Problem 19 - Calculus an Applied Approach

Question

Find the distance between the point and the plane (see figure). The distance \(D\) between a point \(\left(x_{0}, y_{0}, z_{0}\right)\) and the plane \(a x+b y+c z+d=0\) is $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}} $$ $$ (3,2,-1), 2 x-3 y+4 z=24 $$

Step-by-Step Solution

Answer

The distance between the given point and the plane is approximately 5.18 units.

Step 1: Identify values to insert into formula

First, identify values from the point and plane that correspond to plug into the distance formula. The given point is (3,2,-1) and the given plane equation is \(2x - 3y + 4z = 24\). So, we have \(x_0 = 3\), \(y_0 = 2\), \(z_0 = -1\), \(a = 2\), \(b = -3\), \(c = 4\), \(d = -24\).

Step 2: Insert identified values into the formula

Secondly, substitute those values into the distance formula: \(D = \frac{|2 * 3 + (-3) * 2 + 4 *(-1) + (-24)|}{\sqrt{2^2 + (-3)^2 + 4^2}}\).

Step 3: Simplify the expression

The last step is to simplify that expression: After calculating the values in the numerator and the denominator, the equation turns to \(\frac{|6 -6 -4 -24|}{\sqrt{4 + 9 + 16}}\) which simplifies to \(D = \frac{28}{\sqrt{29}} \approx 5.18\)

Key Concepts

Distance Formula

The distance formula is an essential tool in mathematics, especially for finding how far apart different objects are in space. When we talk about the distance from a point to a plane, we're interested in finding the shortest possible path from a specific point to the surface of a plane. This shortest path is a line that is perpendicular to the plane.
In the context of three-dimensional geometry, finding this distance involves a specific formula:

• The given point is denoted as \(x_{0}, y_{0}, z_{0}\).
• The plane is defined by its equation \(a x + b y + c z + d = 0\).
• The formula to calculate the distance \(D\) is: \[D = \frac{\left| a x_{0} + b y_{0} + c z_{0} + d \right|}{\sqrt{a^{2} + b^{2} + c^{2}}}\]

This equation allows us to methodically calculate the distance by substituting the appropriate values for \(a, b, c, d\) and the coordinates of the point. Remember, the numerator considers the evaluation of the plane equation at the given point, while the denominator represents the norm of the normal vector to the plane.

Three-dimensional Geometry

Three-dimensional geometry explores the properties and relations of points, lines, surfaces, and solids in three-dimensional space. In this arena, we can visualize shapes such as cubes, spheres, and planes. Understanding three-dimensional geometry helps us grasp concepts like volume, surface area, and more complex spatial relationships.
When we talk about measuring the distance in this context, we're often concerned with understanding how figures relate to each other spatially, like how far a point is from a particular plane or line. Point and plane distances utilize perpendicular segments from the point to the plane, which may not always be directly visible but can be calculated using vector subtractions or projections.
This is exactly what makes three-dimensional geometry fascinating – it requires one to think beyond the flat surfaces of two-dimensional space and consider depth. By imagining the perpendicular drop from a point to a plane, we translate abstract numbers into a tangible spatial relationship, giving us the language to describe and analyze the world in more complex dimensions.

Plane Equation

The equation of a plane in three-dimensional geometry is straightforward yet incredibly useful: \[\ a x + b y + c z + d = 0 \]. This equation can represent any plane in three-dimensional space, and each term and coefficient interacts to determine the plane’s orientation and position.

• Parameters \(a, b, c\) are coefficients that come from the vector that is normal (or perpendicular) to the plane.
• The coefficient \(d\) helps position the plane relative to the origin.

Understanding the role of these coefficients is crucial. For example, adjusting \(d\) will shift the plane parallel to itself without changing its orientation. Meanwhile, the vector \( (a, b, c) \) determines the tilt or slant of the plane. By altering this vector, you can transform how the plane stands or lies within the coordinate space.
Grasping how these components work together allows us to manipulate planes algebraically, providing a foundation for more intricate studies in geometry and linear algebra. Whether we are checking intersections, reflections, or distances, the plane equation enables a multitude of calculations in describing spatial relations.