Chapter 8.7, Problem 18PPS

To determine

To calculation:

Magnitude and direction of given vector.

Answer to Problem 18PPS

Magnitude of given vector $\stackrel{\rightharpoonup }{CD}$ is $6.71$ units and direction is $206.57°$ .

Explanation of Solution

Given information:

  $\stackrel{\rightharpoonup }{CD}:C\left(10,5\right),D\left(4,2\right)$

Calculation:

Here given vector is $\stackrel{\rightharpoonup }{CD}:C\left(10,5\right),D\left(4,2\right)$ .

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

  $\begin{array}{l}\stackrel{\rightharpoonup }{CD}:C\left(10,5\right),D\left(4,2\right)\\ \langle x_2-x_1,y_2-y_1\rangle \\ \langle \left(4\right)-\left(10\right),\left(2\right)-\left(5\right)\rangle \\ \langle -6,-3\rangle \end{array}$

Now use Pythagoras theorem to find magnitude.

  $\begin{array}{l}\Vert \stackrel{\rightharpoonup }{CD}\Vert =\sqrt{{\left(-6\right)}^2+{\left(-3\right)}^2}\\ \Vert \stackrel{\rightharpoonup }{CD}\Vert =\sqrt{36+9}\\ \Vert \stackrel{\rightharpoonup }{CD}\Vert =\sqrt{45}\\ \Vert \stackrel{\rightharpoonup }{CD}\Vert =\sqrt{9.5}\\ \Vert \stackrel{\rightharpoonup }{FG}\Vert =3\sqrt{5}\\ \Vert \stackrel{\rightharpoonup }{FG}\Vert =6.71\end{array}$

Now to find direction and angle,

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

  $\begin{array}{l}\tan \theta =\left|\frac{y_2-y_1}{x_2-x_1}\right|\\ \tan \theta =\left|\frac{2-5}{4-10}\right|\\ \tan \theta =\left|\frac{-3}{-6}\right|\\ \tan \theta =\frac{1}{2}\end{array}$

Calculate the angle,

  $\begin{array}{l}\theta ={\tan }^{-1}\frac{1}{2}\\ \theta =26.57°\end{array}$

Direction of $\stackrel{\rightharpoonup }{CD}=26.57+180=206.57°$

Hence, Magnitude and direction of given vector $\stackrel{\rightharpoonup }{CD}$ is $6.71$ units and $206.57°$ respectively.