Problem 43 - Advanced Engineering Mathematics

Question

In Problems, find an equation of the plane that contains the given point and is perpendicular to the indicated vector. $$ \left(\frac{1}{2}, \frac{3}{4},-\frac{1}{2}\right) ; 6 \mathrm{i}+8 \mathbf{j}-4 \mathbf{k} $$

Step-by-Step Solution

Answer

The equation of the plane is \( 6x + 8y - 4z = 11 \).

Step 1: Understanding the Problem

We are tasked with finding the equation of a plane. The plane must pass through a given point \( \left(\frac{1}{2}, \frac{3}{4}, -\frac{1}{2}\right) \) and be perpendicular to the given vector \( \mathbf{n} = 6\mathbf{i} + 8\mathbf{j} - 4\mathbf{k} \). The vector \( \mathbf{n} \) is the normal vector to the plane.

Step 2: Review the Standard Plane Equation

The standard equation of a plane with a normal vector \( \mathbf{n} = (a, b, c) \) and passing through a point \( (x_0, y_0, z_0) \) is given by: \( a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \).

Step 3: Identify Components

Here, from the normal vector \( \mathbf{n} = 6\mathbf{i} + 8\mathbf{j} - 4\mathbf{k} \), we identify \( a = 6 \), \( b = 8 \), and \( c = -4 \). Also, the point \( (x_0, y_0, z_0) = \left(\frac{1}{2}, \frac{3}{4}, -\frac{1}{2}\right) \).

Step 4: Substitute Values into the Plane Equation

Substitute \( a = 6 \), \( b = 8 \), \( c = -4 \), and the coordinates \( x_0 = \frac{1}{2} \), \( y_0 = \frac{3}{4} \), \( z_0 = -\frac{1}{2} \) into the standard plane equation: \[ 6(x - \frac{1}{2}) + 8(y - \frac{3}{4}) - 4(z + \frac{1}{2}) = 0 \].

Step 5: Simplify the Equation

Expand and simplify the equation: \[ 6x - 3 + 8y - 6 - 4z - 2 = 0 \] Combine like terms:\[ 6x + 8y - 4z - 11 = 0 \].

Step 6: Final Equation Formulation

The simplified equation of the plane is now clearly presented as: \( 6x + 8y - 4z = 11 \).

Key Concepts

Vector Algebra

Vector algebra is a fundamental tool in mathematics and physics that deals with vectors and operations on them. Vectors are quantities that have both magnitude and direction. In this context, we work with vectors to describe geometric entities such as lines and planes. Vectors are also used to represent physical quantities like force and velocity.

When dealing with planes in three dimensions, vectors are essential. A plane can be defined by a point and a normal vector. Operations in vector algebra, such as addition, scalar multiplication, and dot product, are used to derive the equation of the plane.

• Addition: Combining two vectors to get a resultant vector.
• Scalar Multiplication: Multiplying a vector by a number to change its magnitude.
• Dot Product: A scalar product that gives the cosine of the angle between two vectors, useful in projecting a vector in the direction of another.

Understanding how to manipulate vectors is crucial for deriving the equation of a plane, as seen when the point and normal vector's components are used to formulate the plane equation.

Normal Vector

The normal vector is a special vector that is perpendicular to a surface. In the three-dimensional coordinate system, every plane has a normal vector that uniquely defines its inclination and position in space. It is a key component when establishing the plane's equation.

The components of the normal vector determine the coefficients of the plane equation. For example, in this exercise, the normal vector is given as \( 6\mathbf{i} + 8\mathbf{j} - 4\mathbf{k} \). Hence, the equation of the plane contains these coefficients:

• 6 is the coefficient of \( x \)
• 8 is the coefficient of \( y \)
• -4 is the coefficient of \( z \)

The normal vector is essential because it fixes the plane's orientation in space, and any vector lying on the plane will be orthogonal to this normal vector. This orthogonality condition is fundamental in deriving the plane's equation.

Geometry in Three Dimensions

Geometry in three dimensions is an exciting and complex field that extends our understanding beyond the confines of the flat plane. In three-dimensional space, concepts like points, lines, and planes become much richer and more intricate.

A plane is a flat, two-dimensional surface that extends infinitely in three-dimensional space. It can be uniquely defined by:

• A point through which it passes
• A normal vector that dictates its orientation

To find the equation of a plane, as shown in the exercise, we start by fixing a point on the plane and then find a vector perpendicular to it. The standard equation for a plane is: \( a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \). This formula incorporates the normal vector's coefficients \( (a, b, c) \) and the coordinates of the point \( (x_0, y_0, z_0) \).

Three-dimensional geometry allows us to model real-world structures and phenomena, from architectural designs to complex physical systems. Understanding these spatial concepts is vital for solving geometric problems in higher dimensions.