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 from the point to the plane is \(\frac{8}{3}\).

Step 1: Understand the Problem

We are given a point \((2, 2, 3)\) and a plane equation \(2x + y + 2z = 4\). The task is to find the distance from the point to the plane, which requires using the formula for the distance from a point to a plane.

Step 2: Use the Distance Formula

The distance \(d\) from a point \((x_0, y_0, z_0)\) to a plane \(Ax + By + Cz + D = 0\) is given by:\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]First, rewrite the plane equation into the form \(Ax + By + Cz + D = 0\). We have \(2x + y + 2z - 4 = 0\), so \(A = 2\), \(B = 1\), \(C = 2\), and \(D = -4\). The point given is \((x_0, y_0, z_0) = (2, 2, 3)\).

Step 3: Substitute into the Formula

Substitute the given values into the distance formula:\[ d = \frac{|2(2) + 1(2) + 2(3) - 4|}{\sqrt{2^2 + 1^2 + 2^2}} \]Calculate the numerator and denominator separately.

Step 4: Calculate the Numerator

The numerator is:\( |2(2) + 1(2) + 2(3) - 4| = |4 + 2 + 6 - 4| = |8| = 8 \)

Step 5: Calculate the Denominator

The denominator is:\( \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \)

Step 6: Compute the Distance

Now divide the numerator by the denominator:\[ d = \frac{8}{3} \]

Step 7: Conclusion

Conclude that the distance from the point \((2, 2, 3)\) to the plane \(2x + y + 2z = 4\) is \(\frac{8}{3}\).

Key Concepts

Plane Equation

In geometry, a plane is a flat, two-dimensional surface that extends indefinitely. It's often defined by an equation in three-dimensional space. The general form of a plane equation is given by:\[ Ax + By + Cz + D = 0 \]where:

• \(A, B, C\) are coefficients that determine the plane's orientation, also acting as components of a normal vector perpendicular to the plane.
• \(D\) is a constant that shifts the plane along the direction of the normal vector.
• \(x, y, z\) are variables representing points in the plane.

To convert a plane equation into this standard form for ease of calculations, rearrange all terms to ensure the right side equals zero. For example, given a plane equation like \(2x + y + 2z = 4\), you would restate it as \(2x + y + 2z - 4 = 0\), making it easier to identify \(A = 2\), \(B = 1\), \(C = 2\), and \(D = -4\). Understanding this form is crucial to solving many geometric problems, including calculating distances from points to planes.

Distance Formula

The distance formula determines how far a specific point lies from a plane in three-dimensional space. This formula is essential in many fields, including computer graphics, physics, and engineering. The formula itself is:\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]Here's what each variable means:

• \((x_0, y_0, z_0)\) is the point from which the distance is measured.
• \(A, B, C, D\) are the coefficients from the plane equation.
• The numerator \(|Ax_0 + By_0 + Cz_0 + D|\) represents the absolute value of the plane equation evaluated at the given point.
• The denominator \(\sqrt{A^2 + B^2 + C^2}\) is the magnitude of the normal vector, ensuring the distance remains a scalar value.

This formula yields the shortest distance from the point to the plane, which is an orthogonal, or "straight-line," distance.

Orthogonal Projection

Orthogonal projection refers to the process of projecting a point directly onto a plane along a line that is perpendicular to the plane itself. The concept underlies many practical applications, such as calculating shadows, reflections, or the closest point on a slope in real-world scenarios.Consider the normal vector to the plane, \(\mathbf{n} = \langle A, B, C \rangle\), which implies the direction needed for orthogonal projection from the point. This process essentially drops a "vertical" line from the point to the plane. The result ensures that the line is the shortest possible path from point to plane.By following the line created by this normal vector, using the distance formula as part of the calculations, you ensure each measurement considers only the shortest possible, direct route — this is why the distance calculated is the orthogonal distance. This method underscores why formulas in geometry often emphasize normal vectors to achieve precision.

Vector Calculations

Vectors play a fundamental role in the geometry of planes and distance calculations. Whenever you deal with the equation of a plane or compute distances, vectors simplify and streamline the process.- **Normal Vector:** In the plane equation \(Ax + By + Cz + D = 0\), \(\mathbf{n} = \langle A, B, C \rangle\) is the normal vector. It acts as a "compass," directing all calculations related to the plane's orientation.- **Point Vectors:** The coordinates of a point \((x_0, y_0, z_0)\) can also form a vector, \(\langle x_0, y_0, z_0 \rangle\), representing its position in space.In vector calculations, operations such as finding magnitudes or dot products provide information on distances or projections. For example:

• The magnitude of a vector \(\mathbf{n}\), used in the distance formula's denominator, is calculated with: \(\|\mathbf{n}\| = \sqrt{A^2 + B^2 + C^2}\).
• The dot product operation helps determine projection values, which relate to distances and angles.

Understanding vectors simplifies complex spatial relationships, offering clear pathways to calculating distances, angles, and projections.