Chapter 8.7, Problem 4CYU

To determine

To calculation:

Magnitude and direction of given vector.

Answer to Problem 4CYU

Magnitude of given vector $\stackrel{\rightharpoonup }{WX}$ is $2\sqrt{17}$ units and direction is $346°$ .

Explanation of Solution

Given information:

  $\stackrel{\rightharpoonup }{WX}:W\left(-9,-5\right),X\left(-1,-7\right)$

Calculation:

Here given vector is $\stackrel{\rightharpoonup }{WX}:W\left(-9,-5\right),X\left(-1,-7\right)$ .

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

  $\begin{array}{l}\stackrel{\rightharpoonup }{WX}=W\left(-9,-5\right),X\left(-1,-7\right)\\ \langle x_2-x_1,y_2-y_1\rangle \\ \langle \left(-1\right)-\left(-9\right),\left(-7\right)-\left(-5\right)\rangle \\ \langle 8,-2\rangle \end{array}$

Now use Pythagoras theorem to find magnitude.

  $\begin{array}{l}\Vert \overline{WX}\Vert =\sqrt{8^2+{\left(-2\right)}^2}\\ \Vert \overline{WX}\Vert =\sqrt{68}\\ \Vert \overline{WX}\Vert =2\sqrt{17}\end{array}$

Now to find direction and angle,

  $\tan x=\frac{O_s}{A_s}=\frac{y\left(C\right)}{x\left(C\right)}$

Where,

  $O_s=$ Opposite side

  $A_s=$ adjacent side.

  $C=$ Component.

  $\begin{array}{l}\tan x=\frac{-2}{8}=-0.25°\\ {\tan }^{-1}0.25°=-14°\end{array}$

The angle is in $IV$ Quadrant.

  $\theta =360°-14°=346°$

Hence, Magnitude and direction of given vector $\stackrel{\rightharpoonup }{WX}$ is $2\sqrt{17}$ units and $346°$ respectively.