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: 83


The value of $b$ for which $15\mathbf{i}-3\mathbf{j}$ and $-4\mathbf{i}+b\mathbf{j}$ are orthogonal is $-20$.

Work Step by Step

For two vectors to be orthogonal, their dot product must be zero.Therefore, $\mathbf{u}\cdot \mathbf{v}=0$ Substituting the values of $\mathbf{u}$ and $\mathbf{v}$ from (1) and (2), we get $\left( 15\mathbf{i}-3\mathbf{j} \right)\centerdot \left( -4\mathbf{i}+b\mathbf{j} \right)=0$ …… (3) Solving equation (3) for the value of $b$, we get $\begin{align} & 15\times \left( -4 \right)+\left( -3 \right)b=0 \\ & -60-3b=0 \\ & b=-20 \end{align}$
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.