Problem 69 - Thomas Calculus

Question

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 \(A x+B y+C z=D\) .

Step-by-Step Solution

Answer

The two planes are \(x + y = 3\) and \(y + z = 5\).

Step 1: Identify Direction Vector and Point on the Line

The line is given in parametric form: \(x = 1 + t\), \(y = 2 - t\), \(z = 3 + 2t\). From these, we can identify the direction vector \(\mathbf{v} = \langle 1, -1, 2 \rangle\) and a point on the line \( (1, 2, 3) \).

Step 2: Choose Two Different Normal Vectors

To find two planes that intersect in this line, we need two different vectors, each normal to the direction vector \(\mathbf{v}\). Choose normal vectors \(\mathbf{n_1} = \langle 1, 1, 0 \rangle\) and \(\mathbf{n_2} = \langle 0, 1, 1 \rangle\), which are not parallel to \(\mathbf{v}\).

Step 3: Use Normal Vector to Write Plane Equation

For the first plane, use \(\mathbf{n_1}\). The equation is \(1(x-1) + 1(y-2) + 0(z-3) = 0\), simplifying to \(x + y = 3\).For the second plane, use \(\mathbf{n_2}\). The equation is \(0(x-1) + 1(y-2) + 1(z-3) = 0\), simplifying to \(y + z = 5\).

Step 4: Verify Planes Intersect at Given Line

Check that both planes contain the direction vector \(\langle 1, -1, 2 \rangle\). For \(x + y = 3\), substituting line equations gives \((1 + t) + (2 - t) = 3\). For \(y + z = 5\), substituting gives \((2-t) + (3+2t) = 5\). Both hold true, so both planes contain the line.

Key Concepts

Parametric Equations

In mathematics, especially in linear algebra and calculus, parametric equations are crucial. They express coordinates of the points on a line or curve as functions of a parameter, typically denoted as \(t\). For our problem, the parametric equations provided are for a line:

• \(x = 1 + t\)
• \(y = 2 - t\)
• \(z = 3 + 2t\)

Here, \(t\) is a parameter that varies over all real numbers, and each trio of \((x, y, z)\) represents a point on the line. This form is beneficial for extracting the direction vector and a specific point on the line. From the given equations, a point can be easily deduced by setting \(t = 0\), giving us the point \((1, 2, 3)\). The direction vector of the line, indicated by how each coordinate changes with \(t\), is \(\mathbf{v} = \langle 1, -1, 2 \rangle\). Parametric equations are useful for modeling line intersections because they clearly express both the line's direction and position.

Normal Vector

The normal vector of a plane is an essential concept in the realm of 3D geometry. It is a vector that is perpendicular to the plane itself. To discover planes intersecting at a given line, different normal vectors, which are not parallel to the line's direction vector, must be chosen. In the exercise, the line's direction vector is \(\mathbf{v} = \langle 1, -1, 2 \rangle\). We need two distinct normal vectors for our two planes:

• \(\mathbf{n_1} = \langle 1, 1, 0 \rangle\)
• \(\mathbf{n_2} = \langle 0, 1, 1 \rangle\)

These vectors were selected because they are not parallel to \(\mathbf{v}\), ensuring they create planes intersecting along the line defined by the parametric equations. The normal vector is key in forming the equation of the plane, as it determines the "tilt" or orientation of the plane in space.

Plane Equations

Plane equations are algebraic expressions that identify all points belonging to a plane in three-dimensional space. Typically, plane equations are written as \(Ax + By + Cz = D\), where \(A, B,\) and \(C\) are the components of the normal vector. In our exercise, utilizing the identified normal vectors \(\mathbf{n_1} = \langle 1, 1, 0 \rangle\) and \(\mathbf{n_2} = \langle 0, 1, 1 \rangle\), we form the following plane equations:

• For \(\mathbf{n_1}\): Starting with \(1(x-1) + 1(y-2) + 0(z-3) = 0\), simplify to \(x + y = 3\). This plane incorporates points where the sum of the x and y coordinates yields 3.
• For \(\mathbf{n_2}\): Starting with \(0(x-1) + 1(y-2) + 1(z-3) = 0\), simplify to \(y + z = 5\). This plane encompasses points where the sum of the y and z coordinates equals 5.

The plane equations are verified for intersection along the given line by ensuring they both contain the direction vector \(\langle 1, -1, 2 \rangle\). Substituting the parametric line equations into each plane equation as shown verifies that both planes indeed intersect along the line.