Problem 16 - 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 $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}} $$ $$ (3,2,1), x-y+2 z=4 $$

Step-by-Step Solution

Answer

The distance between the point and the plane is \(D= \frac{ \sqrt{6} } {6} \).

Step 1: Identify the values in the exercise

Firstly, identify the values given in the exercise. The equation of the plane is given as \(x-y+2z=4\), which can be rewritten as \(1x -1y +2z -4 = 0\). From this, \(a=1, b=-1, c=2, d=-4\) for the plane's equation. The point is given as \((3,2,1)\), so it follows that the coordinates of the point are \(x_{0}=3, y_{0}=2, z_{0}=1\).

Step 2: Substitute the values into the distance formula

Plug the values obtained from step 1 into the distance formula \(D=\frac{ |a x_{0}+b y_{0}+c z_{0}+d |} { \sqrt {a^{2}+b^{2}+c^{2}}} \). This gives \(D=\frac{|1*3 -1*2 +2*1 -4|} { \sqrt {(1^2 + -1^2 + 2^2)}}\). After calculation, it results in \(D=\frac{1} { \sqrt {6}}\).

Step 3: Simplify the result

Lastly, simplify this result as much as possible. \(D=\frac{1} { \sqrt {6}}\) can be simplified to \(D= \frac{ \sqrt{6} } {6} \).

Key Concepts

Understanding the Plane Equation

In 3D geometry, a plane is a flat, two-dimensional surface that extends infinitely in all directions. It's defined mathematically using a plane equation, which can be expressed in the form \[ ax + by + cz + d = 0 \]Here, \(a\), \(b\), and \(c\) are coefficients that determine the orientation of the plane in 3D space, while \(d\) influences its position relative to the origin.

• The coefficients \(a\), \(b\), \(c\) represent the components of a normal vector to the plane. This vector is perpendicular to the plane.
• The constant \(d\) can be thought of as a shift from the origin, influencing where the plane is located.

Understanding the plane equation is essential because it allows us to describe the position and orientation of planes in 3D space, which is crucial for calculating distances to points, lines, and other planes.

Applying the Distance Formula

The distance formula for calculating the shortest distance between a point and a plane in 3D space is a beautiful mathematical tool. Given a point \((x_0, y_0, z_0)\) and a plane with equation \(ax + by + cz + d = 0\), the distance \(D\) from the point to the plane is calculated using:\[ D = \frac{ |ax_0 + by_0 + cz_0 + d| }{ \sqrt{a^2 + b^2 + c^2} } \]The numerator of the formula represents the absolute value of the plane equation when the point is substituted into it. This gives us an unscaled measure of distance.

• The absolute value ensures the distance is always a non-negative number.
• The denominator \(\sqrt{a^2 + b^2 + c^2}\) scales the result appropriately based on the plane's orientation normal vector.

By dividing by the normal vector's magnitude, we ensure the distance is the shortest, or perpendicular, distance, making this formula very practical in applications like physics and engineering.

Exploring 3D Geometry

3D geometry is a fascinating area of mathematics that deals with space that has three dimensions: length, width, and height. This concept is central to understanding how we model the real world mathematically.In 3D geometry:

• Points are defined by three coordinates \((x, y, z)\) corresponding to their position along the x, y, and z axes.
• Lines are defined by two points or a point and a direction vector.
• Planes are defined using plane equations, involving coefficients representing normal vectors.

This space is used to solve real-world problems, from computer graphics and architecture to physical simulations. For many students, developing a solid intuition of geometry in three dimensions is the key to advanced problem-solving and mathematical analysis.