Question

The vertices of a triangle are given. Determine whether the triangle is an acute triangle, an obtuse triangle, or a right triangle. Explain your reasoning. $$ (-3,0,0),(0,0,0),(1,2,3) $$

Short answer

The given triangle with vertices (-3,0,0), (0,0,0), (1,2,3) is an obtuse triangle.

Step-by-step solution

Calculate lengths of sides

To calculate the lengths of the sides, use the distance formula: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\). Thereby, length \(a\) between points (0,0,0) and (1,2,3) is calculated as \(a = \sqrt{(1-0)^2 + (2-0)^2 + (3-0)^2} = \sqrt{14}\), length \(b\) between (0,0,0) and (-3,0,0) is \(b = \sqrt{((-3-0)^2 +(0-0)^2 + (0-0)^2)} = 3\), length \(c\) between points (-3,0,0) and (1,2,3) is \(c = \sqrt{((1+3)^2 + (2-0)^2 + (3-0)^2)} = \sqrt{29}\).

Determine the triangle type

With the lengths \(a=\sqrt{14}\), \(b=3\), and \(c=\sqrt{29}\), it’s possible to check whether the triangle is a right triangle by testing the Pythagorean theorem: \(c^2 = a^2 + b^2\). Here, \(\sqrt{29}^2 = \sqrt{14}^2 + 3^2\), which results in \(29 = 14 + 9\), which becomes \(29 = 23\). The result shows the theorem isn't satisfied and therefore it’s not a right triangle. To find if it's an obtuse or acute triangle, apply the rule that for an obtuse triangle, \(c^2 > a^2 + b^2\), and for an acute triangle, \(c^2 < a^2 + b^2\). Here, \(29 > 23\), so the triangle is obtuse.