Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 794: 79

Answer

.

Work Step by Step

Consider the given identity $\mathbf{u}\cdot \mathbf{v}=~~\mathbf{v}\cdot \mathbf{u}$ …… (1) For the given vectors $\mathbf{u}={{a}_{1}}\mathbf{i}+{{b}_{1}}\mathbf{j}$ $\mathbf{v}={{a}_{2}}\mathbf{i}+{{b}_{2}}\mathbf{j}$ Now, take the left side of equation (1). The left-hand side of the given identity represents the dot product of vectors $\mathbf{u}\ \text{ and }\ \mathbf{v}$. It is calculated as follows: $\mathbf{u}\cdot \mathbf{v}={{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}$ …… (2) Now, take the right side of equation (1). The right side of the given identity represents the dot product of $\mathbf{v}\ \text{ and }\ \mathbf{u}$. It is calculated as follows: $\mathbf{v}\cdot \mathbf{u}={{a}_{2}}{{a}_{1}}+{{b}_{2}}{{b}_{1}}$ …… (3) Equations (2) and (3) are identical. So, Left side of equation (1) = Right side ofequation (1) Hence, $\mathbf{u}\cdot \mathbf{v}=~~\mathbf{v}\cdot \mathbf{u}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.