Chapter 11.1, Problem 31E

To determine

Tofind:Thedistance between the points $\left(6,-9,1\right)$ and $\left(-2,-1,5\right)$ .

Answer to Problem 31E

Thedistance between the points $\left(6,-9,1\right)$ and $\left(-2,-1,5\right)$ is $12\text{units}$ .

Explanation of Solution

Given information:

The given points are $\left(6,-9,1\right)$ and $\left(-2,-1,5\right)$ .

Calculation:

Apply the distance between the two points $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ can be calculated by the distance formula in space.

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

Let $\left(x_1,y_1,z_1\right)=\left(6,-9,1\right)$ and $\left(x_2,y_2,z_2\right)=\left(-2,-1,5\right)$ are the two points.

Substitute the values in the above equation to calculate the distance.

  $\begin{array}{c}d=\sqrt{{\left(-2-6\right)}^2+{\left(-1-\left(-9\right)\right)}^2+{\left(5-1\right)}^2}\\ =\sqrt{{\left(-8\right)}^2+8^2+{\left(4\right)}^2}\\ =\sqrt{64+64+16}\\ =\sqrt{144}\\ =12\text{units}\end{array}$

Therefore, the distance between the points $\left(6,-9,1\right)$ and $\left(-2,-1,5\right)$ is $12\text{units}$ .