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Chapter 11: Problem 26

Find equations for the planes. The plane through \(A(1,-2,1)\) perpendicular to the vector from the origin to \(A\)

Short Answer

Expert verified
The equation of the plane is \( x - 2y + z = 6 \).

Step by step solution

01

Understand the additional information

First, identify what the problem asks. It requires finding the equation of a plane through the point \( A(1, -2, 1) \) and perpendicular to a given vector.
02

Determine the normal vector

The plane is perpendicular to the vector from the origin to the point \( A(1, -2, 1) \). The vector from the origin to \( A \) is \( \vec{n} = \langle 1, -2, 1 \rangle \). This vector will be the normal vector to the plane.
03

Use the point-normal form of a plane

The equation of a plane can be written as \( a(x-x_1) + b(y-y_1) + c(z-z_1) = 0 \) where \( (x_1, y_1, z_1) \) is a point on the plane and \( \langle a, b, c \rangle \) is the normal vector to the plane. Here, \( (x_1, y_1, z_1) = (1, -2, 1) \) and the normal vector \( \langle a, b, c \rangle = \langle 1, -2, 1 \rangle \).
04

Substitute into the point-normal form

Substitute the point \( A(1,-2,1) \) and the normal vector \( \langle 1, -2, 1 \rangle \) into the point-normal form: \( 1(x-1) - 2(y+2) + 1(z-1) = 0 \).
05

Simplify the equation

Expand and simplify the equation: \( 1(x-1) - 2(y+2) + 1(z-1) = 0 \) becomes \( x - 1 - 2y - 4 + z - 1 = 0 \) which simplifies to \( x - 2y + z = 6 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Vector
The normal vector is a key concept when discussing the geometry of planes. It provides a direction that is perpendicular to the plane. When a plane is defined, one way to describe its orientation is through a normal vector.
  • A normal vector to a plane is a vector that points directly away from the plane's surface.
  • In mathematical terms, if our normal vector is \( \langle a, b, c \rangle \), this vector is perpendicular to every vector lying in the plane.
  • For example, in this exercise, the vector from the origin to the point \( A(1,-2,1) \) is \( \langle 1, -2, 1 \rangle \). This vector is used as the normal vector for the plane.
The normal vector can also influence the plane's equation. Picking the correct normal vector helps in describing the orientation of the plane accurately.
Point-Normal Form
The point-normal form is an equation format used to express a plane using a point and a normal vector.
  • The equation for a plane in point-normal form is \( a(x-x_1) + b(y-y_1) + c(z-z_1) = 0 \).
  • Here, \( (x_1, y_1, z_1) \) is a known point on the plane, and \( \langle a, b, c \rangle \) is the normal vector.
  • This form is useful because it directly incorporates the normal vector and the point, making it simple to derive the equation of a plane.
In the problem, the point \( A(1,-2,1) \) and the vector \( \langle 1, -2, 1 \rangle \) are plugged into this formula. This results in the initial equation \( 1(x-1) - 2(y+2) + 1(z-1) = 0 \), which can be simplified to the final plane equation.
Coordinate Geometry
Coordinate geometry allows us to describe geometric figures and analyze their properties using a coordinate system.
  • It provides a link between algebraic equations and geometric figures like lines, planes, and circles.
  • In a three-dimensional coordinate system, planes can be represented algebraically with equations derived from vectors and points.
  • A point like \( A(1,-2,1) \) helps serve as a known point for describing the position of the plane in space.
When dealing with coordinate geometry, the position and orientation of geometric figures are expressed in terms of coordinates and vectors. This simplifies complex geometric problems into solvable algebraic equations.
Vector Algebra
Vector algebra is used to handle operations like addition, subtraction, and dot products between vectors.
  • It underpins the mathematical operations conducted on vectors, essential in problems like deriving plane equations.
  • With vectors, you can describe directions and calculate properties such as angles and perpendiculars.
  • In this exercise, vector algebra helps establish the normal vector to the plane from the point and origin, \( \langle 1, -2, 1 \rangle \).
By understanding vector algebra, students can excel in finding and verifying plane equations. It aids in moving between different forms and solving diverse problems in three-dimensional space.

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