Chapter 8.6, Problem 55E

Solution

To prove that the distance $d\left(P,Q\right)$ between the points $P\left(x_1,y_1,z_1\right)$ and $Q\left(x_2,y_2,z_2\right)$ in space is $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2}$ by using the point $R\left(x_2,y_2,z_1\right)$ , the two dimensional distance formula within the plane $z=z_1$ , the one-dimensional distance formula within the line $r=〈x_2,y_2,t〉$ .

Given information:

  $P\left(x_1,y_1,z_1\right)$ and $Q\left(x_2,y_2,z_2\right)$ are the points in space.

Proof:

Construct the diagram for the given information as shown below.

  

In plane $z=z_1$ , the distance between the points $P\left(x_1,y_1,z_1\right)$ and $R\left(x_2,y_2,z_1\right)$ is the distance between the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ in two dimensional space. So, $\overline{PR}=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$ .

The points $Q\left(x_2,y_2,z_2\right)$ and $R\left(x_2,y_2,z_1\right)$ lie on the same one-dimensional line $r=〈x_2,y_2,t〉$ . So, the distance between $Q\left(x_2,y_2,z_2\right)$ and $R\left(x_2,y_2,z_1\right)$ is $\overline{QR}=z_2-z_1$ .

Now, observe that the triangle $PRQ$ is right triangle. So, the length of the hypotenuse $\overline{PQ}$ can be obtained by using the formula, ${\overline{PQ}}^2={\overline{PQ}}^2+{\overline{QR}}^2$ .

Substitute the value of $\overline{PQ}$ and $\overline{QR}$ , and simplify.

  $\begin{array}{c}{\overline{PQ}}^2={\overline{PQ}}^2+{\overline{QR}}^2\\ ={\left(\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}\right)}^2+{\left(z_2-z_1\right)}^2\\ ={\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_2-z_1\right)}^2\\ ={\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2\end{array}$

So, $\overline{PQ}=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2}$ .

Therefore, the distance between the points $P\left(x_1,y_1,z_1\right)$ and $R\left(x_2,y_2,z_1\right)$ is $d\left(P,Q\right)=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2}$ .