Problem 27 - Thomas Calculus

Question

Find the point of intersection of the lines \(x=2 t+1, \quad y=3 t+2\) \(z=4 t+3,\) and \(x=s+2, y=2 s+4, z=-4 s-1,\) and then find the plane determined by these lines.

Step-by-Step Solution

Answer

The intersection point is (1, 2, 3), and the plane equation is \(-20x + 12y + z = -25\).

Step 1: Find Intersecting Point Step 1: Equalize Equations

Since each line is represented parametrically, we need to find values of parameters \(t\) and \(s\) such that both sets of equations give the same \(x, y, z\). Start by setting the equations for \(x\) equal: \(2t + 1 = s + 2\). Simplify it: \(2t = s + 1\).

Step 2: Find Intersecting Point Step 2: Solve Equations for y

Now, set the \(y\) equations equal: \(3t + 2 = 2s + 4\). Simplify it to find a relation between \(t\) and \(s\): \(3t = 2s + 2\).

Step 3: Find Intersecting Point Step 3: Solve System of Equations

Solve the simultaneous equations: 1. \(2t = s + 1\)2. \(3t = 2s + 2\).Substitute \(s = 2t - 1\) from the first equation into the second equation:\(3t = 2(2t - 1) + 2\) which becomes \(3t = 4t - 2 + 2\) or \(t = 0\).Using \(t = 0\) in \(s = 2t - 1\), we find \(s = -1\).

Step 4: Find Intersecting Point Step 4: Calculate the Intersection Point

Use \(t = 0\) in the first set of parametric equations: \(x = 2(0) + 1 = 1\), \(y = 3(0) + 2 = 2\), \(z = 4(0) + 3 = 3\).The point of intersection is \((1, 2, 3)\).

Step 5: Determine Plane Step 1: Find Direction Vectors

Calculate the direction vectors of the lines: The first line has direction vector \(\langle 2, 3, 4 \rangle\).The second line has direction vector \(\langle 1, 2, -4 \rangle\).

Step 6: Determine Plane Step 2: Cross Product of Vectors

To find the plane determined by the lines, compute the cross product of the direction vectors \(\langle 2, 3, 4 \rangle\) and \(\langle 1, 2, -4 \rangle\):\(\mathbf{n} = \langle 3(-4) - 4(2), 4(1) - 2(-4), 2(2) - 3(1)\rangle = \langle -20, 12, 1 \rangle \).

Step 7: Determine Plane Step 3: Equation of the Plane

We use the intersection point \((1, 2, 3)\) and the normal vector \(\langle -20, 12, 1 \rangle\) to form the equation of the plane:\(-20(x - 1) + 12(y - 2) + 1(z - 3) = 0\).Simplify to: \(-20x + 12y + z = -25\).

Key Concepts

Parametric Equations

Parametric equations are a powerful way to represent a line or curve using parameters, often denoted as variables like \( t \) or \( s \). In 3D space, parametric equations for a line generally take the form:

• \( x = at + b \)
• \( y = ct + d \)
• \( z = et + f \)

These represent equations for \( x \), \( y \), and \( z \) in terms of the parameter \( t \).
In our case, two lines are given:- First line: \( x = 2t + 1, \; y = 3t + 2, \; z = 4t + 3 \).- Second line: \( x = s + 2, \; y = 2s + 4, \; z = -4s - 1 \).
To find the point where these lines intersect, we must find values for the parameters \( t \) and \( s \) such that the \( x \), \( y \), and \( z \) equations from both lines are equal. This happens when the same point in space corresponds to the same parameter values for \( x \), \( y \), and \( z \) from each line. By solving these equations simultaneously, as shown in the solution, you obtain the exact point of intersection.

Direction Vectors

Direction vectors provide critical insights into the orientation of lines in space. For line defined by parametric equations, the direction vector is extracted directly from the coefficients associated with the parameter.
For example, in the given equations:- First line's direction vector is \( \langle 2, 3, 4 \rangle \) because \( 2, 3, 4 \) are the coefficients of \( t \).- Second line's direction vector is \( \langle 1, 2, -4 \rangle \) extracted similarly from \( s \).
Direction vectors are essential for describing how a line extends through space and are especially useful in determining relationships between different lines. When comparing two lines, their direction vectors can help in analyzing if the lines are parallel or intersect at some point. In this particular exercise, the direction vectors are later used to find the normal vector for a plane.

Cross Product

The cross product is a mathematical operation that finds a vector perpendicular to two given vectors in three-dimensional space. When dealing with two direction vectors, the cross product helps calculate a normal vector, which is important for creating equations of planes.
Given direction vectors \( \langle 2, 3, 4 \rangle \) and \( \langle 1, 2, -4 \rangle \), the cross product is computed as follows:\[ \mathbf{n} = \langle 3(-4) - 4(2), \; 4(1) - 2(-4), \; 2(2) - 3(1) \rangle \]
Simplifying, we get:\[ \mathbf{n} = \langle -20, 12, 1 \rangle \]
This normal vector \( \mathbf{n} \) is used extensively for defining planes, as it is perpendicular to the plane surface marked out by the intersection of the two lines. Calculating the cross product is an essential step in moving from linear representations to plane equations.

Plane Equation

A plane in three-dimensional space can be described mathematically using a plane equation. This equation is derived using a point on the plane and a normal vector to that plane.
To form a plane equation:- Use the intersection point on the plane, say \((x_0, y_0, z_0)\).- Employ the normal vector \( \mathbf{n} = \langle a, b, c \rangle \).
The general formula for the plane equation is:\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
Plugging in the intersection point \((1, 2, 3)\) and normal vector \( \langle -20, 12, 1 \rangle \), you form:\[ -20(x - 1) + 12(y - 2) + 1(z - 3) = 0 \]
This simplifies to:\[ -20x + 12y + z = -25 \]
Thus, the equation \(-20x + 12y + z = -25\) defines a plane in space, outlining where the two original lines lie. This process highlights the intersection and relationship between linear and planar geometries.