Chapter 8.7, Problem 15PPS

To determine

To calculation:

Magnitude and direction of given vector.

Answer to Problem 15PPS

Magnitude of given vector $\stackrel{\rightharpoonup }{AB}$ is $6$ units and direction is $270°$ .

Explanation of Solution

Given information:

  $\stackrel{\rightharpoonup }{AB}:A\left(9,-2\right),B\left(9,-8\right)$

Calculation:

Here given vector is $\stackrel{\rightharpoonup }{AB}:A\left(9,-2\right),B\left(9,-8\right)$ .

For finding out Magnitude and direction of given vector, initially write vector in component form.

  $\begin{array}{l}\stackrel{\rightharpoonup }{AB}:A\left(9,-2\right),B\left(9,-8\right)\\ \langle x_2-x_1,y_2-y_1\rangle \\ \langle 9-9,\left(-8\right)-\left(-2\right)\rangle \\ \langle 0,-6\rangle \end{array}$

Now use Pythagoras theorem to find magnitude.

  $\begin{array}{l}\Vert \stackrel{\rightharpoonup }{AB}\Vert =\sqrt{{\left(0\right)}^2+{\left(-6\right)}^2}\\ \Vert \stackrel{\rightharpoonup }{AB}\Vert =\sqrt{0+36}\\ \Vert \stackrel{\rightharpoonup }{AB}\Vert =\sqrt{36}\\ \Vert \stackrel{\rightharpoonup }{JK}\Vert =6\end{array}$

Now to find direction and angle,

  $\tan x=\frac{opp}{adj}=\frac{y\left(Component\right)}{x\left(Component\right)}$

  $\begin{array}{l}\tan \theta =\left|\frac{y_2-y_1}{x_2-x_1}\right|\\ \tan \theta =\left|\frac{-8-\left(-2\right)}{9-9}\right|\\ \tan \theta =\left|\frac{-6}{0}\right|\\ \tan \theta =undefined\end{array}$

An undefined slope means that the vector points straight u p or down.

In case this plot on graph, then it sis observed that vector points down so the direction is $270°$

Direction of $\stackrel{\rightharpoonup }{AB}=270°$

Hence, Magnitude and direction of given vector $\stackrel{\rightharpoonup }{AB}$ is $6$ units and $270°$ respectively.