Chapter 9.1, Problem 65SR

To determine

To find: The magnitude and direction of given vector.

Answer to Problem 65SR

The magnitude and direction of given vector is $2\sqrt{13}$ and $146.3°$ .

Explanation of Solution

Given information: The coordinates of J and K are $\left(8,1\right),\left(2,5\right)$ .

Calculation:

The magnitude of the vector is,

  $\begin{array}{c}\mathrm{J}\mathrm{K}=\sqrt{{\left(2-8\right)}^2+{\left(5-1\right)}^2}\\ =\sqrt{52}\\ =2\sqrt{13}\end{array}$

The component form of given vector is,

  $\begin{array}{c}\overline{\mathrm{J}\mathrm{K}}=\langle {\mathrm{x}}_2-{\mathrm{x}}_1,{\mathrm{y}}_2-{\mathrm{y}}_1\rangle \\ =\langle 2-8,5-1\rangle \\ =\langle -6,4\rangle \end{array}$

The direction of the vector is,

  $\begin{array}{c}\tan \mathrm{\theta }=-\frac{4}{6}\\ \mathrm{\theta }={\tan }^{-1}\left(-\frac{4}{6}\right)\\ =146.3°\end{array}$