Problem 68 - University Calculus: Early Transcendentals

Question

Show that the point \(P(3,1,2)\) is equidistant from the points \(A(2,-1,3)\) and \(B(4,3,1)\)

Step-by-Step Solution

Answer

Point \(P(3,1,2)\) is equidistant from points \(A(2,-1,3)\) and \(B(4,3,1)\), with both distances equal to \(\sqrt{6}\).

Step 1: Understand the Problem

We need to prove that the distances from point \(P(3,1,2)\) to points \(A(2,-1,3)\) and \(B(4,3,1)\) are the same. This means calculating the distances \(PA\) and \(PB\) and showing they are equal.

Step 2: Calculate Distance PA

Use the distance formula: \[PA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\]Plug in the points \(P(3,1,2)\) and \(A(2,-1,3)\):\[PA = \sqrt{(3 - 2)^2 + (1 - (-1))^2 + (2 - 3)^2} = \sqrt{1^2 + 2^2 + (-1)^2}\]Calculate each square:\[PA = \sqrt{1 + 4 + 1} = \sqrt{6}\]

Step 3: Calculate Distance PB

Use the same distance formula for \(PB\): \[PB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\]Insert the coordinates for \(P(3,1,2)\) and \(B(4,3,1)\):\[PB = \sqrt{(3 - 4)^2 + (1 - 3)^2 + (2 - 1)^2} = \sqrt{(-1)^2 + (-2)^2 + 1^2}\]Calculate each square:\[PB = \sqrt{1 + 4 + 1} = \sqrt{6}\]

Step 4: Compare the Distances

Since the distance \(PA = \sqrt{6}\) and the distance \(PB = \sqrt{6}\) are equal, we can conclude that point \(P\) is equidistant from points \(A\) and \(B\).

Key Concepts

Equidistant Points

Equidistant points are points that are exactly the same distance from a fixed central point in a given space, which could be a line, plane, or 3D space. In this exercise, the challenge is to determine whether point \(P(3,1,2)\) is equidistant from points \(A(2,-1,3)\) and \(B(4,3,1)\). To prove that \(P\) is equidistant from \(A\) and \(B\), we need to calculate the distances between \(P\) and each of these points and show that these distances are equal. Brevity in calculation and clarity in understanding what the term 'equidistant' represents are keys to solving such problems effectively.
Understanding the notion of equidistant in geometrical terms means finding points that maintain uniform spacing from multiple reference points. This concept is quite useful in various applications, including design and robotics.
Recognizing equidistant points not only aids in solving geometric problems but also enhances comprehension when dealing with symmetrical shapes in diverse coordinate geometries.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a method of determining spatial relationships using a coordinate system. It allows for a detailed understanding of geometric shapes using algebraic equations. Points are represented using coordinates, which are sets of numbers indicating location in space. Here, the point \(P(3,1,2)\) is expressed in 3D space, indicating its position relative to the origin along the x-axis, y-axis, and z-axis.
With the help of coordinate geometry, geometric problems become informative and manageable with algebraic equations. Notably, it makes understanding the concept of equidistant points intuitive by incorporating algebra and geometry.

• The spatial arrangement of points and distances between said points can be solved using straightforward calculations.
• Understanding coordinate geometry principles is crucial when dealing with more complex geometric shapes and figures.

In 3D space, each point’s position allows us to perform calculations—like distance measures—similarly to how we would in 2D, but with an additional coordinate for depth (z-axis). This enables precise distance calculations between points, as demonstrated with points \(P\), \(A\), and \(B\) in the exercise.

Distance Calculation

Distance calculation is fundamental in coordinate geometry, especially in a 3D space, as it helps find the exact distance between two points. The formula used for this calculation in 3D is an extension of the Pythagorean theorem. In 3D, the distance \(d\) between two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) is given by:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\]This equation computes the straight line, or Euclidean distance, between two points. In our exercise, using points \(P(3,1,2)\), \(A(2,-1,3)\), and \(B(4,3,1)\), we apply this formula to find the distances \(PA\) and \(PB\).
The steps include:

• Subtract the corresponding coordinates of the two points to find the differences.
• Square each difference.
• Add these squared differences together.
• Take the square root of the sum to find the distance.

By following these steps, we find that both \(PA\) and \(PB\) equal \(\sqrt{6}\), proving that \(P\) is equidistant from \(A\) and \(B\). Understanding this process is essential for solving problems involving spatial relationships in a clear and systematic manner.