Chapter 10: Problem 28
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
Expert verified
The distance is \( \frac{8}{\sqrt{21}} \).
Step by step solution
01
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 \( D = 2 \).
02
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.
03
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 \( |8| = 8 \).
04
Calculate the Denominator of the Formula
The denominator of the distance formula is calculated as: \[ \sqrt{A^2 + B^2 + C^2} = \sqrt{2^2 + (-1)^2 + 4^2} \] \[ = \sqrt{4 + 1 + 16} \] \[ = \sqrt{21} \]
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.