Chapter 8.7, Problem 34PPS

a.

To determine

To calculation:

The magnitude and direction of the vector.

Answer to Problem 34PPS

The magnitude of the vector is about $5.1$ units and the direction are at an angle of about ${348.7}^{\circ }$ .

Explanation of Solution

Given information:

  

Calculation:

  

The vectors formed by the starting point at the school and the terminal point at the end of the parade is $\langle 5,-1\rangle$ .

  $\begin{array}{l}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(5-0\right)}^2+{\left(\left(-1\right)-0\right)}^2}\\ =\sqrt{{\left(5\right)}^2+{\left(-1\right)}^2}\\ =\sqrt{26}\\ =5.1\end{array}$

Now use the trigonometry to get the direction of the vector.

  $\begin{array}{l}\tan \theta =\frac{1}{5}\\ \theta ={\tan }^{-1}\left(\frac{1}{5}\right)\\ \theta =11.3\end{array}$

The parade started at the origin and terminated below the positive x-axis is quadrant fourth, the direction of the vector is the angle that is makes clockwise from the positive x-axis, which is ${360}^{\circ }-{11.3}^{\circ }$ or ${348.7}^{\circ }$ .

Therefore, the magnitude of the vector is about $5.1$ units and the direction is at an angle of about ${348.7}^{\circ }$ .

b.

To determine

To calculation:

The sum of two vectors and determine the length.

Answer to Problem 34PPS

The total distance of the parade is about the $8.4$ units and the length of parade is $0.084$ .

Explanation of Solution

Given information:

  

  $1$ unit $=\frac{1}{4}$ mile.

Calculation:

  

Use the distance formula to get the length of each leg of the parade.

First get the length of the leg between $\left(5,2\right)$ and $\left(0,0\right)$ .

  $\begin{array}{l}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(5-0\right)}^2+{\left(2-0\right)}^2}\\ =\sqrt{{\left(5\right)}^2+{\left(2\right)}^2}\\ =\sqrt{29}\\ =5.4\end{array}$

Next, find the length of the leg between $\left(5,-1\right)$ and $\left(5,2\right)$ .

  $\begin{array}{l}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(5-5\right)}^2+{\left(2-\left(-1\right)\right)}^2}\\ =\sqrt{{\left(5\right)}^2}\\ =3\end{array}$

So, the total distance of the parade is about $5.4+3$ or $8.4$ units and the length of parade is $0.084$ .