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 - Section 12.3 - The Dot Product - Exercises 12.3 - Page 712: 18


Both two vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ are orthogonal.

Work Step by Step

Let $\overrightarrow{CA}$ and $\overrightarrow{CB}$ be the two vectors. Now, $(-v+(-u)) \cdot (-v+u)=v \cdot v-v \cdot u+u \cdot v -u \cdot u$ and $v \cdot v-v \cdot u+u \cdot v -u \cdot u=|v|^2-|u|^2$ when the both vectors posses same radius of circle then we have $|v|^2 =|u|^2$ This implies that $|v|^2-|u|^2=|v|^2-|v|^2=0$ Hence, we conclude that both two vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ are orthogonal.
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