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Chapter 13: Problem 17

Find the point on the plane \(x+2 y+3 z=13\) closest to the point (1,1,1)

Short Answer

Expert verified
The point is (2, 3, 0).

Step by step solution

01

Understand the Problem

We want to find a point on the plane defined by the equation \(x+2y+3z=13\) that is closest to the given point \((1,1,1)\). This involves minimizing the distance between a point \((x, y, z)\) on the plane and \((1, 1, 1)\).
02

Set up the Distance Function

The distance \(d\) between a point \((x, y, z)\) and \((1, 1, 1)\) is given by the formula: \[ d = \sqrt{(x - 1)^2 + (y - 1)^2 + (z - 1)^2} \] Since minimizing \(d\) is equivalent to minimizing \(d^2\), we focus on the function \[ f(x, y, z) = (x - 1)^2 + (y - 1)^2 + (z - 1)^2 \].

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Key Concepts

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

Distance Minimization
When dealing with problems involving geometry and optimization, **distance minimization** is a crucial concept. It often involves finding the smallest distance between points and lines, planes, or surfaces. In our scenario, the problem is to minimize the distance between a point on a specific plane and a fixed point in space. Considering that the traditional distance formula is derived from the Pythagorean theorem, we get:
  • The distance between two points \( (x, y, z) \) and \( (x_0, y_0, z_0) \) is \( d = \sqrt{(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2} \).
By squaring the distance formula, we effectively transform the problem into a simpler form of minimizing the squared distance:
  • Minimize \( f(x, y, z) = (x-1)^2 + (y-1)^2 + (z-1)^2 \).
This method avoids dealing with the square root function, simplifying calculations while ensuring we find the closest point.
Plane Equation
The **plane equation** is fundamental when examining linear surfaces in three-dimensional space. A basic plane equation takes the form \( ax + by + cz = d \). In our problem, the plane is described by the equation:
  • \( x + 2y + 3z = 13 \).
Here, it describes a flat surface where every point on this surface satisfies this equation. When trying to determine what points lie on a specific plane:
  • Each \(x, y, z\) coordinate must satisfy the equation condition.
  • This function represents a relationship between the variables, depicting how each changes relative to the others while still lying on the plane.
Understanding this equation allows us to identify the subset of space we are analyzing for the shortest distance problem.
Gradient Descent
**Gradient descent** is a popular optimization technique used to minimize functions, commonly seen in machine learning and calculus. In the context of finding the point on a plane closest to another point, gradient descent helps in:
  • Iteratively adjusting the coordinates of the point on the plane to minimize the distance to the target point.
  • It uses the derivative or gradient of the function, guiding movement steps towards points of lower function values.
The basic idea is, given a function like \( f(x, y, z) = (x - 1)^2 + (y - 1)^2 + (z - 1)^2 \):
  • Calculate the gradient \( abla f = (2(x-1), 2(y-1), 2(z-1)) \).
  • Move in the opposite direction of the gradient to decrease the function value.
  • The iterative updates usually look like \( x_{new} = x - \alpha abla f_x \) for a small step size \( \alpha \).
Using this method ensures that we zero in on the optimal point, finding the minimal distance fast and efficiently.
Calculus
**Calculus** forms the backbone of concepts like optimization and distance minimization, providing the tools necessary to handle changes and trends in functions. In this context, calculus aids in:
  • Calculating derivatives and gradients needed to implement methods like gradient descent.
  • Using these calculations to follow paths towards minimal values of a function.
Derivatives allow us to understand the rate at which values change or how we can tweak variables to reach a goal. In our exercise:
  • Finding partial derivatives of \( f(x, y, z) = (x - 1)^2 + (y - 1)^2 + (z - 1)^2 \) gives insights into optimizing the position on the plane.
  • See how \( abla f \) translates into a direction pointing towards lower values.
With calculus, mathematical intuition guides us through setting up and solving a host of optimization problems effectively.

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Most popular questions from this chapter

Find the minimum distance from the cone \(z=\sqrt{x^{2}+y^{2}}\) to the point (-6,4,0).

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Does a function \(f(x, y)\) with continuous first partial derivatives throughout an open region \(R\) have to be continuous on \(R ?\) Give reasons for your answer.

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