University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.3 - The Dot Product - Exercises - Page 616: 18


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

Work Step by Step

Since, we have two vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ and $(-v+(-u)) \cdot (-v+u)=v \cdot v-v \cdot u+u \cdot v -u \cdot u$ Now, $v \cdot v-v \cdot u+u \cdot v -u \cdot u=|v|^2-|u|^2$ Here, both given vectors acquire the same radius of the circle so,$|v|^2 =|u|^2$ . . or, $|v|^2-|u|^2=|v|^2-|v|^2=0$ Thus, the two vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ are orthogonal.
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