Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that \((4,5,3),(1,7,4)\), and \((2,4,6)\) are vertices of an equilateral triangle.

Step-by-Step Solution

Answer

All three sides have distance \( \sqrt{14} \), confirming an equilateral triangle.

Step 1: Determine the Distance Formula

The distance formula between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \] We will use this formula to check if all sides of the triangle are equal.

Step 2: Calculate Distance AB

Calculate the distance between points \( A(4, 5, 3) \) and \( B(1, 7, 4) \):\[ AB = \sqrt{(1 - 4)^2 + (7 - 5)^2 + (4 - 3)^2} \]\[ AB = \sqrt{(-3)^2 + 2^2 + 1^2} \]\[ AB = \sqrt{9 + 4 + 1} \]\[ AB = \sqrt{14} \]

Step 3: Calculate Distance BC

Calculate the distance between points \( B(1, 7, 4) \) and \( C(2, 4, 6) \):\[ BC = \sqrt{(2 - 1)^2 + (4 - 7)^2 + (6 - 4)^2} \]\[ BC = \sqrt{1^2 + (-3)^2 + 2^2} \]\[ BC = \sqrt{1 + 9 + 4} \]\[ BC = \sqrt{14} \]

Step 4: Calculate Distance CA

Calculate the distance between points \( C(2, 4, 6) \) and \( A(4, 5, 3) \):\[ CA = \sqrt{(4 - 2)^2 + (5 - 4)^2 + (3 - 6)^2} \]\[ CA = \sqrt{2^2 + 1^2 + (-3)^2} \]\[ CA = \sqrt{4 + 1 + 9} \]\[ CA = \sqrt{14} \]

Step 5: Compare Distances

Each distance calculated is \( \sqrt{14} \). Since all sides are equal, the points \( A, B, C \) form an equilateral triangle.

Key Concepts

Distance Formula

The distance formula is an essential tool in geometry for measuring the distance between two points, particularly in three-dimensional space. This formula extends the basic idea we learn in 2D geometry to 3D, allowing us to account for a third coordinate (z-axis) in our calculations.
To find the distance between two points

• Given two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) in 3D, use the formula:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\]This equation ensures that we consider the differences across all three axes, combining them into a single measurement of distance. The formula derives from the Pythagorean theorem, which is foundational for calculating distances in any dimensional space.

3D Geometry

Understanding 3D geometry is crucial when analyzing objects like equilateral triangles in space. This branch of mathematics doesn't just deal with flat, canvas-like 2D shapes; it extends to shapes submerged in depth, encapsulating length, width, and height.
Working in 3D allows us to

• Analyze points and angles involving three coordinates.
• Consider the relation of lines and planes in space.

In our equilateral triangle example, each vertex

• Has a unique combination of three coordinates
• Positioning them precisely in space.

Calculations determine the equality of distances between these vertices, a crucial check to verify the type of triangle formed.

Vertices of a Triangle

The vertices of a triangle are the distinct points marking the ends of the triangle's sides. In 3D space, vertices are expressed with three coordinates, reflecting their position in three dimensions.
Here, we examine vertices such as

• \((4,5,3)\), \((1,7,4)\), and \((2,4,6)\)

Each set of coordinates corresponds to a point in the 3D space, defining the triangle's shape and size.
When identifying the type of triangle, the relationship between distances of each pair of vertices helps.
In an equilateral triangle:

• All side lengths are equal.
• Verification involves computing each vertex's distance and confirming their equality.

Studying these distances is fundamental in proving equilateral properties in the triangle's structure.