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 799: 93


The vector is $6\mathbf{i}+6\sqrt{3}\mathbf{j}$.

Work Step by Step

A non-zero vector $\mathbf{v}$ is expressed as $\mathbf{v}=\left\| \mathbf{v} \right\|\cos \theta \mathbf{i}+\left\| \mathbf{v} \right\|\sin \theta \mathbf{j}$ if the magnitude and direction angle from positive x-axis are $\left\| \mathbf{v} \right\|\text{ and }\theta $, Here, $\left\| \mathbf{v} \right\|=12$ and $\theta =60{}^\circ $. So, $\mathbf{v}$ can be expressed as: $\begin{align} & \mathbf{v}=\left\| \mathbf{v} \right\|\cos \theta \mathbf{i}+\left\| \mathbf{v} \right\|\sin \theta \mathbf{j} \\ & =12\cos 60{}^\circ \mathbf{i}+12\sin 60{}^\circ \mathbf{j} \\ & =12\left( \frac{1}{2} \right)\mathbf{i}+12\left( \frac{\sqrt{3}}{2} \right)\mathbf{j} \\ & =6\mathbf{i}+6\sqrt{3}\mathbf{j} \end{align}$ So, the vector $\mathbf{v}$ is $6\mathbf{i}+6\sqrt{3}\mathbf{j}$.
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