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Magazine Chapter 3: Problem 33
Find the distance between the point and the plane. $$(3,1,-2) ; x+2 y-2 z=4$$
Short Answer
Expert verified
The distance is \( \frac{5}{3} \).
Step by step solution
01
Identify the point and plane equation
We have a point \( P(3, 1, -2) \) and a plane given by the equation \( x + 2y - 2z = 4 \). The plane equation is in the form \( Ax + By + Cz = D \) where A=1, B=2, C=-2, and D=4.
02
Write the formula for distance from a point to a plane
The distance \( d \) from a point \( (x_1, y_1, z_1) \) to a plane \( Ax + By + Cz = D \) is given by \(d = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}}.\)
03
Substitute the point and plane coefficients into the formula
Substitute \( x_1 = 3, y_1 = 1, z_1 = -2 \), and the plane coefficients A=1, B=2, C=-2, D=4 into the distance formula: \(d = \frac{|1 \times 3 + 2 \times 1 - 2 \times (-2) - 4|}{\sqrt{1^2 + 2^2 + (-2)^2}}.\)
04
Calculate the numerator
Calculate \( |1 \times 3 + 2 \times 1 - 2 \times (-2) - 4| = |3 + 2 + 4 - 4| = |5| = 5 \).
05
Calculate the denominator
Calculate the denominator \( \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \).
06
Calculate the distance
Substitute the values from Step 4 and Step 5 into the formula to find the distance: \(d = \frac{5}{3}.\)
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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 serves as a model to describe a flat surface in three-dimensional space using a simple algebraic expression. It is written in the standard form: \( Ax + By + Cz = D \). Here, \( A \), \( B \), and \( C \) are constants that determine the plane's orientation, acting like coefficients that relate the position vectors to the plane. The constant \( D \) represents the plane's distance from the origin along the normal vector.
- The coefficients \(A, B, C\) can also be seen as components of a normal vector (perpendicular to the plane).
- The equation denotes that for any point \((x, y, z)\) lying on the plane, the equality holds true.
- Understanding the plane equation helps in calculating geometric concepts like distances or intersections.
Point-Plane Distance Formula
The point-plane distance formula is a way to measure how far a point is from a given plane in three-dimensional space. The formula is derived from the geometry of triangles and vectors. It is expressed as:
\[ d = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}} \]
Here's a breakdown of its components:
\[ d = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}} \]
Here's a breakdown of its components:
- \((x_1, y_1, z_1)\) represents the coordinates of the point from which the distance is measured.
- \(A, B, C\) are the coefficients from the plane's equation used to align the calculation with the plane's orientation.
- The numerator computes the scalar projection of the point onto the normal vector of the plane, reflecting how much the point "extends" towards the plane.
- The denominator, \(\sqrt{A^2 + B^2 + C^2}\), normalizes this projection by the length of the normal vector, ensuring the distance is consistent regardless of scaling.
3D Geometry
3D geometry extends the familiar two-dimensional geometry to a third dimension, adding depth alongside width and height. This extension allows for the study of objects in a more realistic space where they have volume, surface area, and spatial relationships.
- Common elements include points, lines, planes, and figures like cubes, spheres, and pyramids.
- Vectors and coordinate systems (like Cartesian coordinates) are foundational, helping describe the position and movement of points and shapes in space.
- Concepts like vector cross product and dot product are used extensively to handle direction and magnitude problems inherent in 3D.
- Understanding 3D geometry is essential for fields like physics, computer graphics, and engineering, where spatial awareness and precision are key.
