Chapter 9, Problem 18CT

To determine

The vector $v$ in terms of its vertical and horizontal components, 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 18CT

Solution:

The vector $v$ in terms of its vertical and horizontal components is $5\sqrt{2}i-5\sqrt{2}j$ .

Explanation of Solution

Given information:

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

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 $\alpha ={315}^{\text{o}}$ .

Now vector $v$ can be written as $v=10\langle \cos {315}^{\circ }i+\sin {315}^{\circ }j\rangle$ .

  $=10\langle \frac{\sqrt{2}}{2}i+\left(-\frac{\sqrt{2}}{2}\right)j\rangle$

  $=5\sqrt{2}i-5\sqrt{2}j$

Therefore, the vector $v$ in terms of its vertical and horizontal components is $5\sqrt{2}i-5\sqrt{2}j$ .