Chapter 10.1, Problem 45E

To determine

Tofind:The lengths of the sides of triangle with vertices $\left(5,0,0\right)$ , $\left(1,3,-2\right)$ and $\left(-2,-1,4\right)$ , and check the triangle is a right triangle or an isosceles or neither.

Answer to Problem 45E

The length of three sides of the triangle are $\sqrt{29}$ , $\sqrt{61}$ and $\sqrt{66}$ .The given triangle is neither an isosceles nor a right triangle.

Explanation of Solution

Given information:

The vertices of the triangle are $\left(5,0,0\right)$ , $\left(1,3,-2\right)$ and $\left(-2,-1,4\right)$ .

Calculation:

The vertices of the right triangle are $\left(5,0,0\right)$ , $\left(1,3,-2\right)$ and $\left(-2,-1,4\right)$ .The length of the sides of triangle is the distance between the vertices.

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

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

The distance between the points $\left(5,0,0\right)$ and $\left(1,3,-2\right)$ is,

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(1-5\right)}^2+{\left(3-0\right)}^2+{\left(-2-0\right)}^2}\\ =\sqrt{{\left(-4\right)}^2+{\left(3\right)}^2+{\left(-2\right)}^2}\\ =\sqrt{16+9+4}\\ =\sqrt{29}\end{array}$

So, the length of thefirst side of triangle is $\sqrt{29}$ .

The distance between the points $\left(1,3,-2\right)$ and $\left(-2,-1,4\right)$ is,

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(-2-1\right)}^2+{\left(-1-3\right)}^2+{\left(4-\left(-2\right)\right)}^2}\\ =\sqrt{{\left(-3\right)}^2+{\left(-4\right)}^2+{\left(6\right)}^2}\\ =\sqrt{9+16+36}\\ =\sqrt{61}\end{array}$

So, the length of the second side triangle is $\sqrt{61}$ .

The distance between the points $\left(5,0,0\right)$ and $\left(-2,-1,4\right)$ is,

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(-2-5\right)}^2+{\left(-1-0\right)}^2+{\left(4-0\right)}^2}\\ =\sqrt{{\left(-7\right)}^2+{\left(-1\right)}^2+{\left(4\right)}^2}\\ =\sqrt{49+1+16}\\ =\sqrt{66}\end{array}$

So, the length of the third side of triangle is $\sqrt{66}$ .

Therefore, the length of three sides of triangle are $\sqrt{29}$ , $\sqrt{61}$ and $\sqrt{66}$ .

The lengths of all sides of triangle are different. Therefore, the given triangle is not an isosceles triangle.

Let us check the Pythagorean Theorem.

  ${\text{P}}^2+{\text{B}}^2={\text{H}}^2$

No two combination of the lengths of sides of the triangle satisfy Pythagorean theorem.

Therefore, the triangle is neither an isosceles nor a right triangle.