Chapter 9, Problem 16CT

To determine

The unit vector in the direction 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 16CT

Solution:

The unit vector in the direction of $v$ is $u=\langle \frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\rangle$ .

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

Here $v=\langle 5\sqrt{2},-5\sqrt{2}\rangle =\langle a_1,a_2\rangle$ .

If $v=\langle a_1,a_2\rangle$ , then $\Vert v\Vert =\sqrt{a_1{}^2+a_2{}^2}$

  $\Rightarrow \Vert v\Vert =\sqrt{{\left(5\sqrt{2}\right)}^2+{\left(-5\sqrt{2}\right)}^2}$

  $\Rightarrow \Vert v\Vert =\sqrt{25\left(2\right)+25\left(2\right)}$

  $\Rightarrow \Vert v\Vert =\sqrt{50+50}$

  $\Rightarrow \Vert v\Vert =\sqrt{100}$

  $\Rightarrow \Vert v\Vert =10$

For any nonzero vector $v$ , the vector $u=\frac{v}{\Vert v\Vert }$ is a unit vector that has the same direction as $v$ .

  $\Rightarrow u=\frac{\langle 5\sqrt{2},-5\sqrt{2}\rangle }{10}$

  $\Rightarrow u=\langle \frac{5\sqrt{2}}{10},\frac{-5\sqrt{2}}{10}\rangle$

  $\Rightarrow u=\langle \frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\rangle$

The unit vector in the direction of $v$ is $u=\langle \frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\rangle$ .