Chapter 19: Chapter 19

Problem 561

Minimize the distance from a point \(p \in R^{n}\) to the hyperplane \(<\mathrm{x}, \mathrm{a}>+\mathrm{b}=0\) where \(\mathrm{a} \in \mathrm{R}^{n}\) and \(\mathrm{b} \in \mathrm{R}\). (Assume \(\left.\mathrm{a} \neq 0 .\right)\)

Short Answer

Expert verified

The minimum distance \(D(p)\) between the point \(p\) and the hyperplane \( + b = 0\) is given by \(\frac{| - b|}{|a|}\), and the point \(x_0\) on the hyperplane closest to the point \(p\) satisfies \( + b = 0\).

See the step by step solution

Step by step solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Write down the distance formula between point and hyperplane

The formula for the distance between a point \(p\) in \(R^n\) and a hyperplane \( + b = 0\) is given by: \(D(p) = \frac{| + b|}{|a|}\).

Step 2: Find the position vector for the point on the hyperplane closest to \(p\)

To minimize the distance, we want to find a point \(x_0\) on the hyperplane that is closest to \(p\). The vector connecting \(p\) and \(x_0\) is orthogonal to the normal vector of the hyperplane, which is given by the vector \(a\). Let \(x_0\) be the point on the hyperplane closest to \(p\) with the position vector \(x_0\). Then, the vector connecting \(p\) and \(x_0\) is given by \((p - x_0)\). The fact that the vector connecting \(p\) and \(x_0\) is orthogonal to the normal vector of the hyperplane can be written as: \(<(p - x_0), a> = 0\).

Step 3: Solve for \(x_0\)

Using the equation \( + b = 0\) and substituting \(x_0\) for \(x\), we have: \( + b = 0\). Now, using the equation from step 2 and substituting the derived equation for \(x_0\), we have: \(<(p - x_0), a> = - = 0\). Adding \(b\) to both sides, we have \( - + b = b\). Since \( + b = 0\), we can substitute \(-\) for \(b\), and we have: \( + b = b\). Therefore, the point \(x_0\) on the hyperplane closest to \(p\) satisfies: \( + b = 0\).

Step 4: Calculate the minimum distance \(D(p)\) from the point \(p\) to the hyperplane

Now we can use the expression for \(x_0\) to calculate the minimum distance between \(p\) and the hyperplane. We know that: \(D(p) = \frac{| + b|}{|a|}\). Since we have \( + b = 0\), we can substitute \(0\) for \(-b\) and have: \(D(p) = \frac{| - b|}{|a|}\). This expression gives us the minimum distance between the point \(p\) and the hyperplane \( + b = 0\). To summarize, the minimum distance \(D(p)\) is given by \(\frac{| - b|}{|a|}\), and the point \(x_0\) on the hyperplane closest to the point \(p\) satisfies \( + b = 0\).

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Recommended explanations on Math Textbooks

Select your language

Chapter 19: Chapter 19

Minimize the distance from a point \(p \in R^{n}\) to the hyperplane \(<\mathrm{x}, \mathrm{a}>+\mathrm{b}=0\) where \(\mathrm{a} \in \mathrm{R}^{n}\) and \(\mathrm{b} \in \mathrm{R}\). (Assume \(\left.\mathrm{a} \neq 0 .\right)\)

Short Answer

Step by step solution

Unlock all solutions

Step 1: Write down the distance formula between point and hyperplane

Step 2: Find the position vector for the point on the hyperplane closest to \(p\)

Step 3: Solve for \(x_0\)

Step 4: Calculate the minimum distance \(D(p)\) from the point \(p\) to the hyperplane

Access millions of textbook solutions in one place

Most popular questions from this chapter

More chapters from the book ‘Advanced Calculus’

Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 14

Chapter 15

Chapter 16

Chapter 17

Chapter 18

Join over 22 million students in learning with our Vaia App

Recommended explanations on Math Textbooks

Decision Maths

Mechanics Maths

Geometry

Calculus

Pure Maths

Probability and Statistics