Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the point on the plane \(2 x+4 y+3 z=12\) that is closest to the origin. What is the minimum distance?

Step-by-Step Solution

Answer

The minimum distance from the origin to the plane is 4 units.

Step 1: Understanding the Problem

We are given a plane with the equation \(2x + 4y + 3z = 12\) and need to find the point on this plane closest to the origin \((0, 0, 0)\). The minimum distance from a point on a plane to the origin is along the perpendicular (normal line) to the plane.

Step 2: Finding the Plane's Normal Vector

The normal vector \(\mathbf{n}\) of the plane \(2x + 4y + 3z = 12\) is derived from its coefficients: \(\mathbf{n} = \langle 2, 4, 3 \rangle\). This vector is perpendicular to the plane.

Step 3: Parameterize the Normal Line

Let's parameterize the line with the direction given by the normal vector, passing through any point on the plane. For simplicity, use any easy point. When \(x=0, y=0\), then \(3z=12\Rightarrow z=4\), so a point is \((0, 0, 4)\). The line is \(\mathbf{r}(t) = (0,0,4) + t\langle 2, 4, 3 \rangle = (2t, 4t, 4+3t)\).

Step 4: Find Intersection of the Line and Plane

Substitute \(x = 2t\), \(y = 4t\), \(z = 4 + 3t\) into the plane equation \(2x + 4y + 3z = 12\):\[ 2(2t) + 4(4t) + 3(4 + 3t) = 12 \]\[ 4t + 16t + 12 + 9t = 12 \]\[ 29t + 12 = 12 \]\[ 29t = 0 \Rightarrow t = 0 \].

Step 5: Compute the Point Closest to the Origin

Plane and line intersect when \(t = 0\). Substituting \(t = 0\) into \((2t, 4t, 4 + 3t)\) results in the point \((0, 0, 4)\). This point lies on both the plane and the normal line, indicating it is closest to the origin.

Step 6: Calculate Minimum Distance

With the point \((0, 0, 4)\) found, calculate the distance from this point to the origin using the distance formula:\[ d = \sqrt{(0-0)^2 + (0-0)^2 + (4-0)^2} = \sqrt{16} = 4 \].

Key Concepts

Understanding the Plane Equation

A plane equation is a mathematical expression that defines a flat surface in three-dimensional space. In the example of the exercise, we have the plane equation given by:

• \(2x + 4y + 3z = 12\)

This equation tells us how the variables \(x\), \(y\), and \(z\) relate to create a particular plane.
A plane in 3D is characterized by its particular balance of these variables.
If you change any of these values, you push and twist that flat surface around space.
When approaching problems about planes, it's crucial to first simplify and identify points or lines that make calculations easy.
In this task, the point on the plane was found based on assumptions about simplicity, such as setting \(x=0\) and \(y=0\) to find a feasible \(z\). It anchors us without altering the essence of the problem.
This will help guide calculations and locate positions within the intended surface efficiently.

Significance of the Normal Vector

The normal vector is integral to understanding each plane's orientation in space. For any plane equation of the form \(ax + by + cz = d\), the coefficients \(a\), \(b\), and \(c\) form the components of the normal vector:

• \(\mathbf{n} = (a, b, c)\)

In this exercise, the plane equation \(2x + 4y + 3z = 12\) provides us with the normal vector \(\mathbf{n} = \langle 2, 4, 3 \rangle\).

• This vector is essential because it is always perpendicular, or at a right angle, to the plane itself.
• In fact, any line landing orthogonally from the plane uses this vector as its directional guide.

So, when finding the closest point on the plane to the origin, the path taken is collinear with this normal vector.
Utilizing the normal vector helps reduce complications associated with varying coordinate positions on the plane, providing a path where distance calculations to the origin are most straightforward.

Parameterization of Lines through Planes

Parameterization is a method of expressing a line in terms of a single variable, typically \(t\), which acts as a time-like path indicator that guides through points along the line.
This process is hugely beneficial when handling vectors, like our plane's normal vector, to establish a directional line through the plane.

• For the plane represented by \(2x + 4y + 3z = 12\), and utilizing the normal vector \(\mathbf{n} = \langle 2, 4, 3\rangle\), our line is defined as:\(\mathbf{r}(t) = (0,0,4) + t\langle 2, 4, 3\rangle = (2t, 4t, 4 + 3t)\)

With parameterization, you create a route cutting through the plane, where each \(t\) value matches a new point along this line.
It serves as a formulaic recipe for generating every spatial coordinate along the intended path through the given plane.
Mastering parameterization considerably enhances your capability to navigate multidimensional spaces.

Calculating the Minimum Distance from the Origin

To find the minimum distance from a plane to the origin, we utilize the parameterized line that acts along the normal vector.
We've already worked through setting this parameterized line:

• \(\mathbf{r}(t) = (2t, 4t, 4 + 3t)\)

By solving where this line intersects the plane equation, we solve for \(t\) and find the precise intersection point closest to \((0, 0, 0)\).
In this case, as the solution shows, when \(t = 0\), the path meets the plane at

• \((0, 0, 4)\)

Using the distance formula

• \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\)

with \((x_1, y_1, z_1)=(0,0,0)\) and our resulting point \((x_2, y_2, z_2)=(0,0,4)\), the minimum distance becomes \(\sqrt{16}\), or 4 units solidly.
Understanding the neat mathematics behind this process lessens intimidation when faced with similar geometry cases and supports forecasting clean, logical outcomes.