Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Practice Exercises - Page 734: 23



Work Step by Step

Let ${\bf v}=\langle v_{1},v_{2},v_{3}\rangle,\qquad {\bf w}=\langle w_{1},w_{2},w_{3}\rangle$ ${\bf v-2w}=\langle v_{1}-2w_{1},v_{2}-2w_{2},v_{3}-2w_{3}\rangle$ $|{\bf v-2w}|^{2}=(v_{1}-2w_{1})^{2}+(v_{2}-2w_{2})^{2}+(v_{3}-2w_{3})^{2}$ $=(v_{1}^{2}+v_{2}^{2}+v_{3}^{2})+4(w_{1}^{2}+w_{2}^{2}+w_{3}^{2})-4(v_{1}w_{1}+v_{2}w_{2}+v_{3}w_{3})$ $=|{\bf v}|^{2}+4|{\bf w}|^{2}-4({\bf v}\cdot{\bf w})$ We are given $|{\bf v}|, |{\bf w}|$ and we know that ${\bf v}\cdot{\bf w}=|{\bf v}|\cdot |{\bf w}|\cdot\cos\theta\qquad $ $|{\bf v-2w}|^{2}=2^{2}+4\displaystyle \cdot 3^{2}-4\cdot 2\cdot 3\cdot\cos\frac{\pi}{3}$ $|{\bf v-2w}|^{2}=4+36-12=28$ $|{\bf v-2w}|\}=\sqrt{28}=2\sqrt{7}$
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