Chapter 11.1, Problem 29E

To determine

Tofind:Thedistance between the points $\left(3,2,5\right)$ and $\left(7,4,8\right)$ .

Answer to Problem 29E

Thedistance between the points $\left(3,2,5\right)$ and $\left(7,4,8\right)$ is $\sqrt{29}\text{units}$ .

Explanation of Solution

Given information:

The given points are $\left(3,2,5\right)$ and $\left(7,4,8\right)$ .

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 $\left(x_1,y_1,z_1\right)=\left(3,2,5\right)$ and $\left(x_2,y_2,z_2\right)=\left(7,4,8\right)$ are the two points.

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

  $\begin{array}{c}d=\sqrt{{\left(7-3\right)}^2+{\left(4-2\right)}^2+{\left(8-5\right)}^2}\\ =\sqrt{4^2+2^2+3^2}\\ =\sqrt{16+4+9}\\ =\sqrt{29}\text{units}\end{array}$

Therefore, the distance between the points $\left(3,2,5\right)$ and $\left(7,4,8\right)$ is $\sqrt{29}\text{units}$ .