Chapter 6.3, Problem 11E

To determine

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

Answer to Problem 11E

The vectors u and vare not equivalent.

Explanation of Solution

Given information:

The given vectors are u and v.

Initial point for uis $\left(2,2\right)$ .

Terminal point for uis $\left(-1,4\right)$ .

Initial point for vis $\left(-3,-1\right)$ .

Terminal point for vis $\left(-5,2\right)$ .

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.

  $\left(2,2\right),\left(-1,4\right)$

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

Therefore, the magnitude of $u$ is,

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

Consider the given points for vector v.

  $\left(-3,-1\right),\left(-5,2\right)$

The magnitude of $v$ is,

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

Now, calculate the direction of both vectors.

The direction of vector $u$ is,

  $\begin{array}{c}\frac{y_2-y_1}{x_2-x_1}=\frac{4-2}{-1-2}\\ =-\frac{2}{3}\end{array}$

The direction of vector $v$ is,

  $\begin{array}{c}\frac{y_2-y_1}{x_2-x_1}=\frac{2+1}{-5+3}\\ =-\frac{3}{2}\end{array}$

Both vectors have same magnitude but not of same direction.

Therefore, the vectors are not equivalent.