Chapter 10.2, Problem 53E

To determine

To check: The given statement is true or false.

Answer to Problem 53E

The given statement is false.

Explanation of Solution

Given information:

The statement is:

The scaler multiple of a vector $v$ has the same direction as $v$ .

Calculation:

Know that the direction angle $\theta$ of a vector $r=v_1,v_2$ with the positive $x$ -axis,

  $\begin{array}{c}\sin \theta =\frac{v_2}{\left|v\right|}\\ \cos \theta =\frac{v_1}{\left|v\right|}\\ \Rightarrow \tan \theta =\frac{v_2}{v_1}\end{array}$

Know that the magnitude of a vector $r=v_1,v_2$ is equal to:

  $\left|v\right|=\sqrt{v_1^2+v_2^2}$

Let, the mark:

  $v=(1,\sqrt{3})$

Then,

  $\begin{array}{c}\left|v\right|=\sqrt{1^2+{(\sqrt{3})}^2}\\ =\sqrt{1+3}\\ =\sqrt{4}\\ =2\end{array}$

To use the previous result,

  $\begin{array}{c}\tan \theta =\frac{\sqrt{3}}{1}\\ =\sqrt{3}\\ \Rightarrow \theta ={60}^{\circ }\end{array}$

Then,

  $\begin{array}{c}av=-1\cdot (1,\sqrt{3})\\ =(-1,-\sqrt{3})\end{array}$

To use the previous result,

  $\begin{array}{c}\left|av\right|=\sqrt{{(-1)}^2+{(-\sqrt{3})}^2}\\ =\sqrt{1+3}\\ =\sqrt{4}\\ =2\end{array}$

Then,

  $\begin{array}{c}\tan {\theta }_a=\frac{-\sqrt{3}}{-1}\\ =\sqrt{3}\\ \Rightarrow {\theta }_a={240}^{\circ }\end{array}$

Because,

  $\begin{array}{c}\sin {\theta }_a=-\sqrt{3}/2<0\\ \cos {\theta }_a=-1/2<0\end{array}$

According to the previous results, conclude that for the vector $v=(1,\sqrt{3}\text{) }$ and the scaler $a=-1$ the direction angles of the vectors $v$ and $av$ are different, which means that the statement is false.

Therefore, the required given statement is false.