Chapter 10.1, Problem 38E

To determine

Tofind:The distance between the points $\left(1,1,-7\right)$ and $\left(-2,-3,-7\right)$ .

Answer to Problem 38E

The distance between the points $\left(1,1,-7\right)$ and $\left(-2,-3,-7\right)$ is $5$ .

Explanation of Solution

Given information:

The points are $\left(1,1,-7\right)$ and $\left(-2,-3,-7\right)$ .

Calculation:

The points are $\left(1,1,-7\right)$ and $\left(-2,-3,-7\right)$ .

Use the distance formula for two points $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ .

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

Substitute $1$ , $1$ , $-7$ for $x_1$ , $y_1$ and $z_1$ respectively and $-2$ , $-3$ , $-7$ for $x_2$ , $y_2$ and $z_2$ respectively in the distance formula.

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(-2-1\right)}^2+{\left(-3-1\right)}^2+{\left(-7-\left(-7\right)\right)}^2}\\ =\sqrt{{\left(-3\right)}^2+{\left(-4\right)}^2+{\left(0\right)}^2}\\ =\sqrt{9+16+0}\\ =\sqrt{25}\\ =5\end{array}$

Therefore, the distance between the points $\left(1,1,-7\right)$ and $\left(-2,-3,-7\right)$ is $5$ .