Chapter 11.1, Problem 38E

To determine

Tocalculate: The length of the sides of the right triangle.

Answer to Problem 38E

The lengths are $\sqrt{62}$ , $\sqrt{6}$ and $\sqrt{56}$ .

Explanation of Solution

Given information:

The three points are $\left(2,-1,2\right),\left(-4,4,1\right),\left(-2,5,0\right)$ .

Formula used:

The distance between the points $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ is given by

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

Calculation:

Let the points be $A\left(2,-1,2\right),B\left(-4,4,1\right),C\left(-2,5,0\right)$ .

Then,

  $\begin{array}{c}B=\sqrt{{\left(-4-2\right)}^2+{\left(4+1\right)}^2+{\left(1-2\right)}^2}\\ =\sqrt{36+25+1}\\ =\sqrt{62}\\ BC=\sqrt{{\left(-2+4\right)}^2+{\left(5-4\right)}^2+{\left(0-1\right)}^2}\\ =\sqrt{4+1+1}\\ =\sqrt{6}\\ CA=\sqrt{{\left(2+2\right)}^2+{\left(-1-5\right)}^2+{\left(2-0\right)}^2}\\ =\sqrt{16+36+4}\\ =\sqrt{56}\end{array}$

Now,

  $\begin{array}{c}B{C}^2+C{A}^2={\left(\sqrt{6}\right)}^2+{\left(\sqrt{56}\right)}^2\\ =6+56\\ =62\\ =A{B}^2\end{array}$

This shows that the length of the sides of the triangle satisfy the Pythagoras Theorem.