Problem 5 - Algebra and Trigonometry

Question

\(3-6\) Two points \(P\) and \(Q\) are given. (a) Plot \(P\) and \(Q .(b)\) Find the distance between \(P\) and \(Q\) $$ P(-2,-1,0), Q(-12,3,0) $$

Step-by-Step Solution

Answer

The distance between points is approximately 10.77 units.

Step 1: Understand the Problem

We have two points, \(P(-2,-1,0)\) and \(Q(-12,3,0)\). We need to plot these points and calculate the distance between them.

Step 2: Plot the Points on a Coordinate System

Plot point \(P(-2, -1, 0)\) in the 3D coordinate system, meaning \(x = -2\), \(y = -1\), and \(z = 0\). Similarly, plot point \(Q(-12, 3, 0)\), where \(x = -12\), \(y = 3\), and \(z = 0\). Both points lie on the XY-plane since their \(z\) coordinates are zero.

Step 3: Apply the Distance Formula

Use the distance formula for two points in 3D space: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\). Substitute \(P(-2, -1, 0)\) and \(Q(-12, 3, 0)\) into the formula.

Step 4: Calculate the Differences in Coordinates

Calculate \((x_2 - x_1) = (-12) - (-2) = -10\), \((y_2 - y_1) = 3 - (-1) = 4\), and \((z_2 - z_1) = 0 - 0 = 0\).

Step 5: Plug Values into the Formula

Substitute the differences into the distance formula. \(d = \sqrt{(-10)^2 + 4^2 + 0^2}\).

Step 6: Compute the Squares and Sum

Calculate \((-10)^2 = 100\) and \(4^2 = 16\), then sum them: \(100 + 16 + 0 = 116\).

Step 7: Take the Square Root

Find \(d = \sqrt{116}\), which simplifies to \(d \approx 10.77\).

Key Concepts

3D coordinate system

Let's begin with understanding the 3D coordinate system, a mathematical space that allows us to place and locate points using three coordinates. This system extends the basic two-dimensional (2D) plane into an additional dimension, known as the z-axis. Here's how it works:

• The x-axis: Represents the horizontal dimension.
• The y-axis: Represents the vertical dimension.
• The z-axis: Represents the depth, adding a third dimension.

In a 3D coordinate system, each point is represented by an ordered triplet \(x, y, z\). These coordinates help us identify the specific location of a point in the space. The origin, where all axes meet, is the point \(0, 0, 0\). Visualizing points in this 3-dimensional plane can help us understand spaces beyond our daily 2D interactions. Points like \(-2, -1, 0\) and \(-12, 3, 0\) lie on the xy-plane since their z coordinates are zero, indicating that they do not rise above or dip below this plane.

plotting points

When working with a 3D coordinate system, plotting points accurately on this space is crucial for understanding spatial relationships. To plot a point:

• Begin with the x-coordinate: Move along the x-axis for the value of x.


• Next, adjust for the y-coordinate: From your x-position, move parallel to the y-axis.


• Finally, consider the z-coordinate: Move vertically (up or down) based on the value of z.

For the given points \(-2, -1, 0\) and \(-12, 3, 0\), both lie on the xy-plane since the z-component is zero. This means, after plotting the x and y positions, there's no need to adjust vertically. Visualizing these plotted points helps in seeing how they line up relating each other spatially on one plane.

distance between points

Once points are plotted, finding the distance between them helps understand how far apart they are in space, particularly in a 3D coordinate system. This distance is calculated using the distance formula:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]This formula considers the differences in x, y, and z coordinates, summing their squares. Let's apply it to \(-2, -1, 0\) and \(-12, 3, 0\):

• First, the differences: \(x_2 - x_1 = -12 - (-2) = -10\)


• \(y_2 - y_1 = 3 - (-1) = 4\)


• \(z_2 - z_1 = 0 - 0 = 0\)

Substitute these into the formula:\[ d = \sqrt{(-10)^2 + 4^2 + 0^2} = \sqrt{100 + 16 + 0} = \sqrt{116} \]The result simplifies to approximately \10.77\ units. This calculation shows how some straightforward arithmetic and an understanding of the 3D space can quantify the distance between two plotted points.