Problem 56 - Advanced Engineering Mathematics

Question

In Problems, find an equation of the plane that satisfies the given conditions. Contains the lines \(\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-5}{6}\) \(\mathbf{r}=\langle 1,-1,5\rangle+t\langle 1,1,-3\rangle\)

Step-by-Step Solution

Answer

The equation of the plane is \(-x + 4y + z = 0\).

Step 1: Understand the Given Lines

The first line is presented in symmetric form: \( \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-5}{6} \). This can be rewritten as a parametric form: \( x = 1 + 2t, y = -1 - t, z = 5 + 6t \). For the second line \( \mathbf{r} = \langle 1,-1,5 \rangle + t\langle 1,1,-3 \rangle \), the parametric equations are: \( x = 1 + t, y = -1 + t, z = 5 - 3t \).

Step 2: Identify Points and Direction Vectors

From the first line, the direction vector is \( \langle 2, -1, 6 \rangle \) and a point on the line is \( (1, -1, 5) \). From the second line, the direction vector is \( \langle 1, 1, -3 \rangle \) and a point on the line is also \( (1, -1, 5) \).

Step 3: Compute Normal Vector of the Plane

To find the equation of the plane containing both lines, we need a normal vector. This can be obtained by the cross product of the direction vectors \( \langle 2, -1, 6 \rangle \) and \( \langle 1, 1, -3 \rangle \).

Step 4: Calculate Cross Product

The cross product \( \langle 2, -1, 6 \rangle \times \langle 1, 1, -3 \rangle \) is calculated as follows:- \(((-1)(-3)) - (6)(1) = 3 - 6 = -3\)- \((6)(1) - (2)(-3) = 6 + 6 = 12\)- \((2)(1) - ((-1)(1)) = 2 + 1 = 3\)Thus, the normal vector is \( \langle -3, 12, 3 \rangle \).

Step 5: Form the Plane Equation

With the normal vector \( \langle -3, 12, 3 \rangle \) and a point on the plane \( (1, -1, 5) \), the equation of the plane can be constructed: \(-3(x - 1) + 12(y + 1) + 3(z - 5) = 0\).

Step 6: Simplify the Plane Equation

Expanding the equation: - \(-3x + 3 + 12y + 12 + 3z - 15 = 0\)Simplifying gives: \(-3x + 12y + 3z = 0\). Divide through by 3 to simplify further: \( -x + 4y + z = 0 \).

Key Concepts

Cross Product

The cross product is an essential operation in the field of vector algebra and is particularly useful in finding a vector perpendicular to two given vectors. In simpler terms, it helps in deriving a "new" vector normal to a plane defined by two direction vectors. This operation is especially useful when trying to find the equation of a plane.

In the given exercise, to find a normal vector to the plane containing two lines, we performed a cross product of the direction vectors from the lines.

• First Direction Vector: \( \langle 2, -1, 6 \rangle \)
• Second Direction Vector: \( \langle 1, 1, -3 \rangle \)

To calculate the cross product, the determinant approach is often used, involving calculating the components systematically:

• \( (-1)(-3) - (6)(1) = -3 \)
• \( (6)(1) - (2)(-3) = 12 \)
• \( (2)(1) - (-1)(1) = 3 \)

Thus, the resulting vector \( \langle -3, 12, 3 \rangle \) serves as the normal vector for the plane.

Normal Vector

Understanding the normal vector is crucial when determining the equation of a plane. The normal vector is a vector that is perpendicular to the plane, and it influences the plane's orientation in 3D space. Essentially, it is an important descriptor of the plane.

To find the normal vector, as in the exercise, one common method is to utilize the cross product of two non-parallel direction vectors that lie on the plane. From the cross product calculated earlier:

• Normal Vector: \( \langle -3, 12, 3 \rangle \)

The coordinates of the normal vector correspond to the coefficients in the plane equation. This vector ensures that the plane is defined accurately in three-dimensional space, providing a perpendicular 'anchor' for its orientation.

Parametric Equations

Parametric equations are a way to express a geometric figure, like a line, using parameters. Unlike symmetric form, which compactly represents a line, parametric equations allow each coordinate to be expressed individually in terms of one or more parameters.

In the context of the exercise, transforming the symmetric form into parametric equations is essential in finding the direction vector of a line. For example, from the provided symmetric form line \( \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-5}{6} \), we can derive its parametric equations:

• \( x = 1 + 2t \)
• \( y = -1 - t \)
• \( z = 5 + 6t \)

These equations portray not only one specific point on the line but rather an infinite series of points as the parameter \( t \) varies. This form is particularly useful because it simplifies the process of deriving essential elements like points or direction vectors, which are needed in vector operations.

Symmetric Form

The symmetric form of a line is a concise representation important in both algebra and geometry. It describes the line in terms of its relationship to its direction ratios, which is usually very compact and efficient.

This format can be seen in the exercise as \( \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-5}{6} \). Each part of the equation signifies the line's direction ratios (or components of the direction vector):

• Direction for \( x \): 2
• Direction for \( y \): -1
• Direction for \( z \): 6

To extract concrete data, such as parametric equations or direction vectors, one may need to break the symmetric form down into individual parametric components. However, the symmetric form remains a powerful way to visually encode a line's directions and structure, especially when determining intersections or constructing planes.