Problem 41 - Thomas Calculus

Question

In Exercises \(39-44,\) find the distance from the point to the plane. $$ (0,1,1), \quad 4 y+3 z=-12 $$

Step-by-Step Solution

Answer

The distance from the point to the plane is 3.8 units.

Step 1: Understand the Formula for Point to Plane Distance

To find the distance from a point \((x_1, y_1, z_1)\) to a plane given by the equation \(ax + by + cz + d = 0\), use the formula:\[D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\]We need to identify \(a\), \(b\), \(c\), and \(d\) for the plane equation and \((x_1, y_1, z_1)\) for the point.

Step 2: Identify Plane Equation Constants

Rewrite the plane equation \(4y + 3z = -12\) in standard form: \[0x + 4y + 3z + 12 = 0\]Here, \(a = 0\), \(b = 4\), \(c = 3\), and \(d = 12\).

Step 3: Identify Point Coordinates

The point given is \((0, 1, 1)\), so \(x_1 = 0\), \(y_1 = 1\), and \(z_1 = 1\).

Step 4: Substitute into the Distance Formula

Substitute the identified values into the distance formula:\[D = \frac{|0 \cdot 0 + 4 \cdot 1 + 3 \cdot 1 + 12|}{\sqrt{0^2 + 4^2 + 3^2}}\]

Step 5: Calculate the Numerator

Calculate the expression inside the absolute value:\[4 \cdot 1 + 3 \cdot 1 + 12 = 4 + 3 + 12 = 19\]So, inside the absolute value, we have \(19\).

Step 6: Calculate the Denominator

Calculate the square root expression:\[\sqrt{0^2 + 4^2 + 3^2} = \sqrt{0 + 16 + 9} = \sqrt{25} = 5\]

Step 7: Compute the Distance

Divide the absolute value by the denominator:\[D = \frac{19}{5} = 3.8\]Therefore, the distance from the point to the plane is 3.8.

Key Concepts

Plane Equation

The equation of a plane provides a way to define a flat, two-dimensional surface in three-dimensional space. An essential step in many geometry problems, understanding a plane equation is crucial for tackling topics like the distance from a point to a plane. A plane equation in its standard form is:

• \( ax + by + cz + d = 0 \)

This equation represents a standard way to describe a plane using three variables: \( x \), \( y \), and \( z \), with constants \( a \), \( b \), \( c \), and \( d \). These constants are derived from the plane's orientation and position in space.
The problem "find the distance from the point to the plane" first requires rewriting any non-standard plane equations into this standard form. For example, the equation in the original exercise was given as \( 4y + 3z = -12 \). By rewriting it as \( 0x + 4y + 3z + 12 = 0 \), we made it conform to the general form, clarifying its coefficients: \( a = 0 \), \( b = 4 \), \( c = 3 \), and \( d = 12 \). This step lays the foundation for further calculations.

Distance Formula

The distance from a point to a plane can be calculated using a specific distance formula tailored for three-dimensional geometry. This formula is particularly useful for determining the shortest distance between any point in space (given by its coordinates) and a defined plane. The distance formula is:

• \[ D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]

Here's how it works:

• The numerator consists of substituting point coordinates \((x_1, y_1, z_1)\) into the plane equation. The sum is enclosed in absolute value to account for any negative outcomes that can arise from calculations.
• The denominator is the square root of the sum of the squares of the plane's coefficients: \( a \), \( b \), and \( c \).

In our exercise, substituting the values, we found \( | 4 \cdot 1 + 3 \cdot 1 + 12 | = 19 \). For the denominator, we calculated \( \sqrt{0^2 + 4^2 + 3^2} = 5 \). Finally, dividing the absolute value result by this square root gives the distance: \( D = \frac{19}{5} = 3.8 \). These calculations offer a step-by-step approach to measuring proximity in 3D space.

3D Geometry

Three-dimensional geometry, or 3D geometry, extends the principles of two-dimensional shapes into a world with an extra axis. It's like moving from drawing on a piece of paper to sculpturing. In 3D geometry:

• We use three axes: the x-axis, y-axis, and z-axis, to determine positions of points in space.
• Shapes in this space can look like cubes, spheres, tetrahedrons, or planes.
• Understanding the principles of 3D geometry allows individuals to analyze distances and relations between shapes and figures in a more complex spatial environment.

In the context of our plane and point distance problem:

• The plane defined in the exercise exists as an indefinite flat surface in a 3D space characterized by the equation \( 0x + 4y + 3z + 12 = 0 \).
• The point \((0, 1, 1)\) is a specific location to which we calculated the shortest perpendicularly measured distance.

This is a foundational element of 3D geometry, helping us understand spatial relationships that can't be seen in just two dimensions, and a critical skill for fields ranging from architecture to virtual modeling.