Precalculus (6th Edition) Blitzer

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

Chapter 6 - Section 6.6 - Vectors - Exercise Set - Page 784: 85


The speed of the plane is $\text{78 mph}$ and angle is $75.4{}^\circ $.

Work Step by Step

The force $\mathbf{F}$ in vector form is given as $\mathbf{F}=\left\| \left. \mathbf{F} \right\| \right.\cos \theta \mathbf{i}+\left\| \left. \mathbf{F} \right\| \right.\sin \theta \mathbf{j}$ For $\text{N70}{}^\circ \text{E}$, the value of $\theta $ for the plane from x-axis is given as $\begin{align} & \theta =90{}^\circ -70{}^\circ \\ & =20{}^\circ \end{align}$ Put the value $\theta =20{}^\circ $ and $\left\| \left. \mathbf{v} \right\| \right.=320$. The vector $\mathbf{v}$ is given as $\begin{align} & \mathbf{v}=\left\| \left. \mathbf{v} \right\| \right.\cos 20{}^\circ \mathbf{i}+\left\| \left. \mathbf{v} \right\| \right.\sin 20{}^\circ \mathbf{j} \\ & =320\cos 20{}^\circ \mathbf{i}+320\sin 20{}^\circ \mathbf{j} \\ & =300.7\mathbf{i}+109.5\mathbf{j} \end{align}$ The value of $\theta $ for ground speed of the plane is $30{}^\circ $. Put the value $\theta =30{}^\circ $ and $\left\| \left. \mathbf{w} \right\| \right.=370$. $\begin{align} & \mathbf{w}=\left\| \left. \mathbf{w} \right\| \right.\cos 30{}^\circ \mathbf{i}+\left\| \left. \mathbf{w} \right\| \right.\sin 30{}^\circ \mathbf{j} \\ & =370\cos 30{}^\circ \mathbf{i}+370\sin 30{}^\circ \mathbf{j} \\ & =320.4\mathbf{i}+185\mathbf{j} \end{align}$ The resultant vector is given by $\mathbf{w}-\mathbf{v}$. Use the vector addition as follows: $\begin{align} & \mathbf{w}-\mathbf{v}=320.4\mathbf{i}+185\mathbf{j}-\left( 300.7\mathbf{i}+109.5\mathbf{j} \right) \\ & =19.7\mathbf{i}+75.5\mathbf{j} \end{align}$ If the vector is $\mathbf{F}=a\mathbf{i}+b\mathbf{j}$, then magnitude of $\mathbf{F}$ is given by $\left\| \mathbf{F} \right\|$. $\left\| \mathbf{F} \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$ The vector $\mathbf{w}-\mathbf{v}$ is $19.7\mathbf{i}+75.5\mathbf{j}$. Comparing it with the above equation gives $\begin{align} & a=19.7\mathbf{i} \\ & b=75.5\mathbf{j} \end{align}$ Then magnitude of $\mathbf{w}-\mathbf{v}$ is given by $\left\| \mathbf{w}-\mathbf{v} \right\|$. $\begin{align} & \left\| \mathbf{w}-\mathbf{v} \right\|=\sqrt{{{\left( 19.7 \right)}^{2}}+{{\left( 75.5 \right)}^{2}}} \\ & \approx 78 \end{align}$ Then magnitude of $\mathbf{w}-\mathbf{v}$ is $\text{78}$. The speed of the plane is $\text{78 mph}$. If the vector is $\mathbf{F}=a\mathbf{i}+b\mathbf{j}$, then magnitude of $\mathbf{F}$ is given by $\left\| \mathbf{F} \right\|$. $\left\| \mathbf{F} \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$ And the direction angle is given as $\cos \theta =\frac{a}{\left\| \mathbf{F} \right\|}$ The vector $\mathbf{w}-\mathbf{v}$ is $19.7\mathbf{i}+75.5\mathbf{j}$. Then direction angle of $\mathbf{w}-\mathbf{v}$ is given by $\begin{align} & \cos \theta =\frac{19.7}{78} \\ & \theta ={{\cos }^{-1}}\left( \frac{19.7}{78} \right) \\ & \theta =75.4{}^\circ \end{align}$
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