Problem 14 - Calculus an Applied Approach

Question

Find the distance between the point and the plane (see figure). The distance \(D\) between a point \(\left(x_{0}, y_{0}, z_{0}\right)\) and the plane \(a x+b y+c z+d=0\) is $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}} $$ $$ (0,0,0), 8 x-4 y+z=8 $$

Step-by-Step Solution

Answer

The distance between the given point and the plane is \(D = \frac{8}{9}\).

Step 1: Identify the variables in the equation of the plane

From the equation \(8x - 4y + z = 8\), we identify \(a = 8, b = -4, c = 1\) and \(d = -8\). These represent the coefficients in the given plane equation.

Step 2: Identify the coordinates of the point

Given the point \((0,0,0)\), hence \(x_{0} = 0, y_{0} = 0, z_{0} = 0\). These specify the location of the point in three dimensions.

Step 3: Substitute the variables into the formula

Substitute the identified a, b, c, d and \(x_{0}, y_{0}, z_{0}\) into the formula, we get \(D = \frac{\left|8*0 - 4*0 + 1*0 - 8\right|}{\sqrt{8^{2} + (-4)^{2} + 1^{2}}}\).

Step 4: Simplify the equation

Simplify the numerator and the denominator separately, we get \(D = \frac{8}{\sqrt{64 + 16 + 1}}\).

Step 5: Final Calculation

After simplification, we have \(D = \frac{8}{\sqrt{81}} = \frac{8}{9}\).

Key Concepts

Point to Plane Distance Formula

The point to plane distance formula allows us to calculate the shortest distance between any given point in three-dimensional space and a plane. It has significant implications in fields such as geometry, aviation, and computer graphics.

The formula itself is relatively straightforward. For a point with coordinates \( (x_0, y_0, z_0) \) and a plane described by the equation \( ax + by + cz + d = 0 \), the distance \( D \) is given by:
\[ D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} \]
To apply this, we must identify the coefficients \( a \), \( b \), and \( c \) from the plane's equation, and the coordinates of the point. With these values known, substitution into the formula will yield the desired distance.

Equations of a Plane

Understanding the equations of a plane is fundamental to solving problems in three-dimensional geometry. A plane can be defined using the equation \( ax + by + cz + d = 0 \), where \( a \), \( b \), and \( c \) are the components of the normal vector to the plane, and \( d \) is the scalar distance from the origin to the plane along the normal vector.

In three-dimensional space, any plane can be uniquely determined by this normal vector and the scalar \( d \). The normal vector gives us the orientation of the plane, while \( d \) specifies its position relative to the origin. When given a plane's equation, we can analyze the properties of the plane, such as its steepness and orientation, and calculate distances to points.

Coordinate Geometry in Three Dimensions

Coordinate geometry in three dimensions, also known as 3D analytic geometry, is concerned with the study of points, lines, and planes in space. Points are described by their coordinates \( (x, y, z) \) and are the fundamental building blocks of this branch of geometry.

By understanding the relationships between points, we can calculate distances, angles, and the equations of lines and planes. This field extends many concepts from two-dimensional coordinate geometry to three dimensions, allowing for more complex applications and analyses. The systematic approach to solving geometric problems in 3D involves the use of the Cartesian coordinate system to represent spatial relationships mathematically.

Vector Calculus

Vector calculus is an essential part of mathematics that deals with the differentiation and integration of vector fields. It has wide-reaching applications, including physics, engineering, and computer science, especially in the study of motion, forces, and fields.

Vector calculus is typically used in the context of three-dimensional space, where vectors represent quantities with both magnitude and direction. In the point to plane distance problem, vector calculus concepts such as dot product and vector magnitude play a crucial role in deriving and understanding the distance formula.
The use of vectors simplifies many calculations involving points and planes since they provide a concise mathematical method for describing and manipulating these entities in space.