Question

Show that the point \(\mathrm{P}_{1}(2,2,3)\) is equidistant from the points \(\mathrm{P}_{2}(1,4,-2)\) and \(\mathrm{P}_{3}(3,7,5)\)

Short answer

Using the distance formula for three-dimensional coordinates, we calculated the distance between points \(P_1\) and \(P_2\) to be \(\sqrt{30}\) and the distance between points \(P_1\) and \(P_3\) also to be \(\sqrt{30}\). Since both distances are equal, point \(P_1\) is equidistant from points \(P_2\) and \(P_3\).

Step-by-step solution

Calculate distance between P1 and P2

To find the distance between points P1 and P2, we use the distance formula for 3D coordinates: \(D(P_1, P_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\). Plugging in the values for P1 and P2, we get: \(D(P_1, P_2) = \sqrt{(-1)^2 + (2)^2 + (5)^2} = \sqrt{1+4+25}=\sqrt{30}\).

Calculate distance between P1 and P3

Similarly, to find the distance between points P1 and P3, we use the same distance formula for 3D coordinates: \(D(P_1, P_3) = \sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2 + (z_3 - z_1)^2}\). Plugging in the values for P1 and P3, we get: \(D(P_1, P_3) = \sqrt{(1)^2 + (5)^2 + (2)^2} = \sqrt{1+25+4}=\sqrt{30}\).

Check if P1 is equidistant from P2 and P3

In step 1, we calculated the distance between points P1 and P2 and found it to be \(\sqrt{30}\). In step 2, we calculated the distance between points P1 and P3 and found it to be \(\sqrt{30}\) as well. Because both distances are equal, we can conclude that point P1 is equidistant from points P2 and P3, as desired.

Key concepts

Understanding the Distance Formula in 3D

The distance formula is vital for calculating the straight line distance between two points in a three-dimensional space. Especially in geometry or physics, understanding this formula can help solve numerous problems related to space and movement. In 3D, the formula is an extension of the Pythagorean theorem, taking into account three axes: the x, y, and z coordinates.

In the example provided, the distance between two points, \(P_1\) and \(P_2\), is found using the formula \(D(P_1, P_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\). By subtracting the corresponding coordinates of \(P_1\) from \(P_2\), squaring each result, and then summing them up, we have created three squared distances on the x, y, and z axes which, when summed and square-rooted, give us the straight line distance between the two points. This method is applied universally for finding distances in three-dimensional spaces.

Navigating 3D Coordinates

A 3D coordinate system allows us to determine the exact location of points in a three-dimensional space. Each point is identified by three numbers, often referred to as coordinates: \( (x, y, z) \). The x-coordinate represents the position along the horizontal axis, the y-coordinate along the vertical axis, and the z-coordinate represents the depth, offering a third dimension to the traditional two-dimensional plane.

When tackling geometry problems, understanding how to plot these points and visualize them in space can be essential. The coordinates also form the foundation of the distance formula in 3D. As seen in the example, the points \(P_1(2,2,3)\), \(P_2(1,4,-2)\), and \(P_3(3,7,5)\) each have their unique location in space defined by their 3D coordinates. When finding whether \(P_1\) is equidistant from \(P_2\) and \(P_3\), we use these coordinates as inputs into our distance formula, allowing us to calculate and compare the respective distances.

The Role of the Square Root in Distance Calculations

The square root function is crucial when working with the distance formula. It serves as the final step, where after adding the squares of the differences in coordinates, we must square root the sum to find the actual distance. The square of a number represents the area of a square with sides of that length. Consequently, when we use the distance formula, we are essentially summing up the areas of these imaginary squares on each axis. The square root then takes us back from area to length – the true measure of distance.

In our example, after calculating the sum of the squared differences for each pair of points, we use the square root to find that \(P_1\) is \(\sqrt{30}\) units away from both \(P_2\) and \(P_3\). The use of the square root guarantees that we are calculating the direct linear distance between the points, rather than merely the sum of the individual distances on each axis.