Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 9 - Section 9.1 - Vectors in Two Dimensions - 9.1 Exercises - Page 637: 40

Answer

$6\sqrt 2$, $\sqrt 5$, $12\sqrt 2$, $\frac{\sqrt 5}{2}$, $\sqrt {89}$, $\sqrt {65}$, $6\sqrt 2-\sqrt 5$

Work Step by Step

Given $\vec u=\langle -6, 6 \rangle$ and $\vec v=\langle -2, -1 \rangle$, use for magnitude formula, we have: (i) $|\vec u|=|\langle -6, 6 \rangle|=\sqrt {(-6)^2+6^2}=6\sqrt 2$ (ii) $|\vec v|=|\langle -2, -1 \rangle|=\sqrt {(-2)^2+(-1)^2}=\sqrt 5$ (iii) $|2\vec u|=|\langle -12, 12 \rangle|=\sqrt {(-12)^2+12^2}=12\sqrt 2$ (iv) $|\frac{1}{2}\vec v|=|\langle -1, -1/2 \rangle|=\sqrt {(-1)^2+(-\frac{1}{2})^2}=\frac{\sqrt 5}{2}$ (v) $|\vec u+\vec v|=|\langle -6-2, 6-1 \rangle|=\sqrt {(-8)^2+(5)^2}=\sqrt {89}$ (vi) $|\vec u-\vec v|=|\langle -6+2, 6+1 \rangle|=\sqrt {(-4)^2+(7)^2}=\sqrt {65}$ (vii) Use results from (i) and (ii), $|\vec u|-|\vec v|=6\sqrt 2-\sqrt 5$
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