Chapter 6.3, Problem 14E

To determine

To check whether $u$ and $v$ are equivalent.

Answer to Problem 14E

The vectors u and vare equivalent.

Explanation of Solution

Given information:

The given vectors are u and v.

Initial point for uis $\left(8,1\right)$ .

Terminal point for uis $\left(13,-1\right)$ .

Initial point for vis $\left(-2,4\right)$ .

Terminal point for vis $\left(-7,6\right)$ .

Formula used:

The distance formula is,

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

Magnitude of a vector $u=xi+yj$ is

  $\Vert u\Vert =\sqrt{{\left(x\right)}^2+{\left(y\right)}^2}$

Calculation:

Consider the given points for vector u.

  $\left(8,1\right),\left(13,-1\right)$

First find the magnitude of both the vectors from the distance formula.

Therefore, the magnitude of $u$ is,

  $\begin{array}{c}\Vert \text{u}\Vert =\sqrt{{\left(13-8\right)}^2+{\left(-1-1\right)}^2}\\ =\sqrt{25+4}\\ =\sqrt{29}\end{array}$

Consider the given points for vector v.

  $\left(-2,4\right),\left(-7,6\right)$

The magnitude of $v$ is,

  $\begin{array}{c}\Vert v\Vert =\sqrt{{\left(-7+2\right)}^2+{\left(6-4\right)}^2}\\ =\sqrt{25+4}\\ =\sqrt{29}\end{array}$

Now, calculate the direction of both vectors.

The direction of vector $v$ is,

  $\begin{array}{c}\frac{y_2-y_1}{x_2-x_1}=\frac{6-4}{-7+2}\\ =\frac{2}{-5}\\ =-\frac{2}{5}\end{array}$

The direction of vector $u$ is,

  $\begin{array}{c}\frac{y_2-y_1}{x_2-x_1}=\frac{-1-1}{13-8}\\ =\frac{-2}{5}\end{array}$

Both vectors have same magnitude as well as same direction.

Therefore, both the vectors are equivalent.