Chapter 11, Problem 3CLT

To determine

To calculate:

The distance between two points for the following points

  $\begin{array}{l}x=\left(-2,3,-6\right)\\ y=\left(4,-5,1\right)\end{array}$

Answer to Problem 3CLT

The distance between two points $\left(-2,3,-6\right)$ and $\left(4,-5,1\right)$ is $\sqrt{149}$ units.

Explanation of Solution

Given information:

  $\begin{array}{l}x=\left(-2,3,-6\right)\\ y=\left(4,-5,1\right)\end{array}$

Formula Used:

By using the formula:

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Calculation:

The distance between two points:

  $\begin{array}{l}x=\left(x_1,x_2,x_3\right)\\ y=\left(y_1,y_2,y_3\right)\end{array}$

By using the formula:

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Put the value in formula:

  $\begin{array}{l}d=\sqrt{{\left(4+2\right)}^2+{\left(-5-3\right)}^2+{\left(1+6\right)}^2}\\ d=\sqrt{6^2+{\left(-8\right)}^2+7^2}\\ d=\sqrt{36+64+49}\\ d=\sqrt{149}\end{array}$

Therefore, the distance between two points $\left(-2,3,-6\right)$ and $\left(4,-5,1\right)$ is $\sqrt{149}$ units.