Problem 16
Question
u se the methods of this section to find the shortest distance from the origin to the plane \(x+2 y+3 z=12\).
Step-by-Step Solution
Verified Answer
The shortest distance is \(\frac{6\sqrt{14}}{7}\).
1Step 1: Understand the Problem
We need to find the shortest distance from the origin (0, 0, 0) to the plane described by the equation \(x + 2y + 3z = 12\). Notice that this is a plane in 3D space.
2Step 2: Find the Normal Vector of the Plane
The equation of the plane \(x + 2y + 3z = 12\) can be expressed as \( \ax + by + cz = d \\), where \( a = 1, b = 2, c = 3, d = 12 \). The normal vector \( \mathbf{n} \) to the plane is given by \( \langle a, b, c \rangle = \langle 1, 2, 3 \rangle \).
3Step 3: Use Distance Formula from a Point to a Plane
The formula for 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}} \] \Here, the point is the origin, \((0, 0, 0)\).
4Step 4: Substitute Values into the Distance Formula
Substituting the values in the formula: \(a = 1, b = 2, c = 3, x_0 = 0, y_0 = 0, z_0 = 0, d = 12\), we find: \\[ D = \frac{|1 \times 0 + 2 \times 0 + 3 \times 0 - 12|}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{|0 - 12|}{\sqrt{1 + 4 + 9}} = \frac{12}{\sqrt{14}} \]
5Step 5: Simplify the Expression
Simplify the expression for \( D \): \\[ D = \frac{12}{\sqrt{14}} \.\] To express this in an alternative form, multiply by \(\frac{\sqrt{14}}{\sqrt{14}}\): \\[ D = \frac{12\sqrt{14}}{14} = \frac{6\sqrt{14}}{7} \].
6Step 6: Final Answer
The shortest distance from the origin to the plane is \( \frac{6\sqrt{14}}{7} \).
Key Concepts
3D GeometryNormal VectorDistance FormulaPlane Equation
3D Geometry
In mathematics, 3D geometry refers to the study of spaces that have three dimensions: length, width, and height. This is the environment where objects like spheres, cubes, prisms, and planes exist.
In 3D space, every point can be defined using three coordinates—usually denoted as \(x, y, z\). The concept of the plane, which is a flat, two-dimensional surface extending infinitely in 3D space, is foundational.
When dealing with planes, their position and orientation are described by a plane equation like \(ax + by + cz = d\), where \(a, b,\) and \(c\) define the normal vector, and \(d\) determines the plane's particular placement in space. Understanding these basic elements allows us to solve various geometric problems, such as finding distances between points and planes.
In 3D space, every point can be defined using three coordinates—usually denoted as \(x, y, z\). The concept of the plane, which is a flat, two-dimensional surface extending infinitely in 3D space, is foundational.
When dealing with planes, their position and orientation are described by a plane equation like \(ax + by + cz = d\), where \(a, b,\) and \(c\) define the normal vector, and \(d\) determines the plane's particular placement in space. Understanding these basic elements allows us to solve various geometric problems, such as finding distances between points and planes.
Normal Vector
The normal vector plays a crucial role in understanding and manipulating planes in 3D geometry. It is a vector that is perpendicular to, or "normal" to the plane. For a given plane equation \(ax + by + cz = d\), the normal vector is given by \(\).
For example, for the plane given by the equation \(x + 2y + 3z = 12\), the normal vector would be \(\langle 1, 2, 3 \rangle\).
This vector not only helps in comprehending the orientation of the plane but is also essential in calculations like finding the distance from a point to the plane or establishing its angle with other planes or lines. Recognizing this relationship assists in visualizing how the plane "sits" in space.
For example, for the plane given by the equation \(x + 2y + 3z = 12\), the normal vector would be \(\langle 1, 2, 3 \rangle\).
This vector not only helps in comprehending the orientation of the plane but is also essential in calculations like finding the distance from a point to the plane or establishing its angle with other planes or lines. Recognizing this relationship assists in visualizing how the plane "sits" in space.
Distance Formula
To find the shortest distance from a point to a plane, we use a specific distance formula. This is different from the usual Euclidean distance formula applied between two points.
For a point \(P(x_0, y_0, z_0)\) and a plane \(ax + by + cz = d\), the distance \(D\) is calculated by: \[D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}\]
This formula considers the perpendicular distance from the point to the plane, utilizing the coefficients \(a, b,\) and \(c\) from the plane's equation—as these directly relate to the normal vector—and combining them with the point coordinates.
This expression must be evaluated carefully, as it involves both absolute values and square roots, representing the true "straightest" path from the point to the plane.
For a point \(P(x_0, y_0, z_0)\) and a plane \(ax + by + cz = d\), the distance \(D\) is calculated by: \[D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}\]
This formula considers the perpendicular distance from the point to the plane, utilizing the coefficients \(a, b,\) and \(c\) from the plane's equation—as these directly relate to the normal vector—and combining them with the point coordinates.
This expression must be evaluated carefully, as it involves both absolute values and square roots, representing the true "straightest" path from the point to the plane.
Plane Equation
The plane equation in 3D geometry is used to describe a flat, two-dimensional surface in three-dimensional space. It is typically written in the form \(ax + by + cz = d\). In this equation:
This equation's versatility allows for complex spatial interpretations and calculations, aiding tasks like finding intersections with other planes or lines, or measuring distances.
- The coefficients \(a, b,\) and \(c\) correspond to the components of the normal vector, which dictates the plane's orientation in space.
- The constant \(d\) positions the plane relative to the origin.
This equation's versatility allows for complex spatial interpretations and calculations, aiding tasks like finding intersections with other planes or lines, or measuring distances.
Other exercises in this chapter
Problem 16
If \(w=x^{2} y+z^{2}, x=\rho \cos \theta \sin \phi, y=\rho \sin \theta \sin \phi\), and \(z=\rho \cos \phi\), find $$ \left.\frac{\partial w}{\partial \theta}\r
View solution Problem 16
In Problems 7-16, sketch the graph of \(f\). $$ f(x, y)=x^{2} / y, y>0 $$
View solution Problem 16
\(\lim _{(x, y) \rightarrow(0,0)} \frac{x y^{2}}{x^{2}+y^{4}}\)
View solution Problem 16
In Problems 1-16, find all first partial derivatives of each function. \(f(r, \theta)=3 r^{3} \cos 2 \theta\)
View solution