Chapter 10.1, Problem 46E

To determine

Tofind:The lengths of the sides of triangle with vertices $\left(1,-2,-1\right)$ , $\left(3,0,0\right)$ and $\left(3,-6,3\right)$ , and check the triangle is a right triangle or an isosceles or neither.

Answer to Problem 46E

The length of three sides of the triangle are $3$ , $3\sqrt{5}$ and $6$ .The given triangle is a right triangle.

Explanation of Solution

Given information:

The vertices of the triangle are $\left(1,-2,-1\right)$ , $\left(3,0,0\right)$ and $\left(3,-6,3\right)$ .

Calculation:

The vertices of the right triangle are $\left(1,-2,-1\right)$ , $\left(3,0,0\right)$ and $\left(3,-6,3\right)$ .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 $\left(1,-2,-1\right)$ and $\left(3,0,0\right)$ is,

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

So, the length of thefirst side of triangle is $3$ .

The distance between the points $\left(3,0,0\right)$ and $\left(3,-6,3\right)$ is,

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(3-3\right)}^2+{\left(-6-0\right)}^2+{\left(3-0\right)}^2}\\ =\sqrt{{\left(0\right)}^2+{\left(-6\right)}^2+{\left(3\right)}^2}\\ =\sqrt{0+36+9}\\ =\sqrt{45}\\ =3\sqrt{5}\end{array}$

So, the length of the second side triangle is $3\sqrt{5}$ .

The distance between the points $\left(1,-2,-1\right)$ and $\left(3,-6,3\right)$ is,

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

So, the length of the third side of triangle is $6$ .

Therefore, the length of three sides of the triangle are $3$ , $3\sqrt{5}$ and $6$ .

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

Let us check the Pythagorean Theorem.

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

Take the left side of the equation.

  $\begin{array}{c}{\text{P}}^2+{\text{B}}^2={\left(3\right)}^2+{\left(6\right)}^2\\ =9+36\\ =45\\ ={\left(3\sqrt{5}\right)}^2\end{array}$

Hence the lengths of the triangle satisfy the Pythagorean Theorem.

Therefore, the triangle is a right triangle.