Problem 28 - Thomas Calculus

Question

Find the point of intersection of the lines \(x=t, y=-t+2\) \(z=t+1,\) and \(x=2 s+2, y=s+3, z=5 s+6,\) and then find the plane determined by these lines.

Step-by-Step Solution

Answer

The lines do not intersect. Plane equation: \(3x + 4y + 3z = 10\).

Step 1: Equate Parameterized Coordinates for Intersection

Given the equations of the lines, you need to equate the x, y, and z coordinates for both parameterizations to find their point of intersection. This results in the system of equations:1. For x: \( t = 2s + 2 \)2. For y: \( -t + 2 = s + 3 \)3. For z: \( t + 1 = 5s + 6 \)

Step 2: Solve the System of Equations

Use the first and third equations to express both t and s in terms of each other:1. From the first equation: \( t = 2s + 2 \)2. Substitute this into the third equation: \[ 2s + 2 + 1 = 5s + 6 \] Simplifying gives \( 3 = 3s \), so \( s = 1 \).3. Substitute \( s = 1 \) back into the first equation to find \( t \): \( t = 2(1) + 2 = 4 \).

Step 3: Find Intersection Point Coordinates

With \( t = 4 \) and \( s = 1 \), substitute these values into the parameterized equations:- From Line 1: - \( x = 4, y = -4 + 2 = -2, z = 4 + 1 = 5 \).- From Line 2: - \( x = 2(1) + 2 = 4, y = 1 + 3 = 4, z = 5(1) + 6 = 11 \).Both lines do not intersect at a common point. Thus, adjust calculations if necessary, or verify non-intersection.

Step 4: Determine Parametric Vectors for Plane

Identify the direction vectors of each line to find the plane determined by the lines:- Line 1 Direction Vector: \( \mathbf{d_1} = (1, -1, 1) \).- Line 2 Direction Vector: \( \mathbf{d_2} = (2, 1, 5) \).

Step 5: Compute the Normal Vector of the Plane

The normal vector \( \mathbf{n} \) can be found by taking the cross product of \( \mathbf{d_1} \) and \( \mathbf{d_2} \):\[ \mathbf{n} = \mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & -1 & 1 \ 2 & 1 & 5 \end{vmatrix} \]Calculate the cross product:\[ \mathbf{n} = (4 - 1)\mathbf{i} - (1 - 5)\mathbf{j} + (1 + 2)\mathbf{k} = (3, 4, 3) \].

Step 6: Write the Equation of the Plane

Use a point from either line and the normal vector to write the plane's equation. Take the point from Line 1 at \( t = 0 \) which is \( (0, 2, 1) \):Substitute into the plane equation:\[ 3(x - 0) + 4(y - 2) + 3(z - 1) = 0 \].Simplifying gives:\[ 3x + 4y + 3z = 10 \].

Key Concepts

Coordinate Geometry

Coordinate Geometry is a branch of geometry where we use the coordinate system to represent and understand the positions of points, lines, and other geometric shapes. In this exercise, we work specifically with lines and their equations in a three-dimensional space.

When dealing with lines in three dimensions, we often write them in parametric form. A parametric equation expresses the coordinates of the points that make up a geometric figure as functions of a third variable, usually called a parameter. For example, in this exercise, we have two lines described by parameters \( t \) and \( s \), respectively.

• The first line is given by: \( x = t, y = -t + 2, z = t + 1 \).
• The second line is given by: \( x = 2s + 2, y = s + 3, z = 5s + 6 \).

To find where two lines intersect in coordinate geometry, their parameterized coordinates must satisfy the same location in space. This results in solving simultaneous equations to find the values of \( t \) and \( s \) where the lines might intersect. Remember that if different solutions are found for the coordinates, the lines do not intersect in reality.

System of Equations

A System of Equations consists of multiple equations that share common variables. Solving a system is obtaining the values for these variables that satisfy all the equations simultaneously. In our example, we are given a system of equations to find a possible intersection point of two lines in 3D space.

Here, the system corresponding to the x, y, and z coordinates is:

• For the x-coordinates: \( t = 2s + 2 \)
• For the y-coordinates: \( -t + 2 = s + 3 \)
• For the z-coordinates: \( t + 1 = 5s + 6 \)

To solve this system, an effective strategy can be to express one variable in terms of another and then substitute back into the other equations. For instance, from the x-equation, express \( t \) in terms of \( s \): \( t = 2s + 2 \).

Substituting \( t = 2s + 2 \) into the z-coordinate equation produces another relationship. Solve these equations step by step, simplifying until the values for \( s \) and \( t \) can be found. If there is a contradiction or incorrect match between results, it can indicate that the lines do not intersect practically in 3D space.

Plane Equation

Once we determine if lines intersect, or at least how they relate in space, we often require the Plane Equation when dealing with geometrical setups involving lines. A plane is a flat, two-dimensional surface that extends infinitely.

In this exercise, after considering the intersection of the lines, we may still explore how they lie in the 3D space by finding a plane equation. We do this using the directional vectors of each line. A direction vector is essentially a vector that indicates the direction of the line.

• For Line 1, its direction is \( \mathbf{d_1} = (1, -1, 1) \).
• For Line 2, its direction is \( \mathbf{d_2} = (2, 1, 5) \).

To find the normal vector \( \mathbf{n} \) of the plane that these lines might define, we take the cross product of \( \mathbf{d_1} \) and \( \mathbf{d_2} \). The normal vector is perpendicular to the plane surface. Calculating the cross product finds \( \mathbf{n} = (3, 4, 3) \).

Using a point and the normal vector, the general equation of the plane can be written. Subposing a point from the line, the equation simplifies to \( 3x + 4y + 3z = 10 \). This plane houses either line and the respective direction in that 3D environment.