Chapter 8, Problem 14PT

To determine

The magnitude and direction of vector.

Answer to Problem 14PT

Magnitude = $4\text{ units}$

Direction = $180°$

Explanation of Solution

Given:

  $\overrightarrow{JK}:J\left(-6,-4\right)\text{ and }K\left(-10,-4\right)$

Calculation:

The component form will be:

  $\begin{array}{c}\overrightarrow{JK}=\left(x_2-x_1,y_2-y_1\right)\\ =\left(-10-\left(-6\right),-4-\left(-4\right)\right)\\ =\left(-4,0\right)\end{array}$

The distance formula to find the magnitude of JK is:

  $\begin{array}{c}\left|\overrightarrow{JK}\right|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-10-\left(-6\right)\right)}^2+{\left(-4-\left(-4\right)\right)}^2}\\ =\sqrt{4^2+0}\\ =4\end{array}$

The slope of JK will be:

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{-4-\left(-4\right)}{-10+6}\\ =\frac{0}{-4}\\ =0\end{array}$

Therefore, $\overrightarrow{JK}$ is a horizontal line.

$\overrightarrow{JK}$ pointsnegative x-axis. Hence,

Magnitude = $4\text{ units}$

Direction = $180°$