Chapter 9.1, Problem 67SR

To determine

To find: The magnitude and direction of given vector.

Answer to Problem 67SR

The magnitude and direction of given vector is $2\sqrt{122}$ and $257.2°$ .

Explanation of Solution

Given information: The coordinates of A and B are $\left(-1,10\right),\left(1,-12\right)$ .

Calculation:

The magnitude of the vector is,

  $\begin{array}{c}\mathrm{A}\mathrm{B}=\sqrt{{\left(-6-\left(-1\right)\right)}^2+{\left(-12-10\right)}^2}\\ =2\sqrt{122}\end{array}$

The component form of given vector is,

  $\begin{array}{c}\overline{\mathrm{A}\mathrm{B}}=\langle {\mathrm{x}}_2-{\mathrm{x}}_1,{\mathrm{y}}_2-{\mathrm{y}}_1\rangle \\ =\langle -6-\left(-1\right),-12-10\rangle \\ =\langle 2,-22\rangle \end{array}$

The direction of the vector is,

  $\begin{array}{c}\tan \mathrm{\theta }=-\frac{22}{2}\\ \mathrm{\theta }={\tan }^{-1}\left(-11\right)\\ =257.2°\end{array}$