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: 86


The application of vectors can be used to get the angle between the two different vectors subtended at some angle.

Work Step by Step

Assume that the group member should research the forces acting on the object; the forces are: ${{\mathbf{F}}_{1}}=\left( 3\mathbf{i}+4\mathbf{j} \right)\text{ N}$ and ${{\mathbf{F}}_{2}}=\left( \mathbf{i}+\mathbf{j} \right)\text{ N}$ To find out the angle between these two vectors, we use the dot product: ${{\mathbf{F}}_{1}}\cdot {{\mathbf{F}}_{2}}=\left\| {{\mathbf{F}}_{1}} \right\|\left\| {{\mathbf{F}}_{2}} \right\|\cos \theta $ The magnitude of ${{\mathbf{F}}_{1}}$ is, $\begin{align} & {{\mathbf{F}}_{1}}=\sqrt{{{3}^{2}}+{{4}^{2}}} \\ & =5 \end{align}$ The magnitude of ${{\mathbf{F}}_{2}}$ is, $\begin{align} & {{\mathbf{F}}_{2}}=\sqrt{{{1}^{2}}+41} \\ & =\sqrt{2} \end{align}$ Where $\theta $ is the angle between these two vectors: $\begin{align} & \left( 3\mathbf{i}+4\mathbf{j} \right)\cdot \left( \mathbf{i}+\mathbf{j} \right)=5\sqrt{2}\cos \theta \\ & 3\text{ }\mathbf{i}\cdot \mathbf{i}+3\text{ }\mathbf{i}\cdot \mathbf{j}+4\text{ }\mathbf{j}\cdot \mathbf{i}+4\text{ }\mathbf{j}\cdot \mathbf{j}=5\sqrt{2}\cos \theta \\ & \cos \theta =\frac{3+4}{5\sqrt{2}} \\ & \theta =8.13{}^\circ \end{align}$ Thus, the angle between these two vectors is $8.13{}^\circ $.
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