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 713: 22


Proof given below.

Work Step by Step

The diagonal is ${\bf u}+{\bf v}$ The cosine of the angle $\theta_{1}$, between ${\bf u}+{\bf v}$ and ${\bf u}$ is $\displaystyle \cos\theta_{1}=\frac{({\bf u+v})\cdot{\bf u}}{|{\bf u+v}||{\bf u}|}$ The cosine of the angle $\theta_{2}$, between ${\bf u}+{\bf v}$ and ${\bf v}$ is $\displaystyle \cos\theta_{2}=\frac{({\bf u+v})\cdot{\bf v}}{|{\bf u+v}||{\bf v}|}$ The denominators in these expressions are equal, because $|{\bf u}|=|{\bf v}|$. If we prove that the numerators are equal, because cosine is one-to-one on $[0,\pi]$, it would follow that $\theta_{1}=\theta_{2}.$ We will apply Properties of the Dot Product (boxed on p. 709) $({\bf u+v})\cdot{\bf u}={\bf u}\cdot({\bf u+v}) \quad$ ... property 1 =${\bf u}\cdot{\bf u} +{\bf u}\cdot{\bf v}\quad$ ... property 3 = $|{\bf u}|^{2} +{\bf u}\cdot{\bf v}\quad$ ... property 4 = $|{\bf v}|^{2} +{\bf u}\cdot{\bf v}\quad$ ... given in the problem =${\bf v}\cdot{\bf v} +{\bf u}\cdot{\bf v}\quad$ ... property $4$ $={\bf v}\cdot({\bf u+v}) \quad$ ... property $3$ $=({\bf u+v})\cdot{\bf v} \quad$ ... property 1 The numerators are equal as well. Thus, $\theta_{1}=\theta_{2}.$
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