Problem 25
Question
The given three points form a triangle. Determine which triangles are isosceles and which are right triangles. $$ (0,0,0),(3,6,-6),(2,1,2) $$
Step-by-Step Solution
Verified Answer
The triangle is a right triangle but not isosceles.
1Step 1: Calculate the Distances Between Points
First, let's find the distances between each pair of points using the distance formula.\The distance between two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) is:\[d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\]\Calculate the distance between \((0,0,0)\) and \((3,6,-6)\):\[d_{AB} = \sqrt{(3-0)^2 + (6-0)^2 + (-6-0)^2} = \sqrt{9 + 36 + 36} = \sqrt{81} = 9\]\Calculate the distance between \((0,0,0)\) and \((2,1,2)\):\[d_{AC} = \sqrt{(2-0)^2 + (1-0)^2 + (2-0)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\]\Calculate the distance between \((3,6,-6)\) and \((2,1,2)\):\[d_{BC} = \sqrt{(2-3)^2 + (1-6)^2 + (2+6)^2} = \sqrt{1 + 25 + 64} = \sqrt{90}\approx 9.49\]
2Step 2: Determine if the Triangle is Isosceles
An isosceles triangle has at least two equal side lengths.The calculated distances are: \(d_{AB} = 9\), \(d_{AC} = 3\), \(d_{BC} \approx 9.49\). Since no two distances are equal, this triangle is not isosceles.
3Step 3: Check If the Triangle is Right-Angled
A right triangle satisfies the Pythagorean theorem: one side's squared length equals the sum of the other two sides' squared lengths.\Check if \(d_{AB}^2 + d_{AC}^2 = d_{BC}^2\):\[9^2 + 3^2 = 81 + 9 = 90\approx 90 = d_{BC}^2\] \Since \(81 + 9 = 90\) and \(d_{BC}^2 \approx 90\), the triangle is a right triangle with the hypotenuse being the side \(BC\).
Key Concepts
Distance FormulaIsosceles TriangleRight TrianglePythagorean Theorem
Distance Formula
In 3D geometry, the distance formula is used to determine the distance between two points in space. This formula extends the concept of Euclidean distance from a 2D plane to a 3D space. It is calculated by:
\[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \]
This formula gives us a direct way to measure how far apart two points are in three dimensions. For example, the distance between the points \((0,0,0)\) and \((3,6,-6)\) is found by substituting these into the formula:
Thus, the distance is \(\sqrt{81} = 9\). This calculation is an essential step for understanding the spatial relationship between points in 3D geometry.
\[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \]
This formula gives us a direct way to measure how far apart two points are in three dimensions. For example, the distance between the points \((0,0,0)\) and \((3,6,-6)\) is found by substituting these into the formula:
- \((3-0)^2 + (6-0)^2 + (-6-0)^2 = 9 + 36 + 36 = 81\)
Thus, the distance is \(\sqrt{81} = 9\). This calculation is an essential step for understanding the spatial relationship between points in 3D geometry.
Isosceles Triangle
An isosceles triangle in any dimension is defined as a triangle with at least two sides of equal length. This property can be used to classify triangles and identify symmetry. In the given problem, we evaluate the distances between the points to see if any pair of sides are equal:
None of these distances match each other, indicating that the triangle formed by these points is not isosceles. This step helps us understand the possible types of triangles that can be formed and whether they exhibit any symmetry or special properties.
- \(d_{AB} = 9\)
- \(d_{AC} = 3\)
- \(d_{BC} \approx 9.49\)
None of these distances match each other, indicating that the triangle formed by these points is not isosceles. This step helps us understand the possible types of triangles that can be formed and whether they exhibit any symmetry or special properties.
Right Triangle
A right triangle is a type of triangle where one of the angles is exactly 90 degrees. To verify if a triangle is right, we use the Pythagorean theorem as a test. This involves checking that the square of one side's length is equal to the sum of the squares of the other two sides:
\[ c^2 = a^2 + b^2 \]
In our exercise, this means checking if:
\[d_{AB}^2 + d_{AC}^2 = d_{BC}^2\]
Since the equation holds true, this confirms that the triangle is indeed a right triangle. In this configuration, the side \(BC\) acts as the hypotenuse, which is the longest side of a right triangle.
\[ c^2 = a^2 + b^2 \]
In our exercise, this means checking if:
\[d_{AB}^2 + d_{AC}^2 = d_{BC}^2\]
- \(9^2 + 3^2 = 81 + 9 = 90\)
- \(d_{BC}^2 \approx 9.49^2 \approx 90\)
Since the equation holds true, this confirms that the triangle is indeed a right triangle. In this configuration, the side \(BC\) acts as the hypotenuse, which is the longest side of a right triangle.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle in geometry, relating the lengths of sides in a right triangle. It states:
\[ a^2 + b^2 = c^2 \]
Where \(c\) is the hypotenuse. In three-dimensional space, we still apply this theorem to verify the presence of a right triangle. By calculating the square of each side and finding their sum, we determine if the formula holds:
In our context:
The sum \(81 + 9 = 90\) matches \(d_{BC}^2\), proving that our triangle is right-angled. Understanding and applying the Pythagorean theorem in 3D space allows for the identification of right triangles and further exploration of geometric properties in such spaces.
\[ a^2 + b^2 = c^2 \]
Where \(c\) is the hypotenuse. In three-dimensional space, we still apply this theorem to verify the presence of a right triangle. By calculating the square of each side and finding their sum, we determine if the formula holds:
In our context:
- \(d_{AB}^2 = 81\)
- \(d_{AC}^2 = 9\)
- \(d_{BC}^2 \approx 90\)
The sum \(81 + 9 = 90\) matches \(d_{BC}^2\), proving that our triangle is right-angled. Understanding and applying the Pythagorean theorem in 3D space allows for the identification of right triangles and further exploration of geometric properties in such spaces.
Other exercises in this chapter
Problem 25
In Problems \(25-28\), find the direction cosines and direction angles of the given vector. $$ \mathbf{a}=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k} $$
View solution Problem 25
Find the indicated scalar or vector without using \((5),(13)\), or \((15)\). $$ (\mathbf{i}+\mathbf{j}) \times(\mathbf{i}+5 \mathbf{k}) $$
View solution Problem 25
Find a unit vector (a) in the same direction as \(\mathbf{a}\), and (b) in the opposite direction of \(\mathbf{a}\). \(\mathbf{a}=\langle 2,2\rangle\)
View solution Problem 26
In Problems \(25-28\), determine whether the given vectors are linearly independent or linearly dependent. $$ \langle 1,1\rangle,\langle 0,1\rangle,\langle 2,5\
View solution