Problem 67 - Advanced Engineering Mathematics

Question

In Problems, find parametric equations for the line of intersection of the given planes. $$ \begin{array}{r} 4 x-2 y-z=1 \\ x+y+2 z=1 \end{array} $$

Step-by-Step Solution

Answer

The parametric equations are: \(x = \frac{1}{2} - \frac{1}{2}t\), \(y = \frac{1}{2} - \frac{3}{2}t\), \(z = t\).

Step 1: Identify Normal Vectors of the Planes

The planes are given by the equations:1) \(4x - 2y - z = 1\) 2) \(x + y + 2z = 1\)The normal vector to a plane \(ax + by + cz = d\) is \(\langle a, b, c \rangle\). For Plane 1, the normal vector is \(\mathbf{n_1} = \langle 4, -2, -1 \rangle\). For Plane 2, the normal vector is \(\mathbf{n_2} = \langle 1, 1, 2 \rangle\).

Step 2: Find Direction Vector of Line of Intersection

The direction vector of the line of intersection is the cross product of the normal vectors of the two planes. Compute the cross product \(\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2}\).\[ \mathbf{d} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \4 & -2 & -1 \1 & 1 & 2\end{vmatrix} \]\(\mathbf{d} = (-4\mathbf{i}, -9\mathbf{j}, 6\mathbf{k})\), which simplifies to \(\langle -4, -9, 6 \rangle\).

Step 3: Solve for a Point on the Line of Intersection

To find a point on the line of intersection, solve the system of equations given by the planes.Choose one variable to parameterize, such as \(z\). Let \(z = t\).Substitute \(z = t\) into the two equations:1) \(4x - 2y - t = 1\)2) \(x + y + 2t = 1\).From Equation 2, express \(x\) in terms of \(y\) and \(t\):\(x = 1 - y - 2t\).Substitute \(x = 1 - y - 2t\) into Equation 1:\(4(1 - y - 2t) - 2y - t = 1\), leading to \(4 - 4y - 8t - 2y - t = 1\).This gives \(-6y - 9t = -3\).Solving for \(y\), \(y = \frac{1}{2} - \frac{3}{2}t\).Thus, \(x = 1 - (\frac{1}{2} - \frac{3}{2}t) - 2t = \frac{1}{2} - \frac{1}{2}t\).

Step 4: Write the Parametric Equations for the Line

The parametric equations for a line can be written as \(x = x_0 + at\), \(y = y_0 + bt\), \(z = z_0 + ct\) where \(\langle a, b, c \rangle\) is the direction vector.From Step 3, we obtained:\[ x = \frac{1}{2} - \frac{1}{2}t \]\[ y = \frac{1}{2} - \frac{3}{2}t \]\[ z = t \]These are the parametric equations for the line of intersection.

Key Concepts

Normal Vector

Understanding the concept of a normal vector is crucial when dealing with planes in three-dimensional space. A normal vector is a vector that is perpendicular to a surface, such as a plane. For any plane given by the equation \(ax + by + cz = d\), the normal vector can be easily identified as \(\langle a, b, c \rangle\).
This vector describes the direction orthogonal to the plane, which is vital when understanding the orientation of the plane in space.
In the given problem, we have two planes with equations:

• Plane 1: \(4x - 2y - z = 1\)
• Plane 2: \(x + y + 2z = 1\).

Thus, their normal vectors are \(\mathbf{n_1} = \langle 4, -2, -1 \rangle\) and \(\mathbf{n_2} = \langle 1, 1, 2 \rangle\), respectively. These vectors are not only essential for defining the planes but also for finding the line of intersection by using operations like the cross product.

Cross Product

The cross product is a fundamental operation in vector calculus, often used to find a vector that is perpendicular to two given vectors. This operation is essential when determining the direction of the line of intersection between two planes.
The cross product of vectors \(\mathbf{a}\) and \(\mathbf{b}\) is denoted by \(\mathbf{a} \times \mathbf{b}\) and results in a vector. Using the determinants, it can be expressed as: \[\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \4 & -2 & -1 \1 & 1 & 2\end{vmatrix}\] In this exercise, to find the direction vector of the line of intersection of the two planes, we take the cross product of their normal vectors: \(\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2}\).
Calculating this, we get: \[\mathbf{d} = \langle -4, -9, 6 \rangle\]. This vector \(\mathbf{d}\) indicates the direction in which both planes intersect.

System of Equations

A system of equations involves multiple equations that are solved together to find a common set of solutions for their variables. In the context of the problem, we need to solve a system of linear equations to find a point that lies on the line where the two planes intersect.
The given equations are: 1) \(4x - 2y - z = 1\) 2) \(x + y + 2z = 1\).To solve this system, we typically choose one variable to express in terms of the others, creating a parameter. Here, we let \(z = t\). Then, both equations can be manipulated to express \(x\) and \(y\) in terms of \(t\).
This process involves substitution and solving for the remaining variables. Eventually, the solutions are derived as:

• \(x = \frac{1}{2} - \frac{1}{2}t\)
• \(y = \frac{1}{2} - \frac{3}{2}t\)
• \(z = t\)

These forms are integral to writing the parametric equations of the line.

Line of Intersection

When two planes intersect in space, their intersection forms a line. To characterize this line, we need parametric equations. These equations represent the line in terms of a parameter, providing a way to describe the position of any point along the line.
The parametric equations of a line are formulated by combining a point on the line with its direction vector. In the exercise, we found the direction vector from the cross product of the normal vectors, resulting in \(\langle -4, -9, 6 \rangle \).
After finding a specific point on this line through solving the system of equations, we can write the parametric form by utilizing the equations:

• \(x = \frac{1}{2} - \frac{1}{2}t\)
• \(y = \frac{1}{2} - \frac{3}{2}t\)
• \(z = t\)

Here, \(t\) serves as the parameter. As \(t\) changes, it allows us to trace the line through space, effectively describing every possible point of intersection between the planes.