Question

Find the given distances. The distance from the point (2,6,2) to the plane \(2(x-1)-y+4(z+1)=0\)

Short answer

The distance is \( \frac{8}{\sqrt{21}} \).

Step-by-step solution

Recognize the General Plane Equation

The equation of a plane can generally be expressed in the form \( Ax + By + Cz + D = 0 \). For this exercise, rewrite the given plane equation into that form: First, expand the given equation: \[ 2(x-1) - y + 4(z+1) = 0 \] \[ 2x - 2 - y + 4z + 4 = 0 \] Combine like terms to obtain the standard form: \[ 2x - y + 4z + 2 = 0 \] Here, \( A = 2 \), \( B = -1 \), \( C = 4 \), and \( D = 2 \).

Apply the Distance Formula

The distance \( d \) from a point \((x_1, y_1, z_1)\) to a plane \( Ax + By + Cz + D = 0 \) is determined by the formula: \[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \] Substitute the given point \((2, 6, 2)\) and plane coefficients \( A = 2, B = -1, C = 4, D = 2 \) into the formula.

Calculate the Numerator of the Formula

Calculate \( Ax_1 + By_1 + Cz_1 + D \) using the values: \[ 2(2) + (-1)(6) + 4(2) + 2 \] \[ = 4 - 6 + 8 + 2 \] \[ = 8 \] The absolute value of the numerator is \( |8| = 8 \).

Calculate the Denominator of the Formula

The denominator of the distance formula is calculated as: \[ \sqrt{A^2 + B^2 + C^2} = \sqrt{2^2 + (-1)^2 + 4^2} \] \[ = \sqrt{4 + 1 + 16} \] \[ = \sqrt{21} \]

Compute the Distance

Now, substitute the calculated numerator and denominator into the distance formula: \[ d = \frac{8}{\sqrt{21}} \] This fraction can be left in this form or simplified for specific numerical outputs by rationalizing the denominator if needed.

Key concepts

Plane Equation

A plane equation is a crucial element in 3D geometry, helping us understand the spatial relationships of three-dimensional shapes. The standard form of a plane's equation is usually given as \( Ax + By + Cz + D = 0 \). This equation describes a flat surface where each point on it satisfies these mathematical conditions. In the plane equation:

• \( A, B, \) and \( C \) are the coefficients representing the plane's orientation in space.
• \( D \) is the constant term that helps position the plane in the 3D space.
• \( x, y, \) and \( z \) are the coordinates of any point on the plane.

For the given problem, the plane equation starts in an unconventional format: \( 2(x-1) - y + 4(z+1) = 0 \). By expanding and rearranging terms, we can transform it into the standard form: \( 2x - y + 4z + 2 = 0 \). This makes it easier to apply various calculations, such as finding the distance of a point from the plane.

Distance Formula

The distance formula for calculating how far a point is from a plane in 3D space builds on the fundamentals of plane equations. This formula is vital for deriving exact distances:\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]Each part of the formula plays a role:

• The numerator \(|Ax_1 + By_1 + Cz_1 + D|\) calculates the expression of the plane containing the point coordinates \((x_1, y_1, z_1)\).
• This numerator is an absolute value, ensuring that distance is always positive, as distance cannot be negative.
• The denominator \(\sqrt{A^2 + B^2 + C^2}\) normalizes the distance based on the plane's orientation.

This formula is powerful because it simplifies finding distances even in complex geometrical configurations, ensuring both consistency and accuracy.

Distance Calculation

To calculate the distance from a point to a plane, specific steps must be followed, starting with plugging into the distance formula. Here's a breakdown of the calculation:1. **Substitute the Coordinates**: For the point \((2, 6, 2)\) and the coefficients \( A = 2, B = -1, C = 4, D = 2 \) from the plane equation, begin by evaluating the numerator: \[ Ax_1 + By_1 + Cz_1 + D = 2(2) + (-1)(6) + 4(2) + 2 = 8 \] Take the absolute value to ensure positivity: \(|8| = 8\). 2. **Compute the Denominator**: Calculate it using: \[ \sqrt{A^2 + B^2 + C^2} = \sqrt{2^2 + (-1)^2 + 4^2} = \sqrt{21} \]3. **Form the Ratio**: Now, assemble the full formula: \[ d = \frac{8}{\sqrt{21}} \]This result provides the distance as a rational number, although further simplification or decimal conversion is optional, depending on the context's requirements.

3D Geometry

3D Geometry expands on basic geometrical principles into the three-dimensional space we live in. This allows us to explore spatial relationships more richly:

• Dimensions: While 2D involves length and width, 3D includes depth, giving us the Cartesian coordinates \((x, y, z)\).
• Crystal Clear Visualizations: Understanding how planes, lines, and points interact in space enable visualizing more abstract concepts, reflecting real-world scenarios.
• Applications: We use 3D geometry in various fields like architecture, computer graphics, and physics to design and analyze structures.

In the context of our problem, we're engaging with these concepts to determine spatial relationships, specifically how a point relates to a plane, demonstrating the elegance and necessity of 3D calculations.