Chapter 9, Problem 17CT

To determine

The direction angle of $v$ , where $P_1=\left(3\sqrt{2},7\sqrt{2}\right)$ and $P_2=\left(8\sqrt{2},2\sqrt{2}\right)$ .

Answer to Problem 17CT

Solution:

The direction angle of $v$ is $\alpha ={315}^{\text{o}}$ .

Explanation of Solution

Given information:

The points, $P_1=\left(3\sqrt{2},7\sqrt{2}\right)$ , $P_2=\left(8\sqrt{2},2\sqrt{2}\right)$

Consider the points $P_1=\left(3\sqrt{2},7\sqrt{2}\right)$ and $P_2=\left(8\sqrt{2},2\sqrt{2}\right)$ .

Suppose that $v$ is a vector with initial point $P_1=\left(x_1,y_1\right)$ , not necessarily the origin, and terminal point $P_2=\left(x_2,y_2\right)$ . If $v=\overline{P_1{P}_2}$ , then $v$ is equal to the position vector $v=\langle x_2-x_1,y_2-y_1\rangle$ .

Here, $P_1=\left(3\sqrt{2},7\sqrt{2}\right)=\left(x_1,y_1\right)$ and $P_2=\left(8\sqrt{2},2\sqrt{2}\right)=\left(x_2,y_2\right)$

  $\Rightarrow v=\langle 8\sqrt{2}-3\sqrt{2},2\sqrt{2}-7\sqrt{2}\rangle$

  $\Rightarrow v=\langle 5\sqrt{2},-5\sqrt{2}\rangle$

Let $v=\langle 5\sqrt{2},-5\sqrt{2}\rangle$ .

  

The figure shows the vector $v$ with the direction of $\alpha$ .

The direction angle of the vector $v$ is $\alpha$ such that $\tan \alpha =\frac{y}{x}$ .

Since $v=\langle 5\sqrt{2},-5\sqrt{2}\rangle$ ,

  $\tan \alpha =\frac{-5\sqrt{2}}{5\sqrt{2}}$ .

By simplifying right side,

  $\tan \alpha =-1$

Since $0^{\text{o}}\le \alpha <{360}^{\text{o}}$ , the direction angle is $\alpha ={315}^{\text{o}}$ .

Therefore, the direction angle of $v$ is $\alpha ={315}^{\text{o}}$ .