Chapter 6.3, Problem 17E

To determine

To calculate:

The following two vectors are equivalent or not:

  

Answer to Problem 17E

Vectors $u$ and $v$ are equivalent.

Explanation of Solution

Given information:

  

Formula used:

The magnitude of vector is given by the following expression:

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The direction of vector is given by the following expression:

  $d\text{'}=\frac{y_2-y_1}{x_2-x_1}$

Calculation:

For vector $u$ : $\begin{array}{l}\left(x_1,y_1\right)=\left(2,6\right)\\ \left(x_2,y_2\right)=\left(4,5\right)\end{array}$

The magnitude of vector $u$ is calculated as:

  $\begin{array}{l}\Vert u\Vert =\sqrt{{\left(6-2\right)}^2+{\left(5-4\right)}^2}\\ \Vert u\Vert =\sqrt{16+1}\\ \Vert u\Vert =\sqrt{17}\end{array}$

The direction of vector $u$ is given by the following expression:

  $\begin{array}{l}\frac{y_2-y_1}{x_2-x_1}=\frac{5-4}{6-2}\\ \frac{y_2-y_1}{x_2-x_1}=\frac{1}{4}\end{array}$

For vector $\begin{array}{l}\left(x_1,y_1\right)=\left(0,0\right)\\ \left(x_2,y_2\right)=\left(4,1\right)\end{array}$ :

The magnitude of vector $v$ is calculated as:

  $\begin{array}{l}\Vert v\Vert =\sqrt{{\left(4-0\right)}^2+{\left(1-0\right)}^2}\\ \Vert v\Vert =\sqrt{16+1}\\ \Vert v\Vert =\sqrt{17}\end{array}$

The direction of vector $v$ is given by the following expression:

  $\begin{array}{l}\frac{y_2-y_1}{x_2-x_1}=\frac{1-0}{4-0}\\ \frac{y_2-y_1}{x_2-x_1}=\frac{1}{4}\end{array}$

So, the values of the magnitude and direction of vector $u$ and $v$ are same. Therefore both vectors are equivalent.