Problem 47 - Advanced Engineering Mathematics

Question

In Problems, find, if possible, an equation of a plane that contains the given points. $$ (0,0,0),(1,1,1),(3,2,-1) $$

Step-by-Step Solution

Answer

The equation of the plane is \(-5x + 4y - z = 0\).

Step 1: Understand the problem

We need to find the equation of a plane that contains the three given points: (0,0,0), (1,1,1), and (3,2,-1). The equation of a plane can be written as \( ax + by + cz = d \), where \((a, b, c)\) is a normal vector to the plane.

Step 2: Find two vectors on the plane

First, find vectors between the points. Let \( \mathbf{v_1} \) = (1,1,1) - (0,0,0) = (1,1,1) and \( \mathbf{v_2} \) = (3,2,-1) - (0,0,0) = (3,2,-1). These vectors lie on the plane.

Step 3: Find the normal vector using cross product

To find a normal vector to the plane, compute the cross product of the vectors \( \mathbf{v_1} \) and \( \mathbf{v_2} \):\[\mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 1 & 1 \3 & 2 & -1 \\end{vmatrix} \]Evaluate the determinant to find \( \mathbf{n} = (-3 - 2)\mathbf{i} - (1 \cdot (-1) - 1 \cdot 3)\mathbf{j} + (1 \cdot 2 - 1 \cdot 3)\mathbf{k} = (-5, 4, -1) \). This normal vector \( \mathbf{n} \) is perpendicular to the plane.

Step 4: Write the equation of the plane

Using the normal vector \( (-5, 4, -1) \) and the point (0,0,0) on the plane, write the equation of the plane in the form \( ax + by + cz = d \):\(-5x + 4y - z = 0\). Here, \( a = -5 \), \( b = 4 \), \( c = -1 \), and since the origin is part of the plane, \( d = 0 \).

Key Concepts

Normal Vector

In the context of a plane, a normal vector is a vector that is perpendicular to the plane itself. This concept is vital for defining the plane uniquely. Given three points, one way to find a normal vector is to first determine two vectors in the plane itself, derived from these points. Once these vectors are identified, the cross product operation can be employed to compute the normal vector. Consequently, the normal vector is represented in the equation of the plane as the coefficients of the variables. For instance, in the equation \(-5x + 4y - z = 0\), the normal vector is \((-5, 4, -1)\). These values demonstrate the direction that the plane is perpendicular to in three-dimensional space.

Cross Product

The cross product is a binary operation on two vectors in three-dimensional space, resulting in a third vector perpendicular to the plane containing the initial two vectors. If you have vectors \(\mathbf{v_1} = (1, 1, 1)\) and \(\mathbf{v_2} = (3, 2, -1)\), the cross product is calculated using the determinant of a matrix composed of these vectors and unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\). The calculation involves:

• Multiply each pair of components diagonally across the rows.
• Subtract the second product from the first for each unit vector component.

This method results in the normal vector \((-5, 4, -1)\) for the plane containing the original points. The cross product is essential for deriving a normal vector to the plane.

Vector Math

Vector math is a fundamental tool in three-dimensional geometry, especially when dealing with planes and their equations. By using basic vector operations such as subtraction and cross product, you can form vectors from sets of points and find the normal to a plane. For example, given points \((0,0,0)\), \((1,1,1)\), and \((3,2,-1)\), you generate vectors \(\mathbf{v_1}\) and \(\mathbf{v_2}\) as:

• \(\mathbf{v_1} = (1-0, 1-0, 1-0) = (1,1,1)\)
• \(\mathbf{v_2} = (3-0, 2-0, -1-0) = (3,2,-1)\)

These vectors can be manipulated using vector operations to explore spatial relationships, find perpendicularly, and even solve for equations of planes or lines in space.

Determinants

Determinants are a key mathematical tool used to calculate the cross product of two vectors and, by extension, the normal vector of a plane in three-dimensional geometry. When we take the cross product \(\mathbf{v_1} \times \mathbf{v_2}\), we arrange the components of these vectors with the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) into a matrix. The determinant of this matrix is computed, summarizing the method to a simpler set of arithmetic operations:

• The determinant recognizes the area spanned by the two vectors in the plane, giving a vector perpendicular to this area.
• This value is useful to get not only direction but also ensuring it adheres to the right-oriented coordinate system.

Thus, determinants simplify the process, allowing you to obtain the cross product and, ultimately, the equation of the plane.