Chapter 11, Problem 9RE

To determine

Find the lengths of the sides of the right triangle with the given vertices.

Answer to Problem 9RE

This is true and hence, the lengths of the sides of right triangle satisfy the Pythagorean theorem.

Explanation of Solution

Given:

The figure

  

Consider the points $\left(x_1,y_1,z_1\right),\left(x_2,y_2,z_2\right)and\left(x_3,y_3,z_3\right)$ then the lengths of the sides of right triangle are as follows:

  $\begin{array}{l}\overline{AB}={\left(x_2–x_1,y_2{}_{}–y_1,z_2–z_1\right)}_{}\hfill \\ \overline{BC}=\left(x_3–x_2,y_3–y_2,z_3–z_2\right)\hfill \\ \overline{AC}=\left(x_3–x_1,y_3–y_1,z_3–z_1\right)\hfill \end{array}$

Pythagorean theorem states that the square of sum of two sides becomes equal to the square of third side of a right angle triangle that is $\parallel \overrightarrow{AB}{\parallel }^2+\parallel \overrightarrow{BC}{\parallel }^2=\parallel \overrightarrow{AC}{\parallel }^2$ .

Consider the following figure,

  

Consider the following points from the above figure,

  $A\left(3,–2,0\right),B\left(0,3,2\right)andC\left(0,5,–3\right)$

The lengths of the sides of right triangle are as follows:

  $\begin{array}{l}\overline{AB}=\left(0–3,3+2,2–0\right)\hfill \\ =\left(–3,5,2\right)\hfill \\ \overline{BC}=\left(0–0,5–3,–3–2\right)\hfill \\ =\left(0,2,–5\right)\hfill \\ \overline{AC}=\left(0–3,5+2,–3–0\right)\hfill \\ =\left(–3,7,–3\right)\hfill \end{array}$

Therefore calculate the norms of the above lengths that is

  $\begin{array}{cc}\parallel \overrightarrow{AB}\parallel & =\sqrt{{(-3)}^2+{(5)}^2+{(2)}^2}\\ & =\sqrt{9+25+4}\\ & =\sqrt{38}\\ \parallel \overrightarrow{BC}\parallel & =\sqrt{{(0)}^2+{(2)}^2+{(-5)}^2}\\ & =\sqrt{0+4+25}\\ & =\sqrt{29}\\ \parallel \overrightarrow{AC}\parallel & =\sqrt{{(-3)}^2+{(7)}^2+{(-3)}^2}\\ & =\sqrt{9+49+9}\\ & =\sqrt{67}\end{array}$

Hence, the lengths of the sides of right triangle are

  $\left|\overrightarrow{AB}\right|=\sqrt{38},\left|\right|\overrightarrow{BC}|=\sqrt{29}\text{and}|\left|\overrightarrow{AC}\right|=\sqrt{67}$

From the Pythagoras theorem is

  $|{\Vert \overline{AB}|\Vert }^2+|\left|\overrightarrow{BC}\right|{|}^2=\left|\right|\overrightarrow{AC}|{|}^2$

Substitute the values $\left|\overrightarrow{AB}\right||=\sqrt{38},\parallel \overrightarrow{BC}|=\sqrt{29}\text{and}\left|\overrightarrow{AC}\right|=\sqrt{67}$ in equation

  $|{\Vert \overline{AB}|\Vert }^2+|\left|\overrightarrow{BC}\right|{|}^2=\left|\right|\overrightarrow{AC}|{|}^2$ to get,

Therefore,

  $\begin{array}{cc}{(\sqrt{38})}^2+{(\sqrt{29})}^2& ={(\sqrt{67})}^2\\ 38+29& =67\\ 67& =67\end{array}$

This is true and hence, the lengths of the sides of right triangle satisfy the Pythagorean theorem.