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Chapter 12: Problem 69

Find two different planes whose intersection is the line \(x=1+t, y=2-t, z=3+2 t .\) Write equations for each plane in the form \(A x+B y+C z=D\)

Short Answer

Expert verified
The planes are \(5x + 2y + 2z = 17\) and \(x + y - z = -2\).

Step by step solution

01

Analyze the Given Line

The line is given in the parametric form: \(x=1+t\), \(y=2-t\), and \(z=3+2t\). The direction vector for this line is \((1, -1, 2)\) and a point on the line is \((1, 2, 3)\). We will use this point and the direction vector information to form equation of planes.
02

Choose a Direction Vector for Each Plane

To find two planes intersecting along the given line, we select a direction vector from the line. Let's choose \(v_1 = (1, -1, 2)\) and another vector \(v_2\) in the plane which is not parallel to the direction vector, such as \<(1, 1, 0)\>.
03

Form the First Plane Equation

Consider the first direction vector \((1, -1, 2)\) and another vector \(v_3 = (0, -2, 1)\), orthogonal to \<(1, -1, 2)\>. Compute the cross product \(v_1 \times v_3\) to get a normal vector for the first plane.
04

Compute Cross Product for the First Plane

Calculate \ = \<(\begin{vmatrix}-1 & 2\-2 & 1\end{vmatrix}, \begin{vmatrix}2 & 1\1 & 0\end{vmatrix}, \begin{vmatrix}1 & -1\0 & -2\end{vmatrix})> which gives us a normal \(n_1 = (5, 2, 2)\).
05

Write the Equation of the First Plane

Using the normal vector \(n_1 = (5, 2, 2)\) and the point \(P(1, 2, 3)\), form the equation: \(5(x-1) + 2(y-2) + 2(z-3) = 0\). Simplifying, we get \(5x + 2y + 2z = 17\).
06

Form the Second Plane Equation

Choose a second non-parallel vector \(v_4 = (1, 1, 0)\). Compute the cross product \(v_1 \times v_4\) to determine the normal vector for the second plane.
07

Compute Cross Product for the Second Plane

Calculate \ = \<(\begin{vmatrix}-1 & 2\1 & 0\end{vmatrix}, \begin{vmatrix}1 & 2\1 & 0\end{vmatrix}, \begin{vmatrix}1 & -1\1 & 1\end{vmatrix})> which gives a normal \(n_2 = (-2, -2, 2)\).
08

Write the Equation of the Second Plane

Using normal vector \(n_2 = (-2, -2, 2)\) and point \(P(1, 2, 3)\), form the equation: \(-2(x-1) - 2(y-2) + 2(z-3) = 0\). Simplify this to get \(-2x - 2y + 2z = 4\), or equivalently, \(x + y - z = -2\).

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Key Concepts

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

Parametric Equations
A parametric equation is a way of defining a line, curve, or surface using parameters. In the given exercise, the parametric equations for the line are given as \(x=1+t\), \(y=2-t\), and \(z=3+2t\). Here, \(t\) is the parameter. This form makes it easy to express the coordinates of any point on the line as a function of \(t\). This is extremely useful in geometry, as it allows us to work with lines dynamically.
  • They provide a flexible method to represent geometric objects and analyze their properties.
  • The parameter \(t\) can take any real value, capturing the entire line that extends infinitely in both directions.
  • Understanding parametric equations is crucial for solving problems in three-dimensional space, such as finding intersections of lines and planes.
Cross Product
The cross product is a vector operation that takes two vectors in three-dimensional space and produces a new vector that is orthogonal to both. In the exercise, we use the cross product to find a normal vector to the plane. For instance, when computing \(v_1 \times v_3\) where \(v_1 = (1, -1, 2)\) and \(v_3 = (0, -2, 1)\), we calculate it by determining the components of the resulting vector using determinants.
The cross product is defined mathematically as:\[\mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2, \, a_3b_1 - a_1b_3, \, a_1b_2 - a_2b_1)\]
  • This operation is essential when needing a perpendicular direction to form plane equations.
  • A non-zero cross product indicates that the original vectors are not parallel.
  • The magnitude of the cross product gives the area of the parallelogram formed by the two vectors.
Direction Vector
A direction vector is a vector that points in the direction of a line. It is derived directly from the parametric equations of the line. In this exercise, the direction vector for the line is \((1, -1, 2)\), which corresponds to the coefficients of \(t\) in the equations \(x=1+t\), \(y=2-t\), and \(z=3+2t\). The direction vector serves as a critical tool in understanding line orientation and is crucial when forming planes that intersect along the line.
Knowing the direction vector:
  • Helps to understand the orientation and extent of the line in space.
  • It is used to determine the intersection between lines and planes.
  • Provides key insights into manipulating parametric and Cartesian forms of line equations.
The direction vector remains constant for a line, illustrating why it plays a fundamental role in calculation and visualization of the geometric space features.

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Most popular questions from this chapter

Using the definition of the projection of \(\mathbf{u}\) onto \(\mathbf{v},\) show by direct calculation that \(\left(\mathbf{u}-\operatorname{proj}_{\mathbf{v}} \mathbf{u}\right) \cdot \operatorname{proj}_{\mathbf{v}} \mathbf{u}=0\)

In Exercises \(9-14,\) sketch the coordinate axes and then include the vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{u} \times \mathbf{v}\) as vectors starting at the origin. $$ \mathbf{u}=2 \mathbf{i}-\mathbf{j}, \quad \mathbf{v}=\mathbf{i}+2 \mathbf{j} $$

In Exercises \(1-8,\) find \begin{equation} \begin{array}{l}{\text { a. } \mathbf{v} \cdot \mathbf{u},|\mathbf{v}|,|\mathbf{u}|} \\ {\text { b. the cosine of the angle between } \mathbf{v} \text { and } \mathbf{u}} \\ {\text { c. the scalar component of } \mathbf{u} \text { in the direction of } \mathbf{v}} \\ {\text { d. the vector proj, u. }}\end{array} \end{equation} \begin{equation} \mathbf{v}=5 \mathbf{i}+\mathbf{j}, \quad \mathbf{u}=2 \mathbf{i}+\sqrt{17} \mathbf{j} \end{equation}

In Exercises \(53-56,\) find the point in which the line meets the plane. $$ x=1+2 t, \quad y=1+5 t, \quad z=3 t ; \quad x+y+z=2 $$

In Exercises \(53-56,\) find the point in which the line meets the plane. $$ x=1-t, \quad y=3 t, \quad z=1+t ; \quad 2 x-y+3 z=6 $$
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