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 783: 83


a) The vectors $\mathbf{v}$ and $\mathbf{w}$ are given as $\mathbf{v}=137.88\mathbf{i}+115.7\mathbf{j},\ \mathbf{w}=40\mathbf{i}$. b) The resultant vector $\mathbf{v}+\mathbf{w}$ is $177.88\mathbf{i}+115.7\mathbf{j}$. c) Approximate ground speed of plane is $212\text{ mph}$. d) The plane true bearing is $\theta =\text{N}57{}^\circ \text{E}\text{.}$

Work Step by Step

(a) The force $\mathbf{F}$ in vector form is given as $\mathbf{F}=\left\| \left. \mathbf{F} \right\| \right.\cos \theta \mathbf{i}+\left\| \left. \mathbf{F} \right\| \right.\sin \theta \mathbf{j}$ The value of $\theta $ from x-axis is given as $\begin{align} & \theta =90{}^\circ -50{}^\circ \\ & =40{}^\circ \end{align}$ Put the value $\theta =40{}^\circ $ and $\left\| \left. \mathbf{v} \right\| \right.=180$. The vector $\mathbf{v}$ is given as $\begin{align} & \mathbf{v}=\left\| \left. \mathbf{v} \right\| \right.\cos 40{}^\circ \mathbf{i}+\left\| \left. \mathbf{v} \right\| \right.\sin 40{}^\circ \mathbf{j} \\ & =180\cos 40{}^\circ \mathbf{i}+180\sin 40{}^\circ \mathbf{j} \\ & =137.88\mathbf{i}+115.7\mathbf{j} \end{align}$ Put the value $\theta =0{}^\circ $ and $\left\| \left. \mathbf{w} \right\| \right.=40$. $\begin{align} & \mathbf{w}=\left\| \left. \mathbf{w} \right\| \right.\cos 0{}^\circ \mathbf{i}+\left\| \left. \mathbf{w} \right\| \right.\sin 0{}^\circ \mathbf{j} \\ & =40\mathbf{i} \end{align}$ (b) Use the vector addition as follows: $\begin{align} & \mathbf{v}+\mathbf{w}=137.88\mathbf{i}+115.7\mathbf{j}+40\mathbf{i} \\ & =177.88\mathbf{i}+115.7\mathbf{j} \end{align}$ (c) 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 $\mathbf{v}+\mathbf{w}$ is $177.88\mathbf{i}+115.7\mathbf{j}$. Comparing it with the above equation gives $\begin{align} & a=177.88 \\ & b=115.7 \end{align}$ Then magnitude of $\mathbf{v}+\mathbf{w}$ is given by $\left\| \mathbf{v}+\mathbf{w} \right\|$. $\begin{align} & \left\| \mathbf{v}+\mathbf{w} \right\|=\sqrt{{{\left( 177.88 \right)}^{2}}+{{\left( 115.7 \right)}^{2}}} \\ & \approx 212 \end{align}$ Then magnitude of $\mathbf{v}+\mathbf{w}$ is $212$. (d) 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 $\mathbf{v}+\mathbf{w}$ is $177.88\mathbf{i}+115.7\mathbf{j}$. Then direction angle of $\mathbf{v}+\mathbf{w}$ is given by $\begin{align} & \cos \theta =\frac{177.88}{212} \\ & \theta ={{\cos }^{-1}}\left( \frac{177.88}{212} \right) \\ & \theta =33{}^\circ \end{align}$ The direction angle is given by $90{}^\circ -33{}^\circ =57{}^\circ $
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