Problem 43 - University Calculus: Early Transcendentals

Question

find the distance from the point to the plane. $$(0,-1,0), \quad 2 x+y+2 z=4$$

Step-by-Step Solution

Answer

The distance is \( \frac{5}{3} \).

Step 1: Understand the problem

You need to find the distance from the point \((0, -1, 0)\) to the plane described by the equation \(2x + y + 2z = 4\). This is a problem of finding the perpendicular distance from a point to a plane in 3D geometry.

Step 2: Recall the distance formula

The distance \(d\) from a point \((x_0, y_0, z_0)\) to the plane \(Ax + By + Cz + D = 0\) is given by the formula:\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]For our plane, we have \(A=2\), \(B=1\), \(C=2\), and \(D=-4\) (rearranging the plane equation to standard form). For our point, \((x_0, y_0, z_0) = (0, -1, 0)\).

Step 3: Substitute into the distance formula

Substitute the values into the formula:\[ d = \frac{|2(0) + 1(-1) + 2(0) - 4|}{\sqrt{2^2 + 1^2 + 2^2}} = \frac{|-1 - 4|}{\sqrt{4 + 1 + 4}} = \frac{|-5|}{\sqrt{9}} \]

Step 4: Simplify the expression

Calculate the absolute value and the square root:\[ d = \frac{5}{3} \]Thus, the distance from the point to the plane is \( \frac{5}{3} \).

Key Concepts

Perpendicular Distance

When determining the distance from a point to a plane in 3D geometry, it's essential to measure this distance perpendicularly. The shortest path from the point to the plane always lies along the perpendicular. Imagine a line segment that connects the point directly to the plane without veering off to any side. This is the perpendicular distance. It isn’t just about measuring any random line from the point to the plane, but that shortest and most direct line. This concept of perpendicularity ensures accuracy in calculating the true minimum distance.

3D Geometry

3D geometry focuses on the spatial relationships and properties of three-dimensional figures. It involves points, lines, planes, and their interactions in space. In our exercise, we are dealing with a point (0, -1, 0) and a plane represented by the equation 2x + y + 2z = 4. Imagine placing the point in a room and a sheet representing the plane. We want to find how close the point gets to the plane without moving parallel or tangentially with respect to it. This understanding aids in visualizing the problem properly and is crucial for correctly applying the relevant mathematical formulas and principles.

Distance Formula

The distance formula used here simplifies the computation of the perpendicular distance from a point to a plane. The formula is: \[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]This formula makes calculating the distance straightforward and less prone to error. In our example, substituting the values from the problem gives us the numerical distance. With A = 2, B = 1, C = 2, and the constant term D = -4 (adjusted from the plane equation), the process becomes a plug-and-play computation. The numerator involves plugging in the coordinates of the point into the plane equation, while the denominator considers the coefficients as part of a vector that defines the plane's orientation.

Plane Equation

A plane in 3D geometry is often described using an equation of the form Ax + By + Cz + D = 0. For our exercise, the plane equation is given as 2x + y + 2z = 4, which must be rewritten into the standard form to find perpendicular distances. By rearranging, we define the values for A, B, C, and D. These components are derived from the coefficients of the variables and the constant term. Understanding plane equations helps in forming the geometric framework for calculations and aids in visualizing the plane's orientation in space. Each coefficient signifies the contribution of its respective variable to the plane's direction and tilt.