Problem 16
Question
Use 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 from the origin to the plane is \(\frac{12}{\sqrt{14}}\).
1Step 1: Understanding the Problem
We need to find the shortest distance from the origin to the given plane equation \(x+2y+3z=12\). This can be achieved by finding a point on the plane that lies closest to the origin and calculating the distance between that point and the origin.
2Step 2: Formula for Distance from Point to Plane
The distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane with equation \(Ax + By + Cz + D = 0\) is given by the formula: \[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]In our case, the point is the origin \((0, 0, 0)\) and the plane is \(x + 2y + 3z - 12 = 0\). Thus, \(A=1, B=2, C=3, D=-12\).
3Step 3: Substitute Values in the Formula
Substitute the values: \(x_1=0, y_1=0, z_1=0, A=1, B=2, C=3, D=-12\) into the distance formula:\[ d = \frac{|1\cdot0 + 2\cdot0 + 3\cdot0 - 12|}{\sqrt{1^2 + 2^2 + 3^2}} \]
4Step 4: Calculate Numerator and Denominator
Calculate the numerator:\[ |0 + 0 + 0 - 12| = | -12 | = 12 \]Calculate the denominator:\[ \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]
5Step 5: Compute the Distance
Plug the values into the formula:\[ d = \frac{12}{\sqrt{14}} \]The expression \(\frac{12}{\sqrt{14}}\) simplifies to its exact form or can be rationalized if necessary, which we often leave in this form unless further specified.
Key Concepts
Plane EquationShortest DistanceDistance FormulaVector Calculus
Plane Equation
A plane equation is a mathematical representation of a flat surface extending infinitely in three-dimensional space. In vector calculus, the equation of a plane can be written in the standard form:
\[ Ax + By + Cz + D = 0 \] Where:
\[ x + 2y + 3z - 12 = 0 \]This way, we can see that \( A = 1 \), \( B = 2 \), \( C = 3 \), and \( D = -12 \). Understanding the plane equation is crucial as it helps in identifying the precise location and orientation of the plane in 3D space.
\[ Ax + By + Cz + D = 0 \] Where:
- \( A, B, \) and \( C \) are the coefficients that determine the orientation of the plane.
- \( D \) is the constant term which shifts the plane from the origin.
- \( (x, y, z) \) represents any point on the plane.
\[ x + 2y + 3z - 12 = 0 \]This way, we can see that \( A = 1 \), \( B = 2 \), \( C = 3 \), and \( D = -12 \). Understanding the plane equation is crucial as it helps in identifying the precise location and orientation of the plane in 3D space.
Shortest Distance
The shortest distance from a point to a plane is the minimum length needed to travel perpendicular from the point to any location on the plane. This concept is significant in various fields such as physics, engineering, and computer graphics because minimizing distance often aligns with optimizing resources or performance.
To find this distance, imagine dropping a perpendicular line from the point (for instance, the origin in 3D space) to the plane. The point where this line intersects the plane is the closest point on the plane to the given point.
For the origin \((0, 0, 0)\) and a plane described by the equation \( x + 2y + 3z = 12 \), we calculate this shortest distance by using the distance formula, which determines the minimum gap between the point and the plane.
To find this distance, imagine dropping a perpendicular line from the point (for instance, the origin in 3D space) to the plane. The point where this line intersects the plane is the closest point on the plane to the given point.
For the origin \((0, 0, 0)\) and a plane described by the equation \( x + 2y + 3z = 12 \), we calculate this shortest distance by using the distance formula, which determines the minimum gap between the point and the plane.
Distance Formula
The distance formula is a mathematical equation used to determine the shortest distance from a point to a plane. In vector calculus, it is represented as:
\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \] Where:
This formula directly provides the shortest possible distance, using mainly arithmetic operations, which makes it quick and efficient for calculations.
\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \] Where:
- \( (x_1, y_1, z_1) \) is the given point, in this case, the origin \((0, 0, 0)\).
- \( A, B, \) and \( C \) are the coefficients from the plane equation.
- \( D \) is the constant from the plane equation.
This formula directly provides the shortest possible distance, using mainly arithmetic operations, which makes it quick and efficient for calculations.
Vector Calculus
Vector calculus is a branch of mathematics used to study vector fields and the operations on them. It is instrumental in describing physical phenomena like electromagnetic fields, fluid motion, and more generally, any spatial transformations.
In our context, vector calculus provides tools for understanding and working with the geometry of planes and points in 3D space. These tools help us formulate plane equations, find directions, and compute distances efficiently.
A significant part of vector calculus involves vectors \( \mathbf{v} = \langle a, b, c \rangle \), which can represent directions from a point towards the plane along the shortest route, known as the normal vector. This vector is crucial since it aligns perpendicularly to the plane, guiding the computation of the shortest distance.
Utilizing vector calculus means you can solve complex geometric problems like finding distances, edges, and angles, using vectors and vectors operations like dot products, without relying on just coordinate geometry.
In our context, vector calculus provides tools for understanding and working with the geometry of planes and points in 3D space. These tools help us formulate plane equations, find directions, and compute distances efficiently.
A significant part of vector calculus involves vectors \( \mathbf{v} = \langle a, b, c \rangle \), which can represent directions from a point towards the plane along the shortest route, known as the normal vector. This vector is crucial since it aligns perpendicularly to the plane, guiding the computation of the shortest distance.
Utilizing vector calculus means you can solve complex geometric problems like finding distances, edges, and angles, using vectors and vectors operations like dot products, without relying on just coordinate geometry.
Other exercises in this chapter
Problem 15
Find all first partial derivatives of each function. \(F(x, y)=2 \sin x \cos y\)
View solution Problem 15
Sketch the graph of \(\bar{f}\). $$ f(x, y)=e^{-\left(x^{2}+y^{2}\right)} $$
View solution 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
Show that the surfaces \(z=x^{2} y\) and \(y=\frac{1}{4} x^{2}+\frac{3}{4}\) intersect at \((1,1,1)\) and have perpendicular tangent planes there.
View solution