Chapter 9.1, Problem 64SR

To determine

To find: The magnitude and direction of given vector.

Answer to Problem 64SR

The magnitude and direction of given vector is $\sqrt{72}$ and $45°$ .

Explanation of Solution

Given information: The coordinates of R and S are $\left(-3,3\right),\left(-9,9\right)$ .

Calculation:

The magnitude of the vector is,

  $\begin{array}{c}\mathrm{R}\mathrm{S}=\sqrt{{\left(-9\left(-3\right)\right)}^2+{\left(9-3\right)}^2}\\ =\sqrt{72}\end{array}$

The component form of given vector is,

  $\begin{array}{c}\overline{\mathrm{R}\mathrm{S}}=\langle {\mathrm{x}}_2-{\mathrm{x}}_1,{\mathrm{y}}_2-{\mathrm{y}}_1\rangle \\ =\langle -9-\left(-3\right),9-3\rangle \\ =\langle -6,6\rangle \end{array}$

The directionof the vector is,

  $\begin{array}{c}\tan \mathrm{\theta }=\frac{6}{6}\\ \mathrm{\theta }={\tan }^{-1}\left(1\right)\\ =45°\end{array}$