gt$ $4$$\vec{a}$ $+$ $2$$\vec{b}$ $=$ $\lt$ $42$, $0$, $-14$ $\gt$ $|$ $\bf{a}$ $|$ $=$ $9$ $|$ $\bf{a-b}$ $|$ $=$ $\sqrt{43}$

Work Step by Step

$\vec{a}$ $+$ $\vec{b}$ $=$ $\lt$ $8+5$, $1+(-2)$, $(-4+1)$ $\gt$ $=$ $\lt$ $13$, $-1$, $-3$ $\gt$ $4$$\vec{a}$ $+$ $2$$\vec{b}$ $=$ $4$$\lt$ $8$, $1$, $-4$, $\gt$ $+$ $2$ $\lt$ $5$, $-2$, $1$ $\gt$ $=$ $\lt$ $32$, $4$, $-16$ $\gt$ $+$ $\lt$ $10$, $-4$, $2$ $\gt$ $=$ $\lt$ $42$, $0$, $-14$ $\gt$ $|$ $\bf{a}$ $|$ $=$ $\sqrt{8^2+1^2+(-4)^2}$ $=$ $\sqrt{81}$ $=$ $9$ $|$ $\bf{a-b}$ $|$ $=$ $|$ $\lt$ $8-5$, $1-(-2)$, $-4-1$$\gt$ $|$ $=$ $|$ $\lt$ $3$, $3$, $-5$$\gt$ $|$ $=$ $\sqrt{3^2+3^2+(-5)^2}$ $=$ $\sqrt{43}$
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