Chapter 4: Problem 32
Welcher Punkt auf der Ebene \(2 x+3 y+z=14\) hat vom Koordinatenursprung den Kleinsten Abstand \(d ?\)
Short Answer
Expert verified
The point on the plane closest to the origin is \((2, 3, 1)\). The minimum distance is \(\sqrt{14}\).
Step by step solution
01
Understand the Problem
The exercise asks for the point on the plane given by the equation \(2x + 3y + z = 14\) to be identified that is closest to the origin \((0,0,0)\).
02
Formula for Distance from a Point to a Plane
The distance \(d\) from a point \((x_0, y_0, z_0)\) to a plane \(Ax + By + Cz + D = 0\) is given by:\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
03
Rearrange the Plane Equation
Rewrite the plane equation \(2x + 3y + z = 14\) in the form \(Ax + By + Cz + D = 0\):\(2x + 3y + z - 14 = 0\). Thus,\(A = 2\), \(B = 3\), \(C = 1\), \(D = -14\).
04
Substitute Point Coordinates
Substitute the coordinates of the origin \((0,0,0)\) into the distance formula to get:\[ d = \frac{|2(0) + 3(0) + 1(0) - 14|}{\sqrt{2^2 + 3^2 + 1^2}} = \frac{| - 14|}{\sqrt{4 + 9 + 1}} = \frac{14}{\sqrt{14}} = \sqrt{14} \]
05
Find the Closest Point
Using the distance formula, it has been shown that the distance \(d\) from the origin to the plane \(2x + 3y + z = 14\) is \(\sqrt{14}\). This distance represents the minimal distance from the origin to the plane, corresponding to the point \( \left( \frac{2}{\sqrt{14}}d, \frac{3}{\sqrt{14}}d, \frac{1}{\sqrt{14}}d \right) = (2, 3, 1)\), since normalization occurs along the coefficients of \(x, y, z\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
minimal distance
In this exercise, we aim to identify the point on the plane given by the equation \(2x + 3y + z = 14\) that is closest to the origin (0,0,0). To achieve this, we use the concept of 'minimal distance'.
The minimal distance between a point and a plane is found by utilizing the perpendicular distance formula. This comes from the fact that the shortest path (or minimal distance) from any point to a plane is always along the line perpendicular to the plane.
This perpendicular distance formula is important because it allows us to compute the exact shortest distance without needing any guesswork. Once we determine the minimal distance, we can also figure out the exact coordinates of the point on the plane that corresponds to this shortest distance.
The minimal distance between a point and a plane is found by utilizing the perpendicular distance formula. This comes from the fact that the shortest path (or minimal distance) from any point to a plane is always along the line perpendicular to the plane.
This perpendicular distance formula is important because it allows us to compute the exact shortest distance without needing any guesswork. Once we determine the minimal distance, we can also figure out the exact coordinates of the point on the plane that corresponds to this shortest distance.
coordinate geometry
Coordinate geometry, also known as analytic geometry, combines algebra and geometry to solve problems using coordinates and equations.
In this exercise, we deal with a plane represented by its equation: \(2x + 3y + z = 14\). The plane is defined in a three-dimensional coordinate system where every point is determined by its \((x, y, z)\) coordinates.
The key aspect of coordinate geometry here is converting the geometric problem of finding the shortest distance to a mathematical problem. We use the distance formula to measure how far any given point is from the plane. This makes understanding and solving the problem much easier.
In this exercise, we deal with a plane represented by its equation: \(2x + 3y + z = 14\). The plane is defined in a three-dimensional coordinate system where every point is determined by its \((x, y, z)\) coordinates.
The key aspect of coordinate geometry here is converting the geometric problem of finding the shortest distance to a mathematical problem. We use the distance formula to measure how far any given point is from the plane. This makes understanding and solving the problem much easier.
plane equation
The equation of a plane in three-dimensional space is a fundamental concept in geometry. A plane can be described by an equation of the form \(Ax + By + Cz + D = 0\).
In this exercise, our plane's equation is \(2x + 3y + z = 14\). We can rewrite it in the standard form as \(2x + 3y + z - 14 = 0\). Here, \(A = 2\), \(B = 3\), \(C = 1\), and \(D = -14\).
Using these values, we apply the distance formula to find the shortest distance from the origin to the plane. Understanding how to manipulate and use plane equations is crucial for solving such problems.
Not only does this help us understand the spatial relationships in geometry, but it also aids in visualizing and solving real-world problems involving planes and distances.
In this exercise, our plane's equation is \(2x + 3y + z = 14\). We can rewrite it in the standard form as \(2x + 3y + z - 14 = 0\). Here, \(A = 2\), \(B = 3\), \(C = 1\), and \(D = -14\).
Using these values, we apply the distance formula to find the shortest distance from the origin to the plane. Understanding how to manipulate and use plane equations is crucial for solving such problems.
Not only does this help us understand the spatial relationships in geometry, but it also aids in visualizing and solving real-world problems involving planes and distances.