Problem 42 - Thomas Calculus

Question

In Exercises \(39-44,\) find the distance from the point to the plane. $$ (2,2,3), \quad 2 x+y+2 z=4 $$

Step-by-Step Solution

Answer

The distance is \(\frac{8}{3}\).

Step 1: Identify the Components

The point given is \((2, 2, 3)\) and the plane equation is \(2x + y + 2z = 4\). The coefficients \(a = 2\), \(b = 1\), \(c = 2\), and \(d = 4\) are extracted from the plane equation.

Step 2: Use the Distance Formula

The distance \(D\) from a point \((x_0, y_0, z_0)\) to a plane \(ax + by + cz = d\) is given by: \[ D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}} \] Substitute \(x_0 = 2\), \(y_0 = 2\), \(z_0 = 3\), \(a = 2\), \(b = 1\), \(c = 2\), \(d = 4\).

Step 3: Calculate the Numerator

Substitute the point and the coefficients into the numerator of the formula: \[ |2(2) + 1(2) + 2(3) - 4| = |4 + 2 + 6 - 4| = |8| = 8 \]

Step 4: Calculate the Denominator

Compute the denominator using the coefficients: \[ \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

Step 5: Compute the Final Distance

Combine the results from Steps 3 and 4 in the distance formula: \[ D = \frac{8}{3} \] Thus, the distance from the point to the plane is \(\frac{8}{3}\).

Key Concepts

Plane Equation

A plane in three-dimensional space can be visualized as a flat, infinitely expansive surface. To mathematically describe this surface, we use a plane equation, which typically takes the form of \( ax + by + cz = d \). Here, \( a \), \( b \), and \( c \) are the coefficients that define the plane's orientation in space, while \( d \) represents the plane's offset from the origin. For instance, in the given problem, the plane equation is \( 2x + y + 2z = 4 \). This tells us:

• Parallel alignment with the standard 3D axes is determined by the coefficients \( a = 2 \), \( b = 1 \), and \( c = 2 \).
• The value \( d = 4 \) signifies that the plane is shifted four units from the origin along the normal vector defined by these coefficients.

Understanding the equation helps visualize how a plane cuts through three-dimensional space and defines how to measure perpendicular distance from a point to this plane.

Distance Formula

To find the shortest distance from a point to a plane, we use a specialized distance formula. This formula is essential in various fields such as computational geometry and physics. The formula to calculate the distance \( D \) from a point \((x_0, y_0, z_0)\) to a plane \( ax + by + cz = d \) is:

• \[ D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}} \]

The numerator, \( |ax_0 + by_0 + cz_0 - d| \), represents the absolute difference between the weighted sum of coordinates and the plane's offset. The denominator, \( \sqrt{a^2 + b^2 + c^2} \), ensures the calculation is with respect to the plane's normal vector's length, providing the true perpendicular distance. By plugging in the values of the point \( (2, 2, 3) \) and the given plane equation coefficients, we simplify and solve for \( D \). This formula underscores the importance of understanding plane coefficients and their geometric implications.

Numerator and Denominator Calculations

Calculating the distance from a point to a plane involves obtaining values for both the numerator and the denominator in the distance formula. Let's break down each part to understand better:

Numerator Calculation

The numerator of the distance formula, \( |ax_0 + by_0 + cz_0 - d| \), calculates the direct perpendicular offset of a point from the plane. Substituting our point \((2, 2, 3)\) and the plane coefficients \( a = 2 \), \( b = 1 \), \( c = 2 \), \( d = 4 \), we calculate:

• Substituting in: \( 2(2) + 1(2) + 2(3) - 4 \)
• Simplified as: \( 4 + 2 + 6 - 4 = 8 \)
• Result is \( |8| = 8 \)

Denominator Calculation

For the denominator, \( \sqrt{a^2 + b^2 + c^2} \), this quantifies the length of the plane's normal vector. It is important for normalization to ensure the distance measurement is accurate:

• Substituting in: \( \sqrt{2^2 + 1^2 + 2^2} \)
• Simplified calculation: \( \sqrt{4 + 1 + 4} = \sqrt{9} \)
• Result is \( 3 \)

By dividing the numerator by the denominator, \( \frac{8}{3} \), we find the precise distance from the point to the plane, ensuring correct interpretation of these calculations in three-dimensional space.