Problem 39 - University Calculus: Early Transcendentals

Question

find the distance from the point to the plane. $$(2,-3,4), \quad x+2 y+2 z=13$$

Step-by-Step Solution

Answer

The distance is 3 units.

Step 1: Understand the Problem

We need to find the distance from the point \((2, -3, 4)\) to the plane given by the equation \(x + 2y + 2z = 13\).

Step 2: Use the Distance Formula

The distance \(D\) from a point \((x_0, y_0, z_0)\) to a plane \(Ax + By + Cz + D = 0\) is given by:\[D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}\]For our plane \(x + 2y + 2z - 13 = 0\), we identify \(A = 1\), \(B = 2\), \(C = 2\), and \(D = -13\).

Step 3: Substitute Point Coordinates

Substitute the coordinates of the point \((2, -3, 4)\) into our formula:\[D = \frac{|1(2) + 2(-3) + 2(4) - 13|}{\sqrt{1^2 + 2^2 + 2^2}}\]

Step 4: Compute the Numerator

Calculate the absolute value in the numerator:\[1(2) + 2(-3) + 2(4) - 13 = 2 - 6 + 8 - 13 = -9\]Take the absolute value: \(|-9| = 9\).

Step 5: Compute the Denominator

Calculate the denominator:\[\sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\]

Step 6: Find the Distance

Substitute the computed values back into the distance formula:\[D = \frac{9}{3} = 3\]The distance from the point \((2, -3, 4)\) to the plane is 3 units.

Key Concepts

Distance Formula

The distance formula allows us to calculate how far one entity is from another in a given set of dimensions. In the context of three-dimensional geometry, this formula is specifically used to find the distance from a point to a plane. The formula used is:

• \[ D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

Here, \(A\), \(B\), and \(C\) are coefficients from the plane's equation, and \(D\) is the constant term when the plane equation is rearranged to the general form. The absolute value in the numerator ensures that the distance is always a positive value. This formula essentially determines how far the given point lies along the normal (a perpendicular line) to the plane.

Plane Equation

A plane in three-dimensional space can be described by a linear equation involving three variables, \(x\), \(y\), and \(z\). The general form of a plane equation is:

• \( Ax + By + Cz + D = 0 \)

In our exercise, the plane is given by the equation \(x + 2y + 2z = 13\). To fit it into the general form, rearrange it as \(x + 2y + 2z - 13 = 0\).

• Here, \(A = 1\), \(B = 2\), \(C = 2\), and \(D = -13\).

These coefficients and constant are crucial for understanding the plane's orientation and position in space. They are used in calculating distances and understanding geometric properties such as intercepts and parallelism in three-dimensional geometry.

Three-Dimensional Geometry

Three-dimensional geometry extends our understanding of shapes and spaces beyond flat, two-dimensional planes. In 3D, we interact with objects that have depth, width, and height. This allows us to describe and solve more complex spatial problems.

• Points are represented as coordinates \((x, y, z)\).
• Planes are represented by equations like \(Ax + By + Cz + D = 0\).
• Lines can intersect, run parallel, or be skew in 3D space.

Understanding these elements is essential when working with planes and points in 3D. In such exercises, one determines not just the position of points relative to planes but also interprets how these planes divide space.

Absolute Value

In mathematics, the absolute value of a number is the non-negative value of that number without regard to its sign. It is represented as \(|x|\). In terms of real-world distance calculations, absolute value ensures that distance is always a positive quantity.

• For example, \(|-9| = 9\).

When calculating the distance between a point and a plane, the absolute value is used to make sure the computed distance is never negative, thereby reflecting the reality that distances cannot be negative as they represent the measure of separation between two physical points or planes.