Chapter 6.3, Problem 19E

To determine

To calculate:

The following two vectors are equivalent or not:

  $\begin{array}{l}u:\text{}\left(x_1,y_1\right)=\left(2,4\right),\left(x_2,y_2\right)=\left(-1,4\right)\\ v:\text{}\left(x_1,y_1\right)=\left(-3,-1\right),\left(x_2,y_2\right)=\left(-5,2\right)\end{array}$

Answer to Problem 19E

Vectors $u$ and $v$ are not equivalent.

Explanation of Solution

Given information:

  $\begin{array}{l}u:\text{}\left(x_1,y_1\right)=\left(2,4\right),\left(x_2,y_2\right)=\left(-1,4\right)\\ v:\text{}\left(x_1,y_1\right)=\left(-3,-1\right),\left(x_2,y_2\right)=\left(-5,2\right)\end{array}$

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 or slope value is given by the following expression:

  $m=\frac{y_2-y_1}{x_2-x_1}$

Calculation:

The magnitude of vector $u$ is calculated as:

  $\begin{array}{l}\Vert u\Vert =\sqrt{{\left(2+1\right)}^2+{\left(2-4\right)}^2}\\ \Vert u\Vert =\sqrt{9+4}\\ \Vert u\Vert =\sqrt{13}\end{array}$

The magnitude of vector $v$ is calculated as:

  $\begin{array}{l}\Vert v\Vert =\sqrt{{\left(-3+5\right)}^2+{\left(-1-2\right)}^2}\\ \Vert v\Vert =\sqrt{4+9}\\ \Vert v\Vert =\sqrt{13}\end{array}$

The slope value of vector $u$ is calculated as:

  $\begin{array}{l}{m}_u=\frac{y_2-y_1}{x_2-x_1}\\ m_u=\frac{4-2}{-1-2}\\ m_u=-\frac{2}{3}\end{array}$

The slope value of vector $v$ is calculated as:

  $\begin{array}{l}{m}_v=\frac{y_2-y_1}{x_2-x_1}\\ m_v=\frac{2+1}{-5+3}\\ m_v=-\frac{3}{2}\end{array}$

So, the values of the magnitudes of vector $u$ and $v$ are same but slope value or direction is not same therefore both vectors are not equivalent.