## University Calculus: Early Transcendentals (3rd Edition)

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