Chapter 10.2, Problem 9E

To determine

To find: The magnitude of the vector and the direction of angle $\theta$ .

Answer to Problem 9E

The answer:

The magnitude of the vector $|(-5,0)|=5$

And

The angle $\theta ={180}^{\circ }$ .

Explanation of Solution

Given information:

The positive $x$ -axis $\left(0\le \theta <{360}^{\circ }\right)$ .

The vector $(-5,0)$ .

Calculation:

Know that the magnitude of a vector $(a,b)$ is equal to:

  $|(a,b)|=\sqrt{a^2+b^2}$

Know that for the direction angle $\theta$ of a vector $(a,b)$ with the positive $x$ -axis,

  $\begin{array}{c}\sin \theta =\frac{b}{|(a,b)|}\\ \cos \theta =\frac{a}{|(a,b)|}\\ \Rightarrow \tan \theta =\frac{b}{a}\end{array}$

To use the previous results, the magnitude of the vector $(-5,0)$ is equal to:

  $\begin{array}{c}|(-5,0)|=\sqrt{{(-5)}^2+0^2}\\ =\sqrt{25}\\ =5\end{array}$

While for the direction angle $\theta$ of a vector $(-5,0)$ with the positive $x$ -axis,

  $\begin{array}{c}\tan \theta =\frac{0}{-5}\\ =0\\ \Rightarrow \theta ={180}^{\circ }\end{array}$

Because,

  $\cos \theta <0$

Therefore, the required magnitude of the vector is $|(-5,0)|=5$ and the direction of angle $\theta$ is

  $\theta ={180}^{\circ }$ .