## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 799: 93

#### Answer

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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