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Chapter 1: Problem 12

Write an equation for the plane containing points \(P=(1,1,-2), Q=(0,2,1)\), and \(R=(-1,-1,0)\) in both standard and general forms

Short Answer

Expert verified
Standard form: \(8x + 4y = 12\), General form: \(8x + 4y + 0z - 12 = 0\).

Step by step solution

01

Find Two Vectors in the Plane

Points \( P \), \( Q \), and \( R \) define a plane in space. First, we need to find two vectors in the plane by subtracting coordinates of these points. Let鈥檚 find vector \( \mathbf{v_1} = \vec{PQ} \) and vector \( \mathbf{v_2} = \vec{PR} \).\( \mathbf{v_1} = Q - P = (0 - 1, 2 - 1, 1 - (-2)) = (-1, 1, 3) \)\( \mathbf{v_2} = R - P = (-1 - 1, -1 - 1, 0 - (-2)) = (-2, -2, 2) \)
02

Find the Normal Vector Using the Cross Product

To find the equation of the plane, we need a normal vector \( \mathbf{n} \). This can be found using the cross product of \( \mathbf{v_1} \) and \( \mathbf{v_2} \):\( \mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & 1 & 3 \ -2 & -2 & 2 \end{vmatrix} \)Solve the determinant:\( \mathbf{n} = (1 \cdot 2 - 3 \cdot -2) \mathbf{i} - (-1 \cdot 2 - 3 \cdot -2) \mathbf{j} + (-1 \cdot -2 - 1 \cdot -2) \mathbf{k} \)\( = (2 + 6) \mathbf{i} - (-2 + 6) \mathbf{j} + (2 - 2) \mathbf{k} \)\( = 8\mathbf{i} + 4\mathbf{j} + 0\mathbf{k} \)So, the normal vector is \( \mathbf{n} = (8, 4, 0) \).
03

Write the Standard Form of the Plane Equation

The standard form of the equation of a plane with a normal vector \( \mathbf{n} = (a, b, c) \) passing through point \( P(x_0, y_0, z_0) \) is given by:\( a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \)Plug in \( \mathbf{n} = (8, 4, 0) \) and point \( P = (1, 1, -2) \):\( 8(x - 1) + 4(y - 1) + 0(z + 2) = 0 \)Simplified, it becomes:\( 8x + 4y = 12 \)
04

Write the General Form of the Plane Equation

The general form of the plane is written as:\( ax + by + cz + d = 0 \)From our standard form equation, \( 8x + 4y = 12 \), rearrange it to find \( d \):\( 8x + 4y - 12 = 0 \)Thus, the general form of the equation is:\( 8x + 4y + 0z - 12 = 0 \)

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Key Concepts

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

Cross Product
The cross product is a powerful tool in vector algebra that helps find a vector perpendicular to two given vectors. This concept is fundamental when finding the normal vector to a plane. For two vectors \( \mathbf{A} \) and \( \mathbf{B} \) in three-dimensional space, their cross product \( \mathbf{A} \times \mathbf{B} \) results in a third vector that is orthogonal (or normal) to both.
  • The cross product is calculated using the determinant of a matrix composed of the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), \( \mathbf{k} \), and the components of \( \mathbf{A} \) and \( \mathbf{B} \).
  • This operation is unique to three-dimensional spaces and forms the essence of vector cross-product properties, such as its distributivity, anticommutativity, and scalar triple products.
Using the vectors \( \mathbf{v_1} = (-1, 1, 3) \) and \( \mathbf{v_2} = (-2, -2, 2) \) from points on a plane, the cross product gives the normal vector \( \mathbf{n} = (8, 4, 0) \), which is a key step in determining the plane's equation.
Normal Vector
A normal vector is vital for defining the orientation of a plane in three-dimensional space. It is perpendicular to every vector lying on the plane and plays a crucial role in forming the plane's equation. If a plane contains the vectors \( \mathbf{v_1} \) and \( \mathbf{v_2} \), the normal vector \( \mathbf{n} \) is obtained through their cross product as illustrated before.
  • A normal vector is essential in articulating the plane equation because it supplies the coefficients for \( x \), \( y \), and \( z \) in the equation.
  • The direction of the normal vector indicates which side of the plane faces a viewer, though any scalar multiple of the normal vector can also serve as a normal vector of the plane.
In our example, the normal vector derived was \( (8, 4, 0) \), providing us with a basis to construct the plane's equation in standard and general forms.
Standard Form
The standard form of a plane equation is a neat way to encapsulate its geometrical properties using a normal vector and a known point on the plane. It is expressed as:\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]where \( (a, b, c) \) is the normal vector, and \( (x_0, y_0, z_0) \) is a point on the plane.
  • The point-slope formula of a plane relies on the normal vector to determine the coefficients \( a, b, \) and \( c \).
  • The solution then involves substituting a specific point such as \( P(1, 1, -2) \) into the equation along with the normal vector.
Upon substitution, the standard form becomes \( 8(x - 1) + 4(y - 1) + 0(z + 2) = 0 \), which simplifies to \( 8x + 4y = 12 \). This form highlights the relationships between coordinates on the plane.
General Form
The general form of a plane equation offers a more conventional algebraic expression commonly used across various applications. It derives from the standard form rearrangement, expressed as:\[ ax + by + cz + d = 0 \]
  • This form simplifies integration with other equations and geometric figures, facilitating broader usage in algebra and geometry.
  • From our example, moving from the standard form \( 8x + 4y = 12 \) to general form involves moving this to \( 8x + 4y - 12 = 0 \).
  • Here, \( d \, (-12) \) is adjusted unless the plane directly intersects the origin \((0,0,0)\).
The general form equation \( 8x + 4y + 0z - 12 = 0 \) appeals due to its straightforward utility in various scenarios, including intersections and transformations.

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Most popular questions from this chapter

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