/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Find the distance from the point... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 11

Find the distance from the point (2,1,-2) to the plane $$6(x-1)+2(y-3)+3(z+4)=0$$

Short Answer

Expert verified
The distance from the point \((2,1,-2)\) to the plane \(6(x-1)+2(y-3)+3(z+4)=0\) is \(\boxed{\frac{8}{7}}\).

Step by step solution

01

Convert the given plane equation into standard form

The given plane equation is \(6(x-1)+2(y-3)+3(z+4)=0\). To convert it to the standard form, we can expand the equation and simplify it: \begin{align*} 6x - 6 + 2y - 6 + 3z + 12 &= 0 \\ 6x + 2y + 3z + 0 &= 0 \end{align*} Now, the equation is in the standard form: \(Ax + By + Cz + D = 0\), with \(A = 6\), \(B = 2\), \(C = 3\), and \(D = 0\).
02

Apply the distance formula

Now that we have the plane equation in the standard form, let's apply the distance formula from the analysis. With our point \(P=(2,1,-2)\), we have \(x_1 = 2\), \(y_1 = 1\), and \(z_1 = -2\). We also have the values of \(A\), \(B\), \(C\), and \(D\) from the previous step. \[d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}\] Plug in the values and calculate the distance: \begin{align*} d &= \frac{|6(2) + 2(1) + 3(-2) + 0|}{\sqrt{6^2 + 2^2 + 3^2}} \\ &= \frac{|12 + 2 - 6|}{\sqrt{36 + 4 + 9}} \\ &= \frac{8}{\sqrt{49}} \\ d &= \frac{8}{7} \end{align*} The distance from the point \((2,1,-2)\) to the plane \(6(x-1)+2(y-3)+3(z+4)=0\) is \(\boxed{\frac{8}{7}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Plane Equation Standard Form
Understanding the standard form of a plane equation is crucial in linear algebra, especially when attempting to solve problems involving distances and positions in three-dimensional space. A plane in three-dimensional space can be defined by the equation in standard form:
\[ Ax + By + Cz + D = 0 \]
Each variable represents a coordinate axis, with A, B, and C being the coefficients that determine the orientation of the plane, and D is a constant that displaces the plane from the origin.

For the exercise under consideration, we start with the non-standard form:
\[ 6(x-1) + 2(y-3) + 3(z+4) = 0 \]
To move to the standard form, we expand and rearrange terms, result in
\[ 6x + 2y + 3z = 0 \]
This is a cleaner expression where A=6, B=2, and C=3 while D=0. It clearly defines a plane in 3D space and is essential for calculating distances from points to the plane.
Distance Formula
The distance formula is an elegant method derived from the Pythagorean theorem that allows us to calculate the shortest distance between a point and a plane. For a given point \( P(x_1, y_1, z_1) \) and a plane in standard form \( Ax + By + Cz + D = 0 \), the distance \( d \) can be found using the formula:
\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
This formula encapsulates a vector projection concept, where the numerator represents the absolute magnitude of the vector from the point to any point on the plane, and the denominator normalizes this magnitude by the length of the plane's normal vector. By applying this formula to the given values in our exercise, we sequentially substitute the point coordinates as well as the A, B, C, and D coefficients, eventually finding the precise shortest distance.
Linear Algebra Applications
Linear algebra provides us with a set of tools and methods to solve various types of problems in mathematics and other fields such as physics, engineering, computer science, and economics. It's applied when dealing with lines, planes, and vectors in both two and three dimensions.

In the exercise provided, we use linear algebra concepts to transform a geometric problem into an algebraic one. This translation from geometric language (points and planes in space) to algebraic expressions (equations and formulas) is the hallmark of linear algebra applications. Moreover, the way we use the distance formula to find the shortest path from a point to a plane is an excellent example of using linear algebra in solving real-world optimization problems 鈥� in this case, finding the minimum distance which could be applied in fields like computer graphics, navigation, or structural design.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Give an example of a nonzero vector \(\mathbf{x} \in \mathbb{R}^{2}\) for which $$\|\mathbf{x}\|_{\infty}=\|\mathbf{x}\|_{2}=\|\mathbf{x}\|_{1}$$

Let \(S\) be a subspace of an inner product space \(V\) Let \(\left\\{\mathbf{x}_{1}, \ldots, \mathbf{x}_{n}\right\\}\) be an orthogonal basis for \(S\) and let \(\mathbf{x} \in V .\) Show that the best least squares approximation to \(\mathbf{x}\) by elements of \(S\) is given by $$\mathbf{p}=\sum_{i=1}^{n} \frac{\left\langle\mathbf{x}, \mathbf{x}_{i}\right\rangle}{\left\langle\mathbf{x}_{i}, \mathbf{x}_{i}\right\rangle} \mathbf{x}_{i}$$

Let \(A\) be an \(m \times n\) matrix of rank \(n\) and let \(P=\) \(A\left(A^{T} A\right)^{-1} A^{T}\) (a) Show that \(P \mathbf{b}=\mathbf{b}\) for every \(\mathbf{b} \in R(A)\). Explain this property in terms of projections. (b) If \(\mathbf{b} \in R(A)^{\perp},\) show that \(P \mathbf{b}=\mathbf{0}\) (c) Give a geometric illustration of parts (a) and (b) if \(R(A)\) is a plane through the origin in \(\mathbb{R}^{3}\)

Use mathematical induction to show that if \(Q \in\) \(\mathbb{R}^{n \times n}\) is both upper triangular and orthogonal, then \(\mathbf{q}_{j}=\pm \mathbf{e}_{j}, j=1, \ldots, n\)

Let \(x_{1}, x_{2}, \ldots, x_{n}\) be distinct points in the interval [-1,1] and let $$A_{i}=\int_{-1}^{1} L_{i}(x) d x, \quad i=1, \ldots, n$$ where the \(L_{i}\) 's are the Lagrange functions for the points \(x_{1}, x_{2}, \ldots, x_{n}\) (a) Explain why the quadrature formula \\[ \int_{-1}^{1} f(x) d x=A_{1} f\left(x_{1}\right)+A_{2} f\left(x_{2}\right)+\cdots+A_{n} f\left(x_{n}\right) \\] will yield the exact value of the integral whenever \(f(x)\) is a polynomial of degree less than \(n\) (b) Apply the quadrature formula to a polynomial of degree 0 and show that \\[ A_{1}+A_{2}+\cdots+A_{n}=2 \\]
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.